CHAPTER 19CHAPTER 19
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SOLUTION
Simple harmonic motion. x x tm n= +sin( )w f
where w p pn
m
T
x
= = =
= = ¥ -
2 2
0 162 83
5 5 10 3
..
srad/s
mm m
Then x t= ¥ +-5 10 62 833 sin( . )f
Differentiate to obtain velocity and acceleration.
v x x t
a x x t
n m n
n m n
= = +
= = - +
&&&
w w f
w w f
cos( )
sin( )2
Maximum velocity. | | ( . )( )max&x xn m= = ¥ -w 62 83 5 10 3 &xmax .= 0 314 m/s �
Maximum acceleration. | | ( . ) ( )max&&x xn m= = ¥ -w2 2 362 83 5 10 &&x amax max .= = 19 74 m/s2 �
PROBLEM 19.1
Determine the maximum velocity and maximum acceleration of a particle which moves in simple harmonic motion with an amplitude of 5 mm and a period of 0.1 s.
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SOLUTION
Simple harmonic motion. x x tm n= +sin( )w f
where w p p pn nf= = =2 2 40 80( )( ) rad/s
Differentiate to obtain velocity and acceleration.
v x x t
v x
a x x t
a x
n m n
m n m
n m n
m n m
= = +=
= = - +
=
&
&&
w w fw
w w f
w
cos( )
sin( )2
2
Use information given to obtain amplitude and maximum velocity.
xa
mm
n
= = =w p2 2
60
800 000950
m/sm
2
( ). xm = 0 950. mm �
v xm n m= = =w p( )( . ) .80 0 000950 2387 m/s vm = 239 mm/s �
PROBLEM 19.2
Determine the amplitude and maximum velocity of a particle which moves in simple harmonic motion with a maximum acceleration of 60 m/s2 and a frequency of 40 Hz.
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PROBLEM 19.3
A particle moves in simple harmonic motion. Knowing that the amplitude is 300 mm and the maximum acceleration is 5 m/s2, determine the maximum velocity of the particle and the frequency of its motion.
SOLUTION
Simple harmonic motion. x x t x
x t
x
m n m
n
n
= + == +=
sin( ) .
( . ) sin( ) ( )
( . )( )c
w fw f
w
0 300
0 300
0 3
m
m
& oos( ) ( )
( . )( ) sin( )
( . )
w f
w w fn
n n
m
t
x t
a
+
= - +
=
m/s
(m/s)
| | m/s
&& 0 3
0 3
2
(( )wn ma2 5= m/s2
Natural frequency. w
ww
nm
n
n
a
f
2
0 3
5
0 316 667
4 082
= = =
=
=
| |
( . )
( )
( . ).
.
m
m/s
mrad/s
rad/s
22
nn
2p
fn = =( . )
( ).
4 082
20 6497
rad/s
rad/cycleHz
p fn = 0 650. Hz �
Maximum velocity. v xm m n= =w ( . )( . )0 3 4 082m rad/s vm = 1 225. m/s �
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PROBLEM 19.4
A 5 kg block is supported by the spring shown. If the block is moved vertically downward from its equilibrium position and released, determine (a) the period and frequency of the resulting motion, (b) the maximum velocity and acceleration of the block if the amplitude of its motion is 50 mm.
SOLUTION
(a) Simple harmonic motion. x x tm n= +sin( )w f
Natural frequency. w
w
w
t pw
n
n
n
nn
k
mk= =
=
= =
=
12000
12000
15
20 2 28 284
2
N/m
N/m
kg
rad/s
( )
.
t pn = =2
28 2840 22214
.. s t n = 0 222. s �
fnn
= = =1 1
0 222144 50
t .. Hz �
(b) x
x tm = =
= +50 0 05
0 05 28 284
mm m.
. sin ( . )f
Maximum velocity. v xm m n= =w ( . )( . )0 05 28 284 vm = 1 414. m/s �
Maximum acceleration. a xm m n= =w2 0 05( . )(28.284)2 am = 40 0. m/s2 �
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PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.5
A 32-kg block is attached to a spring and can move without friction in a slot as shown. The block is in its equilibrium position when it is struck by a hammer, which imparts to the block an initial velocity of 250 mm/s. Determine (a) the period and frequency of the resulting motion, (b) the amplitude of the motion and the maximum acceleration of the block.
SOLUTION
(a) x x t
k
m
m n
n
n
nn
= +
= =¥
=
=
sin( )
.
w f
w
w
t pw
12 10
19 365
2
3 N/m
32 kg
rad/s
t p
t
n
n
=
=
2
19 3650 324
.. s fn
n
= = =1 1
0 3243 08
t .. Hz �
(b) At t x= =0 00, , &x v0 0 250= = mm/s
Thus, x xm n0 0 0= = +sin( ( ) )w f
and f = 0
&x v x x
v x
x
m n n m n
m
m
0 0
0
0 0
0 250 19 365
= = + == =
=
w w wcos( ( ) )
. ( . )m/s rad/s
(( . )
( . )
.
0 250
19 365
12 91 10 3
m/s
rad/s
mxm = ¥ - xm = 12 91. mm �
a xm m n= = ¥ -w2 312 91 10( . m)(19.365 rad/s)2 am = 4 84. m/s2 �
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PROBLEM 19.6
A simple pendulum consisting of a bob attached to a cord oscillates in a vertical plane with a period of 1.3 s. Assuming simple harmonic motion and knowing that the maximum velocity of the bob is 400 mm/s, determine (a) the amplitude of the motion in degrees, (b) the maximum tangential acceleration of the bob.
SOLUTION
(a) Simple harmonic motion. q q w f
w pt
p
w
q q w w
= +
= =
=
=
m n
nn
n
m n n
tsin( )
( )
( . )
.
cos(
2 2
1 3
4 833
s
rad/s& tt
v l l
m m n
m m m n
+
=
= =
f
q q w
q q w
)&
&
qwm
m
n
v
l= (1)
For a simple pendulum, wng
l=
Thus, lg
ln
= =
=w2 2
9 81
4 833
0 4200
.
( . )
.
m/s
rad/s
m
2
Amplitude.
From Eq. (1), qwm
m
n
v
l= =
( . )
( . )( . / )
0 4
0 42 4 833
m/s
m rad s
qm = 0 19706. rad qm = ∞11 29. �
(b) Maximum tangential acceleration. a lt = &&q Maximum tangential acceleration occurs when &&q is maximum.
&&&&
q q w w f
q q w
q w
= - +
=
=
=
m n n
m n
t m n
t
t
a l
a
2
2
2
0 4
sin( )
( )
( ) ( .
max
max
max 2200 0 19706 4 833 2m rad rad/s)( . )( . ) ( ) .at m = 1 933 m/s2 �
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PROBLEM 19.7
A simple pendulum consisting of a bob attached to a cord of length l = 800 mm oscillates in a vertical plane. Assuming simple harmonic motion and knowing that the bob is released from rest when q = ∞6 , determine (a) the frequency of oscillation, (b) the maximum velocity of the bob.
SOLUTION
(a) Frequency. wng
l= =
( . )
( . )
9 81
0 8
m/s
m
2
ww
p p
n
nnf
=
= =
3 502
2
3 502
2
.
( . )
rad/s
rad/s fn = 0 557. Hz �
(b) Simple harmonic motion. q q w f= +m ntsin( )
where qm = ∞ =6 0 10472. rad
Maximum velocity. &q q w w f= +m n ntcos( )
&&&
q q w
q q wm m n
m m m n
m
v l l
v
=
= = =
= ¥ -
( . )( . )( . )
.
0 8 0 10472 3 502
293 4 10 3
m
mm/s vm = 293 mm/s �
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PROBLEM 19.8
An instrument package A is bolted to a shaker table as shown. The table moves vertically in simple harmonic motion at the same frequency as the variable-speed motor which drives it. The package is to be tested at a peak acceleration of 50 m/s2. Knowing that the amplitude of the shaker table is 60 mm, determine (a) the required speed of the motor in rpm, (b) the maximum velocity of the table.
SOLUTION
In simple harmonic motion,
a x n
n
n
n
max max
( . )
( . )
.
=
=
==
w
w
ww
2
2
2 2
50 0 06
833 33
28 8675
m/s m
rad/s
r
2
aad/s
Hz (cycles per second)
fnn=
=
=
wp
p
228 8675
24 5945
.
.
(a) Motor speed. ( . )( )4 5945 60rev/s s/min speed rpm= 276 �
(b) Maximum velocity. v x nmax max ( . )( . )= =w 0 06 28 8675m rad/s vmax .= 1 732 m/s �
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PROBLEM 19.9
The motion of a particle is described by the equation x t t= +5 2 4 2sin cos , where x is expressed in meters and t in seconds. Determine (a) the period of the motion, (b) its amplitude, (c) its phase angle.
SOLUTION
For simple harmonic motion x x tm n= +sin( )w f
Double angle formula (trigonometry): sin( ) (sin )(cos ) (sin )(cos )A B A B B A+ = +
Let A t Bn= =w f,
Then x x t
x x t x t
x x
m n
m n m n
m
= += +=
sin( )
(sin )(cos ) (sin )(cos )
( cos
w fw f f wf))(sin ) ( sin )(cos )w f wn m nt x t+
Given x t t= +5 2 4 2sin cos
Comparing, w fn mx= =2 5cos (1)
xm sinf = 4 (2)
(a) t pw
p pnn
= = =2 2
2( )rad/ss t = 3 14. s �
(b) Squaring Eqs. (1) and (2) and adding,
x x
x x
m m
m m
2 2 2 2 2 2
2 2 2 2
4 5
41
cos sin
(cos sin )
f f
f f
+ = +
+ = = m2 xm = 6 40. m �
(c) Dividing Eq. (2) by Eq. (1), tanf = 4
5 f = ∞38 7. �
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PROBLEM 19.10
An instrument package B is placed on the shaking table C as shown. The table is made to move horizontally in simple harmonic motion with a frequency of 3 Hz. Knowing that the coefficient of static friction is ms = 0 40. between the package and the table, determine the largest allowable amplitude of the motion if the package is not to slip on the table. Given answers in both SI and U.S. customary units.
SOLUTION
Maximum allowable acceleration of B.
ms = 0 40.
+ SF ma= : F ma
mg ma
a g a g
m m
s m
m s m
=== =
mm 0 40.
Simple harmonic motion. f
a x
g x
x
nn
n
m m n
m
m
= =
=
=
=
= ¥
32
6
0 40 6
1 1258 1
2
2
Hz
rad/s
rad/s
wp
w p
w
p. ( )
. 00 3- g
Largest allowable amplitude.
SI: xm = ¥ = ¥- -1 1258 10 9 81 11 044 103 3. ( . ) . m xm = 11 04. mm �
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PROBLEM 19.11
A 32-kg block attached to a spring of constant k = 12 kN/m can move without friction in a slot as shown. The block is given an initial 300-mm displacement downward from its equilibrium position and released. Determine 1.5 s after the block has been released (a) the total distance traveled by the block, (b) the acceleration of the block.
SOLUTION
(a) x x t
k
m
m n
n
n
n
= +
= = ¥
=
=
sin( )
(
(
.
w f
w
w
t
12 10
32
19 365
2
3 N/m)
kg)
rad/s
ppw
p
tn
n
=
=
( )
( . )
.
2
19 365
0 3245 s
Initial conditions. x x
x
x x
x
m
m n
m
( ) . ( )
. sin( )
( ) cos( )
0 0 3 0 0
0 3 0
0 0 0
2
= == += = +
=
m, &
&f
w f
f p
==
= +ÊËÁ
ˆ¯
=
=
0 3
0 3 19 3652
0 3245
1 5 0
.
( ) ( . )sin . , .
( . ( .
x t t
x
np t s
s) 33 19 365 1 52
0 2147
1 5 0 3 19 36
)sin ( . )( . ) .
( . ) ( . )( .
+ÈÎÍ
˘˚
= -
=
p m
s&x 55 19 365 1 52
4 057)cos ( . )( . ) .+ÈÎÍ
˘˚
= Øp m/s
In one cycle, block travels ( )( .4 0 3 m) 1.2 m=
To travel 4 cycles, it takes (4 cyc)(0.3245 s/cyc) 1.2980 s=
at t = 1 5. s. Thus, total distance traveled is
4 1 2 0 6 0 3 0 2147 5 49( . ) . ( . . ) .+ + - = m �
(b) &&x( . ) ( . )( . ) sin ( . )( . ) .1 5 0 3 19 365 19 365 1 52
80 52= - +ÈÎÍ
˘˚
=p m/s22 Ø �
14
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PROBLEM 19.12
A 2 kg block is supported as shown by a spring of constant k = 400 N/m, which can act in tension or compression. The block is in its equilibrium position when it is struck from below by a hammer, which imparts to the block an upward velocity of 2.5 m/s. Determine (a) the time required for the block to move 100 mm upward, (b) the corresponding velocity and acceleration of the block.
SOLUTION
Simple harmonic motion. x x tm n= +sin( )w f
Natural frequency. wnk
mk= =, 400 N/m
w
wf
f
n
n
mx x
x x
=
= == = +==
400
2
10 2 14 1421
0 0 0
0
0
N/m
rad/s.
( ) sin( )
( )& mm n
m m
x
x x
x
w cos( )
( ) .
. ( . ) .
( .
0 0
0 2 5
2 5 14 1421 0 17678
0 1
+== ==
& m/s
m
77678 14 1421)sin( . )(t m/s) (1)
(a) Time at mm ( m)x x= =100 0 1.
0 1 0 17678 14 1421
14 14210 04252
1 0 10 17678
. . sin( . )
sin
..
..
=
=( )
=-
t
t s
st = 0 046. �
Note: Since w is in rad/s, convert the argument of sin-1 to radians.
(b) Velocity and acceleration.
&&&
&
x x t
x x t
t
x
m n n
m n n
=
= -==
w w
w w
cos( )
sin
.
( . )( .
2
0 04252
0 17678 14 14221 14 1421 0 04252
2 0615
)cos[( . )( . )]
.&x = m/s v = ≠2 06. m/s �
&&x = -
= -
( . )( . ) sin[( . )( . )]0 1768 14 1421 14 1421 0 04252
20
2
m/s2 a = Ø20 0. m/s2 �
15
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.13
In Problem 19.12, determine the position, velocity, and acceleration of the block 0.90 s after it has been struck by the hammer.
SOLUTION
Simple harmonic motion. x x tm n= +sin( )w f
Natural frequency. w
w
w
n
n
n
k
mk= =
=
= =
,
.
400
400
2
10 2 14 1421
N/m
N/m
kg
rad/s
x x
x x x
x
m
m n
m
( ) sin( )
( ) cos( ) ( ) .
. (
0 0 0
0
0 0 0 0 2 5
2 5
= = +== + ==
ff
w& & m/s
114 1421 0 17678
0 17678 14 1421
. ) .
( . )sin ( . )(
x
x tm =
= m
m/s)
Simple harmonic motion. x x t
x x t
x x t
m n
m n n
m n n
= += +
= - +
sin( )
cos( )
sin( )
w fw w f
w w f
&&& 2
Note: arguments of sin and cos are in radians.
At 0.90 s: x = =( . )sin[( . )( . )] .0 17678 14 1421 0 90 0 02843 m x = ≠0 0284. m �
&x = = -( . )( . )cos[( . )( . )] .0 17678 14 1421 14 1421 0 90 2 4675 m/s v = ≠2 47. m/s �
&&x = - = -( . )( . ) sin[( . )( . )] .0 17678 14 1421 14 1421 0 90 5 68592 m/s2 a = Ø5 69. m/s2 �
16
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PROBLEM 19.14
The bob of a simple pendulum of length l = 800 mm is released from rest when q = + ∞5 . Assuming simple harmonic motion, determine 1.6 s after release (a) the angle q, (b) the magnitudes of the velocity and acceleration of the bob.
SOLUTION
q q w f w
w
= + = =
=
m n n
n
tg
lsin( )
.
..
9 81
0 83 502
m/s
m rad/s
Initial conditions: q p
q
q p q f
q q
( )( )( )
( )
( ) sin( )
( )
0 55
180
0 0
05
1800
0 0
= ∞ =
=
= = +
= =
rad&
&m
mm n
m
t
w f
f p
q p
q p p
cos( )
sin .
0
25
1805
1803 502
2
+
=
=
= +ÊËÁ
ˆ¯
rad
(a) At t = 1 6. s: q p p= +ÈÎÍ
˘˚
5
1803 502 1 6
2sin ( . )( . ) q = =0 06786. rad 3.89° �
(b)
17
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PROBLEM 19.14 (Continued)
&q q w w f p p= + = ÊËÁ
ˆ¯
+ÈÎÍ
˘m n ntcos( ) ( . )cos ( . )( . )
5
1803 502 3 502 1 6
2 ˚=&q( .1 6 s) 0.19223 rad/s
v l= = =&q ( .0 800 m)(0.19223 rad/s) 0.1538 m/s �
&&q q w w fp p
= - +
= -ÊËÁ
ˆ¯
+
m n nt2
25
1803 502 3 502 1 6
sin( )
( . ) sin ( . )( . )22
0 8319
0 8
2 2
ÈÎÍ
˘˚
= -
= +
= = -
&&
&&
q
q
.
( ) ( )
( . (
rad/s
m) 0.
2
a a a
a l
t n
t 88319 rad/s m/s
m)(0.19223 rad/s)
2 2
2
) .
( .
= -
= = =
0 6655
0 82a ln&q 00 02956
0 6655 0 02956 0 66622 2
.
( . ) ( . ) .
m/s
m/s
2
2a = + = a = 0 666. m/s2 �
18
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.15
A 5-kg collar rests on but is not attached to the spring shown. It is observed that when the collar is pushed down 180 mm or more and released, it loses contact with the spring. Determine (a) the spring constant, (b) the position, velocity, and acceleration of the collar 0.16 s after it has been pushed down 180 mm and released.
SOLUTION
(a) x x t
x x
x x
x
m n
m
m
m
= += + == = +
=
=
sin( )
sin( ) .
cos( )
w ff
f
f p
0
0
0 0 180
0 0
2
m
&
00 180
0 1802
.
. sin
m
x tn= +ÊËÁ
ˆ¯
w p
When the collar just leaves the spring, its acceleration is gØ and v = 0.
&x t
v t
n n
n n
n
= +ÊËÁ
ˆ¯
= = +ÊËÁ
ˆ¯
( . ) cos
( . ) cos
0 1802
0 0 0 1802
w w p
w w p
w tt
a g t
g
n n
+ÊËÁ
ˆ¯
=
= - = - +ÊËÁ
ˆ¯
- = -
p p
w w p2 2
0 1802
0 180
2( . )( ) sin
( . )(( ).
..
(
w w
w
w
w
n n
n
n
n
k
m
k m
2
2
9 81
0 1807 382
5
=
=
=
=
=
m/s
m rad/s
kg)(
2
77.382 rad/s)2
= 272 5. N/m k = 273 N/m �
19
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.15 (Continued)
(b) wp
n
x t
=
= +ÈÎÍ
˘˚
7 382
0 180 7 3822
.
. sin ( . )
rad/s
At st = 0 16. :
Position.
x = +ÈÎÍ
˘˚
=0 1802
0 06838. . sin (7.382)(0.16) mp
x = 68 4. mm below equilibrium position �
Velocity.
&x = +ÈÎÍ
˘˚
= -
( . )( . )cos ( . )( . )
.
0 180 7 382 7 382 0 162
1 229
p
m/s v = ≠1 229. m/s �
Acceleration.
&&x = - +ÈÎÍ
˘˚
= -
( . )( . ) sin ( . )( . )
.
0 180 7 382 7 382 0 162
3 726
2 p
m/s2 a = ≠3 73. m/s2 �
20
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.16
An 8-kg collar C can slide without friction on a horizontal rod between two identical springs A and B to which it is not attached. Each spring has a constant of 600 N/m. The collar is pushed to the left against spring A, compressing that spring 20 mm, and released in the position shown. It then slides along the rod to the right and hits spring B. After compressing that spring 20 mm, the collar slides to the left and hits spring A, which it compresses 20 mm. The cycle is then repeated. Determine (a) the period of the motion of the collar, (b) the position of the collar 1.5 s after it was pushed against spring A and released. (Note: This is a periodic motion, but not a simple harmonic motion.)
SOLUTION
(a) For either spring, t p
t p
t
n km
n
n
C
=
=
=
2
2
0 7255
6008
N/mkg
s.
Complete cycle is 1234, 4321.
Time from 1 to 2 is ( / ),t n 4 which is the same as time from 3 to 4, 4 to 3 and 2 to 1.
Thus, the time during which the springs are compressed is 4 4 0 7255( / ) .t tn n= = s.
Velocity at 2 or 3.
Use conservation of energy. v T V kx
V
T mv
1 1 12
1
2 22
0 01
2
1
2600
0 120
1
2
1
= = = =
=
= =
(
.
N/m)(0.020 m)
J
2
228 4
0
0 0 120 4 0 1732
22
2 22
2
1 1 2 2 22
2
( )
. .
kg)(
m
v T v
V
T V T V v v
=
=
+ = + + = = //s
Time from 2 to 3 is t2 30 020
0 17320 11545- = =( .
( ..
m)
m/s) s
and is the same as the time from 3 to 2.
Thus, total time for a complete cycle is t tc n t= += +
-2
0 7255 2 0 115452 3
. ( . )
= 0 9564. t c = 0 956. s �
21
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.16 (Continued)
(b) From (a), in 0.9564, the spring A is again fully compressed. Spring B is compressed the second time in 1.5 cycles or ( . )( . ) .1 5 0 9564 1 4346= s. At 1.5 s, the collar is still in contact with spring B moving to the left and is at a distance Dx from the maximum deflection of B equal to
D
D
x
x
= - -ÈÎÍ
˘˚
= -=
20 202
0 72551 5 1 4346
20 16 877
3 123
cos.
( . . )
.
.
p
mm
Thus, collar C is 60 3 123 56 877- =. . mm from its initial position.
56 9. mm from initial position �
22
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.17
A 35-kg block is supported by the spring arrangement shown. The block is moved vertically downward from its equilibrium position and released. Knowing that the amplitude of the resulting motion is 45 mm, determine (a) the period and frequency of the motion, (b) the maximum velocity and maximum acceleration of the block.
SOLUTION
(a) Determine the constant k of a single spring equivalent to the three springs
P k
k
== + +=
d d d d16 8 8
32 kN/m
Natural frequency.
w
w
n
n
k
m=
= ¥ = ◊
=
32 10
351 1
30 237
3 N/m
kg N kg 1 m/s
rad/s
2( )
.
t pw
pn
n
= = =2 2
30 230 208
.. s �
fnn
= =14 81
t. Hz �
(b) x x t x x
x
m n m
n
= + = ===
sin( ) .
.
.
w fw
0 0 045
30 24
0 045
m
rad/s
sin(30.24tt + f)
&x t= +( . )( . )cos( . )0 045 30 24 30 24 f vmax .= 1 361 m/s �
&&x t= - +( . )( . ) sin( . )0 045 30 237 30 242 f amax .= 41 1 m/s2 �
23
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.18
A 35-kg block is supported by the spring arrangement shown. The block is moved vertically downward from its equilibrium position and released. Knowing that the amplitude of the resulting motion is 45 mm, determine (a) the period and frequency of the motion, (b) the maximum velocity and maximum acceleration of the block.
SOLUTION
(a) Determine the constant k of a single spring equivalent to the two springs shown.
d d d= + = + =
= + =
1 2 16 161 1
16
1
168
P P P
k
kk
kN/m kN/m
kN/m
Period of the motion.
t p pn k
m
= = =¥
2 20 416
8 1035
3. s �
fnn
= = =1 1
0 4162 41
t .. Hz �
(b) w p pn nf= = =2 2 2 41 15 12( . ) . rad/s
x t
x t
= += +
0 045
0 045 15 12 15 12
. )
( . )( . )cos( . )
sin(15.12 ff& vmax .= 0 680 m/s �
&&x t= - +( . )( . ) sin( . )0 045 15 12 15 122 f amax .= 10 29 m/s2 �
24
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.19
A 15 kg block is supported by the spring arrangement shown. If the block is moved from its equilibrium position 40 mm vertically downward and released, determine (a) the period and frequency of the resulting motion, (b) the maximum velocity and acceleration of the block.
SOLUTION
Determine the constant k of a single spring equivalent to the three springs shown.
Springs 1 and 2: d d d= +¢
= +1 21 1
1
1
2
, andP
k
P
k
P
k
Hence, ¢ =+
kk k
k k1 2
1 2
where ¢k is the spring constant of a single spring equivalent of springs 1 and 2.
Springs ¢k and 3 Deflection in each spring is the same.
So P P P P k P k P k= + = = ¢ =1 2 1 2 3, , ,and d d d
Now k k k
k k kk k
k kk
k
d d d= ¢ +
= ¢ + =+
+
=+
+ =
3
31 2
1 23
4 2 4
4 2 43 2 4 7
( )( . )
.. . kN/m ==
=
= = = =
4700
15
4700
15313 33 17 70122
N/m
kg
rad/s
m
k
mn nw w. .
25
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.19 (Continued)
(a) Period. t pwn
n
= 2 t n = 0 355. s �
Frequency. fnn=
wp2
fn = 2 82. Hz �
(b) Simple harmonic motion.
x t= 0 04. sin( ) in mw j+ xm = 40 mm �
&x t= +( . )( . )cos( )0 04 17 7012 w f
v xm m= =& 0 70805. m/s vm = 0 708. m/s �
&&x t= - +( . )( . ) sin( )0 04 17 7012 2 w f
a xm m= =&& 12 533. m/s2 am = 12 53. m/s2 �
26
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.20
A 5-kg block, attached to the lower end of a spring whose upper end is fixed, vibrates with a period of 6.8 s. Knowing that the constant k of a spring is inversely proportional to its length, determine the period of a 3-kg block which is attached to the center of the same spring if the upper and lower ends of the spring are fixed.
SOLUTION
Equivalent spring constant.
¢ = + =k k k k2 2 4 (Deflection of each spring is the same.)
For case 1 , t
w pt
p
w
w
n
nn
n
n
k
m
k m
1
11
2
1
1 12
6 8
2 2
6 80 924
5 0
=
= = =
=
= =
.
..
( )( .
s
rad/s
9924 4 26892) .= N/m
For case 2 , wnk
m22
2
4 4 4 2689
35 6918= = =( )( . ). (rad/s)2
w
t pw
pn
nn
2
2
2 3857
2 2
2 3857
=
= =
.
.
rad/s
t n2 2 63= . s �
27
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.21
A 15 kg block is supported by the spring arrangement shown. The block is moved from its equilibrium position 20 mm vertically downward and released. Knowing that the period of the resulting motion is 1.5 s, determine (a) the constant k, (b) the maximum velocity and maximum acceleration of the block.
SOLUTION
Since the force in each spring is the same, the constant ¢k of a single equivalent spring is
1 1
2
1 1
2 5¢= + + ¢ =
k k k kk
k (See Problem 19.19.)
.
(a) t p pn k
m
k12
1 52 2
1 52 5= = = Ê
ËÁˆ¯¢
. ;.
( )( . ) s m
k =
=
( )
( . )( )( . )
.
2
1 515 2 5
657 974
2
2
p s
kg
N/m k = 658 N/m �
(b) x x t
x x t
v x
m n
m n n
m n
= += +=
sin( )
cos( )
max
w fw w fw
&
w pt
pn
n
mx
v
= = =
= ==
2 2
1 54 189
20 0 02
0 02 4 1
..
.
( . )( .max
srad/s
mm m
m 889 rad/s) vmax .= 0 0838 m/s �
&&x x tm n n= - +w w f2 cos( )
| | ( . )( . )maxa xm n= =w2 20 02 4 189 m | | .maxa = 0 351 m/s2 �
28
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.22
Two springs of constants k1 and k2 are connected in series to a block A that vibrates in simple harmonic motion with a period of 5 s. When the same two springs are connected in parallel to the same block, the block vibrates with a period of 2 s. Determine the ratio k k1 2/ of the two spring constants.
SOLUTION
Equivalent springs.
Series: kk k
k ks =+1 2
1 2
Parallel: k k kp = +1 2
t pw
p t pw
p
tt
ss
km
pp
k
m
s
p
p
s
s p
k
k
k k
= = = =
Ê
ËÁ
ˆ
¯˜ = Ê
ËÁˆ¯
= =+
2 2 2 2
5
2
2 21
;
22 1 22
1 2
1 2 12
1 2 22
1
1 2
1 2
6 25 2
4
( )
( )
( . )( )
( .
k kk k
k k
k k
k k k k k k
k
+
=+
= + +
=225 4 25 4
22
222
22) ( . )k k km -
k
k1
2
2 125 3 516 0 250 4 00= =. . . .m or k
k1
2
0 250 4 00= . .or �
29
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.23
The period of vibration of the system shown is observed to be 0.6 s. After cylinder B has been removed, the period is observed to be 0.5 s. Determine (a) the weight of cylinder A, (b) the constant of the spring.
SOLUTION
m m
k
mk
A1 1
11
11
12
1
1 5 0 6
2 2 2
0 63 333
= + =
= = = =
= =
. .
..
t
t pw
w pt
p p
w
s
rad/s
mm mA1 12 21 5 3 333w p= + . ( . ) (1)
m m
k
mk m m
A2 2
22
2
22
22 2
2
0 5
2 2 2
0 54
= =
= = = =
= = =
t
t pw
w pt
p p
w w
.
.
s
rad/s2
AA( )4 2p (2)
(a) Equating the expressions found for k in Eqs. (1) and (2):
m mA A+ =1 5 3 333 42 2. ( . ) ( )p p
( . )( . )
. .
.
(
11 111 1 5 16
4 889 16 6665
3 40898
3
m m
m
m
W m g
A A
A
A
A A
+ ==== =
kg
.. )( . )40898 9 81 WA = 33 4. N �
(b) Eq. (1): k = +3 40898 1 5 3 333 2. . ( . )p rad/s
k = 538 22. N/m k = 538 N/m �
30
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.24
The period of vibration of the system shown is observed to be 0.8 s. If block A is removed, the period is observed to be 0.7 s. Determine (a) the mass of block C, (b) the period of vibration when both blocks A and B have been removed.
SOLUTION
m m
W
k
mk m
C1 1
11
12
11
6 0 8
2 2
0 8
2
0 8
= + =
= = =
= =
kg s
s rad/s
tp
tp p
w w
.
. .
; 112
2
62
0 8= + Ê
ËÁˆ¯
( ).
mCp
(1)
m m
k
mk m
C2
2
22
22 2
2
3 0 7
2 2
0 7
2
0 7
= + =
= = =
= =
kg s
srad/s
2
2
t
w pt
p p
w w
.
. .
; == + ÊËÁ
ˆ¯
( ).
mC 32
0 7
2p (2)
Equating the expressions found for k in Eqs. (1) and (2):
( ).
( ).
.
.
m m
m
m
C C
C
C
+ ÊËÁ
ˆ¯
= + ÊËÁ
ˆ¯
++
= ÊËÁ
ˆ¯
62
0 83
2
0 7
6
3
0 8
0 7
2 2p p
22
; :solvefor mC mC = 6 80. kg �
w pt
w w pt
33
32
32
3
2
2
2
=
= = =ÊËÁ
ˆ¯
k
mk m m
CC C; (3)
Equating expressions for k from Eqs. (2) and (3),
( ).
m mC C+ ÊËÁ
ˆ¯
=ÊËÁ
ˆ¯
32
0 7
22
3
2p p
t
Recall mC = 6 8. : kg ( . ).
.6 8 32
0 76 8
22
3
2
+ ÊËÁ
ˆ¯
=ÊËÁ
ˆ¯
p pt
t t3
23
0 7
6 8
9 8 0 70 833
.
.
.;
..
ÊËÁ
ˆ¯
= = t3 0 583= . s �
31
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.25
The period of vibration of the system shown is observed to be 0.2 s. After the spring of constant k2 4= kN/m is removed and block A is connected to the spring of constant k1, the period is observed to be 0.12 s. Determine (a) the constant k1 of the remaining spring, (b) the weight of block A.
SOLUTION
Equivalent spring constant for springs in series.
kk k
k ke =+1 2
1 2( )
For and k k1 2 , t p p= =+
2 2
1 2
1 2
km
k km k k
e
A A
( )( )( )
For alone,k1 ¢ =t p2
1kmA
(a) tt
tt
tt
¢=
+=
+
¢ÊËÁ
ˆ¯
= +
¢=
( )( )
( )
.
.
k k k
k k
k k
k
k k k
1 2 1
1 2
1 2
2
2
2
1 2
0 2
0 12==
=
1 6667
42
.
k kN/m
( )( . )4 1 6667 421 kN/m s kN/= +k m k1 7 11= . kN/m �
(b) kkmA
1 7 11112
1
= ¢ =. kN / m t p
mk
W m g
k
W
A A A
A
=¢
=
= =
=
( )
( );
. .
( .
tp
21
2
1
2
7 1111 7111 1
9 81
kN/ N/m
m/
m
ss s N/m2 2
2
25 4453 0 12 7111 1
2
) . ( . ) ( . )
( )
= N
p WA = 25 4. N �
32
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.26
The 50-kg platform A is attached to springs B and D, each of which has a constant k = 2 kN/m. Knowing that the frequency of vibration of the platform is to remain unchanged when a 40-kg block is placed on it and a third spring C is added between springs B and D, determine the required constant of spring C.
SOLUTION
Frequency of the original system.
Springs B and D are in parallel. k k k
k
m
e B D
ne
A
n
= + = = =
= =
2 2 4 4000
4000
2
( ) kN/m kN/m
N/m
50 kg2
N / m
w
w == 80 2( )rad/s
Frequency of new system.
Springs A, B, and C are in parallel. ¢ = + + = +k k k k ke B D C c( )( )2 200 in N/m
( )( )
( )
( )( )
¢ =¢
+=
++
¢ =+
w
w
w
ne
A B
c
nc
n
k
m m
k
k
2
2
2000
50 40
2000
90
kg kg
22 2
802000
905200 5 2
= ¢
=+
= =
( )
( )
.
wn
c
C
k
k N/m kN / m kC = 5 20. kN/m �
33
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.27
From mechanics of materials it is known that when a static load P is applied at the end B of a uniform metal rod fixed at end A, the length of the rod will increase by an amount d = PL AE/ , where L is the length of the undeformed rod, A is its cross-sectional area, and E is the modulus of elasticity of the metal. Knowing that L = 450 mm and E = 200 GPa and that the diameter of the rod is 8 mm, and neglecting the mass of the rod, determine (a) the equivalent spring constant of the rod, (b) the frequency of the vertical vibrations of a block of mass m = 8 kg attached to end B of the same rod.
SOLUTION
(a) P k
PL
AE
PAE
L
kAE
L
Ad
A
e
e
=
=
= ÊËÁ
ˆ¯
=
= =¥
= ¥
-
d
d
d
p p2 3 2
4
8 10
4
5 027 10
( )
.
m
--
-
=
= ¥
=¥ ¥
5 2
9 2
5 2 9 2
0 450
200 10
5 027 10 200 10
0
m
m
N/m
m N m
L
E
ke
.
( . )( / )
( .. )450 m
ke = ¥22 34 106. N/m ke = 22 3. MN/m �
(b) fn
km
e
=
=¥
2p
p
22 3 108
6
2
.
= 265 96. Hz fn = 266 Hz �
34
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.28
From mechanics of materials it is known that for a cantilever beam of constant cross section, a static load P applied at end B will cause a deflection d B PL EI= 3 3/ , where L is the length of the beam, E is the modulus of elasticity, and I is the moment of inertia of the cross-sectional area of the beam. Knowing that L = 3 m, E = 230GPa, and I = 5 ¥ 106 mm4, determine (a) the equivalent spring constant of the beam, (b) the frequency of vibration of a 250 kg block attached to end B of the same beam.
SOLUTION
(a) Equivalent spring constant. kP
P k
PL
EI
PEI
L
eB
e B
B
B
=
=
=
= ÊËÁ
ˆ¯
dd
d
d
3
3
33
kEI
L
k
e
e
=
=¥ ¥ ¥
=
3
3 230 10 5 10 10
3
127 778 10
3
9 6 12
3
3
( )( )( )
( )
.
N/m m2 4-
¥ NN/m ke = 127 8. kN/m �
(b) Natural frequency.
f
k
n
km
e
e
=
=2
N/
p127 778 103. ¥ m
f
f
n
n
=
=
¥( . )
.
127 778 103
3 5981
N/m250 kg
2 Hz
p fn = 3 60. Hz �
35
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.29
A 40-mm deflection of the second floor of a building is measured directly under a newly installed 4000-kg piece of rotating machinery, which has a slightly unbalanced rotor. Assuming that the deflection of the floor is proportional to the load it supports, determine (a) the equivalent spring constant of the floor system, (b) the speed in rpm of the rotating machinery that should be avoided if it is not to coincide with the natural frequency of the floor-machinery system.
SOLUTION
(a) Equivalent spring constant.
W k
kW
e s
e
=
=
= =
d
d( )( . )
( .
4000 9 81
0 04981 103
m)N / m¥ ke = 981 kN/m �
(b) Natural frequency. f
f
n
km
n
e
=
=
===
¥( )
2
Hz
Hz cycle/s
r
N/m)
p
p
(
.
981 104000
3
22 4924
1 1
60
kg
ppm
Speed Hzrpm
Hzrpm
=
=
( . )( )
.
2 492460
149 5 �
36
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.30
The force-deflection equation for a nonlinear spring fixed at one end is F x= 5 1 2/ where F is the force, expressed in newtons, applied at the other end and x is the deflection expressed in meters. (a) Determine the deflection x0 if a 120-g block is suspended from the spring and is at rest. (b) Assuming that the slope of the force-deflection curve at the point corresponding to this loading can be used as an equivalent spring constant, determine the frequency of vibration of the block if it is given a very small downward displacement from its equilibrium position and released.
SOLUTION
(a) Deflection x0 . mg
mg
F mg
x
===
=
( . )( . )
.
/
0 120 9 81
1 177
5
2
01 2
kg m/s
N
x0
21 177
5
0 0554
= ÊËÁ
ˆ¯
=
.
. m x0 55 4= . mm �
Equivalent spring constant.
At x0 , dF
dxx
dF
dx
x
x
ÊËÁ
ˆ¯
= =
ÊËÁ
ˆ¯
=
- -
0
0
5
2
5
20 0554
10 618
01 2 1 2( ) ( . )
.
/ /
NN/m
N/mke = 10 618.
(b) Natural frequency. fn
km
e
=
=
2
2
10 6180 120
p
p
( . )( .
N/m kg)
fn = 1 4971. Hz fn = 1 497. Hz �
37
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.31
If h = 700 mm and d = 500 mm and each spring has a constant k = 600 N/m, determine the mass m for which the period of small oscillations is (a) 0.50 s,(b) infinite. Neglect the mass of the rod and assume that each spring can act in either tension or compression.
SOLUTION
x xd
h
F kx kd
hx
s
s
=
= =2 2 2
+ S SM M Fd mgx mx hA A= - = -( ) : ( )eff 2 &&
2
20
2
2
2
22
2
kd
hx d mgx mxh
xkd
mh
g
hx
kd
mhn
ÊËÁ
ˆ¯
- = -
+ -È
ÎÍ
˘
˚˙ =
= -
&&
&&
w gg
h
k
m
d
h
g
hn
È
ÎÍ
˘
˚˙
= ÊËÁ
ˆ¯
-w222
(1)
Data: d
h
k
===
0 5
0 7
600
.
.
m
m
N/m
38
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.31 (Continued)
(a) For st = 0 5. : tn n
n= = =20 5
24
pw
pw
w p; .
Eq. (1): ( )( ) .
.
.
.4
2 600 0 5
0 7
9 81
0 72
2
p = ÊËÁ
ˆ¯
-m
m = 3 561. kg m = 3 56. kg �
(b) For infinite:t = tn
n= =20
pw
w
Eq. (1): 02 600 0 5
0 7
9 81
0 7
2
= ÊËÁ
ˆ¯
-( ) .
.
.
.m
m = 43 69. kg m = 43 7. kg �
39
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.32
Denoting by dST the static deflection of a beam under a given load, show that the frequency of vibration of the load is
fg= 1
2p dST
Neglect the mass of the beam, and assume that the load remains in contact with the beam.
SOLUTION
kW
mW
g
k
m
gn
W
Wg
=
=
= = =
d
wd
d
ST
ST
ST2 Fg
nn= =
wp p d2
1
2 ST
�
40
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.33*
Expanding the integrand in Equation (19.19) of Section 19.4 into a series of even powers of sinf and integrating, show that the period of a simple pendulum of length l may be approximated by the formula
t pq
= +ÊËÁ
ˆ¯
2 12
1
42l
gmsin
where qm is the amplitude of the oscillations.
SOLUTION
Using the Binomial Theorem, we write
1
11
2
11
2
22
2
21 2
2
- ( )= - Ê
ËÁˆ¯
ÈÎÍ
˘˚˙
= +
-
sin sinsin sin
sin
q f
qf
q
m
m
m
/
222sin f + ◊ ◊ ◊ ◊
Neglecting terms of order higher than 2 and setting sin ( cos ),2 12 1 2f f= - we have
tq
f fp
nml
gd
l
g
= + -( )ÈÎÍ
˘˚
ÏÌÓ
¸˝˛
= +
Ú4 11
2 2
1
21 2
4 11
4
2
0
2
2
sin cos
sinqq q
f fp m m d
2
1
4 222
0
2-Ï
ÌÓ
¸˝˛Ú sin cos
= + ÊËÁ
ˆ¯
-ÈÎÍ
˘˚˙
= +
41
4 2
1
8 22
42
1
4
2 2
0
2l
g
l
g
m mfq
fq
f
p
p
sin sin sin
si
/
nn2
2 20
q pmÊËÁ
ˆ¯
+ÈÎÍ
˘˚˙ t p
qn
ml
g= +Ê
ËÁˆ¯
2 11
4 22sin �
41
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.34*
Using the formula given in Problem 19.33, determine the amplitude qm for which the period of a simple pendulum is 1
2 percent longer than the period of the same pendulum for small oscillations.
SOLUTION
For small oscillations, ( )t pnl
g0 2=
We want t t
p
n n
l
g
=
=
1 005
1 005 2
0. ( )
.
Using the formula of Problem 19.33, we write
t tq
tq
n nm
n
m
= +ÊËÁ
ˆ¯
=
= - =
( ) sin
. ( )
sin [ . ] .
02
0
2
11
4 2
1 005
24 1 005 1 0 022
20 02
28 130
sin .
.
q
q
m
m
=
= ∞ qm = ∞16 26. �
42
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.35*
Using the data of Table 19.1, determine the period of a simple pendulum of length l = 750 mm (a) for small oscillations, (b) for oscillations of amplitude qm = ∞60 , (c) for oscillations of amplitude qm = ∞90 .
SOLUTION
(a) t pnl
g= 2 (Equation 19.18 for small oscillations):
t pn =
=
20 750
9 81
1 737
.
.
.
m
m/s
s
2
t n = 1 737. s �
(b) For large oscillations (Eq. 19.20),
tp
p
p
nk l
g
k
= ÊËÁ
ˆ¯
Ê
ËÁˆ
¯
=
22
21 737( . )s
For qm = ∞60 , k = 1 686. (Table 19.1)
tpn ( )
( . )( . )
.
602 1 686 1 737
1 864
∞ =
=
s
s t n ( ) .60 1 864= s �
(c) For qm = ∞90 , k = 1 854. tpn = =
2 1 854 1 7372 05
( . )( . ).
ss �
43
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.36*
Using the data of Table 19.1, determine the length in mm of a simple pendulum which oscillates with a period of 2 s and an amplitude of 90°.
SOLUTION
For large oscillations (Eq. 19.20),
tp
pnk l
g= Ê
ËÁˆ¯
Ê
ËÁˆ
¯2
2
for qm = ∞90
k = 1 854. (Table 19.1)
( ) ( )( . )( ).
( ) ( . )
[( )( . )]
2 2 1 854 29 81
2 9 81
4 1 854
2
2
sm/s
s m/s
2
2
=
=
=
l
l
00 71349. m l = 713 mm �
44
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.37
The 5-kg uniform rod AC is attached to springs of constantk = 500 N/m at B and k = 620 N/m at C, which can act in tension or compression. If the end C of the rod is depressed slightly and released, determine (a) the frequency of vibration, (b) the amplitude of the motion of Point C, knowing that the maximum velocity of that point is 0.9 m/s.
SOLUTION
F k x
k
F k x
k
B B B B
B B
C C C C
C
= += += += +
( ( ) )
( . ( ) )
( ( ) )
( . (
dq d
dq
ST
ST
ST
0 7
1 4 ddST ) )C
Equation of motion. + S SM MA A= ( ) :eff
( . )[ [( . ( ) ) ] . [ ( . ( ) )] ( . )(0 7 0 7 1 4 1 4 0 7k mg k IB B C Cq d q d a+ - + + = - -ST ST mmat ) (1)
But in equilibrium ( ),q = 0 + SM k mg kA B B C C= = - +0 0 7 1 4. [ ( ) ] . ( )d dST ST (2)
Substituting Equation (2) into Equation (1),
I ma k kt B Ca q q+ + + =0 7 0 7 1 4 02 2. ( . ) ( . )
Kinematics ( ):a q= && at = =0 7 0 7. .a q&&
[ ( . ) ] [( . ) ( . ) ]
( )( .
I m k k
I ml
B C+ + + =
=
=
0 7 0 7 1 4 0
1
121
125 1 4
2 2 2
2
&&q q
kg mm
kg m
)
.
2
20 8167= ◊
45
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.37 (Continued)
(. ) ( . )( )
.
( . ) ( . ) ( . )(
7 0 49 5
2 45
0 7 1 4 0 49 5
2 2
2
2 2
m
k kB C
=
= ◊
+ =
m kg
kg m
m2 000 1 96 620
245 1215 2
1460 2
0 8167 2 45
N/m m N/m
N m
2) ( . )( )
.
.
[ . . ]
+= += ◊
+ &&&
&&
&&
q q
q q
q q
+ =
+◊
◊=
+ =
1460 2 0
1460 2
3 2670
447 0
.
( . )
( . )
(
N m
kg m
N/k
2
gg m s 2◊ = - )
(a) Natural frequency. wn ==
-447
21 14
s
rad/s
2
.
fnn= = =
wp p2
21 14
23 36
.. Hz �
(b) Maximum velocity. q q w f
q q w w f
= +
= +m n
m n n
t
t
sin( )
( )( )cos( )&
Maximum angular velocity. &q q wm m n=
Maximum velocity at C. ( ) .
( . ) ( )( )
( . )
( . )( . )
& &xC m m
m n
m
==
=
=
1 4
1 4
0 9
1 4 21 14
qq w
q
m
m/s
m rad/s
00 03041. rad
Maximum amplitude at C. ( ) ( . )( )
( . )( . )
xC m m==
1 4
1 4 0 03041
m
m
q
( ) .xC m = 0 0426 m ( ) .xC m = 42 6 mm �
46
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.38
The uniform rod shown weighs 7.5 kg and is attached to a spring of constant k = 800 N/m If end B of the rod is depressed 10 mm and released, determine (a) the period of vibration, (b) the maximum velocity of end B.
SOLUTION
k
m
==
800
7 5
N/m
kg.
where F k x
k
ma mr
I
= += +
= = =
=
( )
( . )
( . )( . )
dq d
a q q
a
ST
ST
m
0 5
7 5 0 12515
161
1
&& &&
227 5 0 75
45
128
2( . )( . ) &&
&&
q
q=
(a) Equation of motion.+ S SM M mg F I maC C= - = +( ) : ( . ) ( . ) ( . )eff m m m0 125 0 5 0 125a
mg k( . ) ( . )( . ) ( . )0 125 0 5 0 545
128
15
160 25- + = +q d q qST m && &&
But in equilibrium, we have + mg k( . ) ( . )0 125 0 5 0- =dST
47
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.38 (Continued)
Thus, - = +ÈÎÍ
˘˚
- =
+
k( . )
( )( . )
(
0 545
128
15
128
800 0 515
32
42
2
2
q q
q q
q
&&
&&
&& 66 667 0. )q =
Natural frequency and period.
ww
t pw
p
n
n
n
2 426 667
20 656
2 2
20 656
==
= =
.
.
.
rad/s
rad/s t = 0 304. s �
(b) At end B. xm = 10 mm = 0.01 m
v xm m n===
w( . )( . )
.
0 01 20 656
0 20656
m rad/s
m/s vm = 0 207. m/s �
48
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
SOLUTION
Spring deflection. x x x
x r
x
rx x
F k x k x
A A
A
A
S A
= +
=
=
== + = +
0 0
0
0
0
0
2
2
/
/
( ) ( )
q
q
d dST ST
S SM M rk x rw rmx IC = - + + ∞ = +( ) : ( ) sineff ST2 2 150 0d q&& && (1)
But in equilibrium, x0 0=
SM rk rwC = = - + ∞0 2 15dST sin (2)
Substituting Eq. (2) into Eq. (1) and noting that q q= =x
r
x
r0 0, && &&
rmx Ix
rrkx
I mr
mrx rkx
xk
m
&& &&
&&
&&
00
0
2
0 0
0
4 0
1
23
24 0
8
3
+ + =
=
+ =
+ ÊËÁ
ˆ¯
=x0 0
+
PROBLEM 19.39
A 15 kg uniform cylinder can roll without sliding on a 15° incline. A belt is attached to the rim of the cylinder, and a spring holds the cylinder at rest in the position shown. If the center of the cylinder is moved 50 mm down the incline and released, determine (a) the period of vibration, (b) the maximum acceleration of the center of the cylinder.
49
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.39 (Continued)
Natural frequency. wnk
m= = =8
3
8 6000
3 1532 6599 1( )( )
( )( ).
N / m
kgs-
(a) Period. t pw
pn
n
= =2 2
32 6599. t n = 0 1924. s �
(b) x x tm n0 0= +( ) sin( )w f
At t = 0, x x
x x t
t
x
m n n
m n
0 0
0 0
0
0 05 0
0
0
= == +==
.
( ) cos( )
( ) cos
m && w w f
w f
Thus, f p
f
=
== = =
20
0 0 05 10 0 0
t
x x xm m( ) . ( ) sin ( ) ( ) m
( ) .
( ) sin( )
( ) ( )
( )
max max
x
x x t
a x
x
m
m n n
0
0 02
0 0
0
0 05=
= - +=
= -
m
&&&&
w w f
mm nw2
1 20 05 32 6599
53 333
= -
=
-( . )( . )
.
m s
m/s2 ( ) .maxa0 53 3= m/s2 �
50
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.40
A 7.5 kg slender rod AB is riveted to a 6 kg uniform disk as shown. A belt is attached to the rim of the disk and to a spring which holds the rod at rest in the position shown. If end A of the rod is moved 20 mm down and released, determine (a) the period of vibration, (b) the maximum velocity of end A.
SOLUTION
Equation of motion. + S SM M mgL
kxr I mL L
IB B AB= - = + ÊËÁ
ˆ¯
ÊËÁ
ˆ¯
+( ) : coseff Disk2 2 2q a a a (1)
where x r= +q dST and from statics, mgL
k r2
= dST
Assuming small angles (cos ),q ª 1 Equation (1) becomes
mgL
kr kr I mL
I
ImL
I
AB
AB
2 2
4
22
2
q q d a- - = + ÊËÁ
ˆ¯
+Ê
ËÁ
ˆ
¯˜
+ +
ST Disk
Disk
ÊÊ
ËÁˆ
¯+ =&&q qkr2 0
Data: m = 7 5. kg
m
L
r
k
disk
mm m
mm m
kN/m N/m
== == == =
6
900 0 9
250 0 25
6 6000
kg
.
.
51
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.40 (Continued)
I mL
I m r
AB =
=
= ◊
=
=
1
121
127 5 0 9
0 50625
1
21
26
2
2
2
( . )( . )
.
(
kg m2
Disk Disk
))( . )
.
0 25
0 1875
2
= ◊kg m2
0 506251
47 5 0 9 0 1875 6000 0 25 02 2. ( . )( . ) . ( )( . )+ +È
Î͢˚
+ =&&q q
2 2125 375 0. &&q q+ = or &&q q+ =169 492 0.
(a) Natural frequency and period. wn2 2169 492= . ( )rad/s
w
t pw
pn
n
=
= =
13 0189
2 2
13 0189
.
.
rad/s
t = 0 483. s �
(b) Maximum velocity. v xm n m= =w ( . )( . )13 0189 0 02rad/s m vm = 260 m/s �
52
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.41
An 8-kg uniform rod AB is hinged to a fixed support at A and is attached by means of pins B and C to a 12-kg disk of radius 400 mm. A spring attached at D holds the rod at rest in the position shown. If Point B is moved down 25 mm and released, determine (a) the period of vibration, (b) the maximum velocity of Point B.
SOLUTION
(a)
Equation of motion. S SM M F kA A S= = +( ) : ( . )eff ST0 6q d
+ 0 6 1 2 0 6 1 2. ( ) . ( ) . ( )( ) . ( )( )m g F m g I I m a m aR S D R D R t D D t B- + = + + +a (1)
At equilibrium ( ),q = 0 F kS = dST
SM m g k m gA R D= = - +0 0 6 1 2. ( ( )) .dST (2)
Substituting Eq. (2) into Eq. (1), ( ) . ( ) . ( ) ( . )I I m a m a kR D R t D D t B+ + + + =a q0 6 1 2 0 6 02
a q
q
q
=
=
=
= =
=
&&&&&&
( ) .
( ) .
( )( . )
.
a
a
I m l
t B
t D
R R
0 6
1 2
1
12
1
128 1 2
0 960
2 2
kg m◊
53
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.41 (Continued)
I m RD D= = = ◊
+ + +
1
2
1
212 0 4 0 960
0 960 0 960 0 6 8 1
2 2
2
( )( . ) .
[ . . ( . ) ( ) (
kg m
.. ) ( )] ( . ) ( )
( . )
2 12 0 6 800 0
288
22 080
2 2&&&&
q q
q q
+ =
+ ◊◊
= N m
kg m2
(a) Natural frequency and period.
w
t pw
p
n
nn
=
=
= =
288
22 083 6116
2 2
3 6116
..
.
rad/s
t n = 1 740. s �
(b) Maximum velocity at B.
( ) ( . )( )
..
..
sin
max maxv
yB
mB
m
m
=
=
= =
=
1 2
1 20 025
1 20 02083
&q
q
q
q q
rad
(( )
cos( )
( . )( . ) .max
w f
q q w w f
q q w
n
m n n
m n
t
t
+
= +
= = =
&& 0 02083 3 612 0 075224 rad/s
( ) ( . )( ) ( . )( . )max maxvB = =1 2 1 2 0 07524&q m rad/s ( ) .maxvB = 0 09029 m/s ( ) .maxvB = 90 3 mm/s �
54
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.42
Solve Problem 19.41, assuming that pin C is removed and that the disk can rotate freely about pin B.
SOLUTION
(a)
Note: This problem is the same as Problem 19.41, except that the disk does not rotate, so that the effective moment IDa = 0.
Equation of motion. S SM M F kA A S= = +( ) : ( . )eff ST0 60 d
+ ( . )( ) . ( . )( )( ) . ( )( )0 6 1 2 0 6 1 2m g F m g I m a m aR S D R R t D D t B- + = + +a (1)
At equilibrium ( ),q = 0 F kS = dST
+ SM m g m gA R D= = - +0 0 6 1 2. ( ) .dST (2)
Substituting Eq. (2) into Eq. (1), I m a m a kR R t D D t Ba q+ + + =0 6 1 2 0 6 02. ( ) . ( ) ( . )
a q
q
q
=
=
=
= =
=
&&&&&&
( ) .
( ) .
( )( . )
.
a
a
I m l
t B
t D
R R
0 6
1 2
1
12
1
128 1 2
0 960
2 2
kg m◊
55
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.42 (Continued)
[ . ( . ) ( ) ( . ) ( )] ( . ) ( )
(
0 960 0 6 8 1 2 12 0 6 800 0
288
2 2 2+ + + =
+ ◊
&&&&
q q
q N m))
.21 120
2 kg m◊=q
(a) Natural frequency and period.
w
t pw
p
n
nn
=
=
= =
288
21 123 693
2 2
3 693
..
.
rad/s
t n = 1 701. s �
(b) Maximum velocity at B.
( ) ( . )( )
.
.
..
sin(
max maxvB
mB
m
=
= = =
=
1 2
1 2
0 025
1 200 02083
&q
qm
q q w
rad
nn
m n n
m n
t
t
+
= +
===
f
q q w w f
q q w
)
cos( )
( . )( . )
.
max
&&&&
0 02083 3 693
0 076994
1 2
1 2 0 07694
0 09233
rad/s
m/s
( ) ( . )( )
( . )( . )
.
max maxvB ===
&q
( ) .maxvB = 92 3 mm/s �
56
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.43
A belt is placed around the rim of a 240-kg flywheel and attached as shown to two springs, each of constant k = 15 kN/m. If end C of the belt is pulled 40 mm down and released, the period of vibration of the flywheel is observed to be 0.5 s. Knowing that the initial tension in the belt is sufficient to prevent slipping, determine (a) the maximum angular velocity of the flywheel, (b) the centroidal radius of gyration of the flywheel.
SOLUTION
Denote the initial tension by T0.
Equation of motion. + S SM M T r T r I
T k r r T kr I
A B0 0
0 0
= - + =
- + + - =
( ) :
( ) ( )
eff&&
&&q
q q q q
&&q q
w
+ =
=
20
2
2
22
kr
I
kr
In (1)
Data: m k
r
nn
= == =
= = = =
240 15
0 5 0 45
2 2 2
0 54
kg kN/m
s m
rad/s
t
t pw
w pt
p p
. .
;.
(a) Maximum angular velocity. If Point C is pulled down 40 mm and released,
q q
q q w
m
m m n
=
= ÊËÁ
ˆ¯
= ¥
=
= ¥
-
-
max
.
( .
40
450
88 89 10
88 92 10
3
3
mm
mm
rad
r
&aad rad/s)( )4p &qm = 1 117. rad/s �
57
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.43 (Continued)
(b) Centroidal radius of gyration.
w
p
nkr
I
I
I
22
22
2
2
42 15 0 45
38 47
=
=
= ◊
( )( )( . )
.
rad/skN/m m
kg m
or since I mk
k
=
= ◊
2
2240 28 47( ) .kg kg m2
k = 0 4004. m k = 400 mm �
58
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.44
A 75-mm-radius hole is cut in a 200-mm-radius uniform disk, which is attached to a frictionless pin at its geometric center O. Determine (a) the period of small oscillations of the disk, (b) the length of a simple pendulum which has the same period.
SOLUTION
Equation of motion.
S SM M0 0= ( ) :eff + - = - -m g I I mH D H H( . )sin ( . )0 1 0 1 2q q q q&& &&
m t R
t
t
m t r
t
D
H
=
==
=
==
r p
r ppr
r p
r p
2
2
2
2
0 2
0 04
0 075
0 00
( )( . )
( . )
( )( . )
( . 55625
1
2
1
20 04 2
800 10
1
21
2 2
6
2
)
( . )(. )
pr
pr
pr
t
I m R t
t
I m r
D D
H H
= =
= ¥
=
=
-
220 005625 0 75
15 82 10
2
6
( . )( . )
.
pr
pr
t
t= ¥ -
59
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.44 (Continued)
Small angles. sinq qª
[( . ( . ) ( . )]
( . )
800 10 15 82 10 0 1 0 005625
0 005625
6 6 2¥ - ¥ -+
- -p p p r qt &&pp r qt ( . )( . )9 81 0 1 0=
727 9 10 5 518 10 06 3. .¥ + ¥ =- -&&q q
(a) Natural frequency and period.
w
w
t pw
n
n
nn
23
6
5 518 10
727 9 107 581
2 753
2
= ¥¥
==
=
-
-.
..
. rad/s
t pn = =2
2 7532 28
.. s �
(b) Length and period of a simple pendulum.
t p
tp
p
n
n
l
g
l g
l
=
= ÊËÁ
ˆ¯
= ÈÎÍ
˘˚
2
2
2 753
29 81
2
2( . )( . )m/s2 l = 1 294. m �
60
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.45
Two small weights w are attached at A and B to the rim of a uniform disk of radius r and weight W. Denoting by t0 the period of small oscillations when b = 0, determine the angle b for which the period of small oscillations is 2 0t .
SOLUTION
a q
a q
=
= =
=
&&&&a r r
IW
gr
t
D1
22
Equation of motion.
S SM M wr wrw
gra I
wr
C C t= - - + = +
- -
( ) : sin( ) sin( )
[sin( ) si
eff b q b q a
b q
2
nn( )] sin cos
sin
( cos )
b q q bq q
q b q
+ = -ª
+ÊËÁ
ˆ¯
+
2
2
222 2
wr
w
gr
W
gr wr&& == 0
Natural frequency. w b b
b t pw
p
t
n W Ww
gr
n
wg
w r
g
r
Ww
=+( ) =
+
= = =
=
+
2
2
4
4
02 2
2
2
00
44
cos cos
( )
( )
pp t pbcos
( ) ( )4
0 44
24
+ +
= =Ww
Wwrg
r
(1)
cos b = ÊËÁ
ˆ¯
=1
2
1
4
2
b = ∞75 5. �
61
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.46
Two 50 g weights are attached at A and B to the rim of a 1.5 kg uniform disk of radius r = 100 mm . Determine the frequency of small oscillations when b = ∞60 .
SOLUTION
a q
a q
=
= =
=
&&&&a r r
IW
gr
t
D1
22
Equation of motion.
S SM M wr wrw
gra I
wr
C C t= - - + = +
- -
( ) : sin( ) sin( )
[sin( ) si
eff b q b q a
b q
2
nn( )] sin cos
sin
( cos )
b q q bq q
q b q
+ = -ª
+ÊËÁ
ˆ¯
+
2
2
222 2
wr
w
gr
W
gr wr&& == 0
Natural frequency. w b bn W W
w
wg
w r
g
r=
+( ) =+
2
2
4
42
cos cos
( ) (1)
Data: W
w
m
mr
n
= =¥
=
= = ∞
= ∞+
-1 5
50 1030
100 60
4 9 81 60
4
3
.
( )( . )cos
(
mm = 0.1 m b
w330 0 1
2 4022
2
2 4022
2
)( . ).
.
=
= =
rad/s
fnwn
p p fn = 0 382. Hz �
62
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.47
For the uniform square plate of side b = 300 mm,determine (a) the period of small oscillations if the plate is suspended as shown, (b) the distance c from O to a Point A from which the plate should be suspended for the period to be a minimum.
SOLUTION
(a) Equation of motion.
S SM M
I mb
a OG
OG b
a b
t
t
0 0
21
6
2
2
2
2
= =
=
=
=
=Ê
ËÁˆ
¯
( ) :
( )( )
eff a q
a
q
&&
&&
+ ( )(sin )( ) ( ) sinOG mg OG ma Itq a q q= - - ª
b m b mb b mg
b
2
2
2
2
1
6
2
20
1
2
1
6
2Ê
ËÁˆ
¯
Ê
ËÁˆ
¯+ +
Ê
ËÁˆ
¯=
+ÊËÁ
ˆ
&& &&q q q
( )¯
+Ê
ËÁˆ
¯=m mg&&q q2
20
63
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.47 (Continued)
&& &&q q q+( )( ) = + =
22
23
03 2
40
g
b
g
bor
Natural frequency and period.
w
w
t pw
n
n
nn
g
b02
0
00
3 2
4
3 2 9 81
4 0 334 68
5 889
2
= = =
=
=
( . )
( )( . ).
. rad/s
t n0 1 067= . s �
(b) Suspended about A. Let e OG c= -( )
a et = a
Equation of motion.
+S SM MA A= ( ) :eff mge ema I me I
m e b mge
tsin ( )q a a
q q
= - - = - +
+ÊËÁ
ˆ¯
+ =
2
2 21
60&&
Frequency and period. w
t pw
p
t p
n
nn
n
eg
e b
e b
eg
ge
b
e
22 1
62
22
2
2 2 16
2
22 2
4 4
4
6
=+
= =+
= +Ê
ËÁˆ
¯
( )
For t n to be minimum, d
dee
b
e+
Ê
ËÁˆ
¯=
2
60
16
0 66
2
2 60 29886
0 29886 300
2
2
2
2- = = =
= - = - =
=
b
e
b
ee
b
c OG e bb
b
c
.
( . )( mm)) c = 89 7. mm �
64
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.48
A connecting rod is supported by a knife-edge at Point A; the period of its small oscillations is observed to be 0.87 s. The rod is then inverted and supported by a knife-edge at Point B and the period of small oscillations is observed to be 0.78 s. Knowing that r ra b+ = 250 mmdetermine (a) the location of the mass center G, (b) the centroidal radius of gyration k .
SOLUTION
Consider general pendulum of centroidal radius of gyration k .
Equation of motion. + S SM M mgr mr r mk0 02= - = +( ) : sin ( )eff q q q&& &&
&&q q++
ÈÎÍ
˘˚
=gr
r k2 20sin
For small oscillations, sin ,q qª we have
&&q q
w
t pw
p
++
ÈÎÍ
˘˚
=
=+
= = +
gr
r kgr
r k
r k
gr
n
n
2 2
22 2
2 2
0
22
For rod suspended at A, t pAa
a
r k
gr=
+2
2 2
g r r kA a at p2 2 2 24= +( ) (1)
For rod suspended at B, t pBb
b
r k
gr=
+2
2 2
g r r kB b bt p2 2 2 24= +( ) (2)
65
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.48 (Continued)
(a) Value of ra .
Subtracting Eq. (2) from Eq. (1), g r g r r r
g r g r r r r r
A a B b a b
A a B b a b a b
t t p
t t p
2 2 2 2 2
2 2 2
4
4
- = -( )- = + -( )( )
Applying the numerical data with r ra b+ = =250 0 25mm m.
( . )( . ) ( . )( . ) ( . )( )
.
9 81 0 87 9 81 0 78 4 0 25
7 4252 5
2 2 2r r r r
ra b a b
a
- = --
p.. . ( )9684 9 8696r r rb a b= -
3 9012 2 4444 0 6266
0 25 0 6266 0 15369
. . .
. . .
r r r r
r r rb a b a
a a a
= == + = m ra = 153 7. mm �
rb = -0 25 0 15369. . rb = 0 09631. m rb = 96 31. mm �
(b) Centroidal radius of gyration.
From Eq. (1), 4 4
9 81 0 87 0 15369 4 0 15369 0
2 2 2 2 2
2 2 2
p p
p
k g r rA a a= -
= - =
t
( . )( . ) ( . ) ( . ) ..
. .
20867
0 072703 72 7
2m
m mmk k= =
66
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.49
For the uniform equilateral triangular plate of side l = 300 mm, determine the period of small oscillations if the plate is suspended from (a) one of its vertices, (b) the midpoint of one of its sides.
SOLUTION
For an equilateral triangle, h b A bh b= = =3
2
1
2
3
42,
I bh l l l
I hb
l
x
y
= =Ê
ËÁˆ
¯=
= - ÊËÁ
ˆ¯
=Ê
ËÁˆ
1
36
1
36
3
2
3
96
21
12 2
1
6
3
2
33
4
3
¯ÊËÁ
ˆ¯
=
= + =
= = = =
ll
I I I l
kI
Al
z x
z
2
3
96
3
481
12
1
120 300 0
34
44
2 2 2( . ) ..0075 2m
Let r be the distance from the suspension point to the center of mass.
Equation of motion. + SM0 = + S( ) : sin ( )M mge r mr I0 eff - = +q q q&& &&
where I mk= 2
For small angle q, - = +
++
=
mrg m r k
gr
r k
q q
q q
( )2 2
2 20
&&&&
Natural frequency: wngr
r k2
2 2=
+
67
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.49 (Continued)
(a) Plate is suspended from a vertex.
r h l
r
n
= = =
=
=
=
2
3
2
3
3
2
2
3
3
20 3
0 173205
0 03
9 81 0 1732
2
2
( . )
.
.
( . )( .
m
m
m2
w 005
0 03 0 007545 31
6 7313
2
)
. ..
.+
=
=
=
w
t pw
n
n
rad/s
t = 0 933. s �
(b) Plate is suspended at the midpoint of one side.
r h l
r
n
= =Ê
ËÁˆ
¯=
=
=
=
1
3
1
3
3
2
1
3
3
20 3
0 0866025
0 0075
9 81
2 2
2
( . )
.
.
( .
m
m
w ))( . )
. ..
.
0 0866025
0 0075 0 007556 64
7 5258+
=
=
=
w
t pw
n
n
rad/s
2 t = 0 835. s �
68
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.50
A uniform disk of radius r = 250 mm is attached at A to a 650-mm rod AB of negligible mass, which can rotate freely in a vertical plane about B. Determine the period of small oscillations (a) if the disk is free to rotate in a bearing at A, (b) if the rod is riveted to the disk at A.
SOLUTION
I mr
mm
a lt
=
= =
= =
=
1
21
20 250
320 650
2
2( . )
.a a
a q&&(a) The disk is free to rotate and is in curvilinear translation.
Thus, I a = 0
S SM MB B= ( ) :eff +- = ªmgl lmatsin sinq q q
ml mgl
g
In
n
n
2
2
0
9 81
0 650
15 092
3 885
2
&&q q
w
w
t
- =
= =
==
=
.
.
.
.
m/s
m
rad/s
2
ppw
p
n
= 2
3 885. t n = 1 617. s �
69
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.50 (Continued)
(b) When the disk is riveted at A, it rotates at an angular acceleration a.
S SM MB B= ( ) :eff
+- = + =mgl I lma I mrtsinq a 1
22
1
20
9 81 0 650
2 2
2
222
mr ml mgl
gl
ln r
+ÊËÁ
ˆ¯
+ =
=+( )
=
&&q q
w
( . )( . )m/s m2
00 2502
220 650
14 053
3 749
2
2
3 749
. ( . )
.
.
.
( ) +ÈÎ
˘˚
==
=
=
w
t pw
p
n
nn
rad/s
t n = 1 676. s �
70
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.51
A small collar weighing 1 kg is rigidly attached to a 3 kg uniform rod of length L = 1 m Determine (a) the distance d to maximize the frequency of oscillation when the rod is given a small initial displacement, (b) the corresponding period of oscillation.
SOLUTION
Equation of motion.
S SM MA A= ( )eff :
+
- - = + +WL
W d I mL
a m d aR C R R t R C t C2 2sin sin ( ) ( )q q a
sinq qª a q a q a q= = = = =&& && &&, ( ) , ( )aL L
a d dt R t C2 2
I mL
m d m gL
m gdR R C R C+ ÊËÁ
ˆ¯
+Ê
ËÁ
ˆ
¯˜ + +Ê
ËÁˆ¯
=2 2
02
2 &&q q
I m L
I mL m L
m Lm d m g
Lm
R R
R RR
RC R C
=
+ ÊËÁ
ˆ¯
=
+Ê
ËÁˆ
¯+ +
1
12
2 3
3 2
2
2 2
22 &&q ggdÊ
ËÁˆ¯
=q 0
71
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.51 (Continued)
&&
&&
q q
q
++( )+( ) =
= =
++( )
+
L mm
L mm
C
R
C
R
C
R
d g
d
m
mL
d
2
32
12
13
20
1
31
13
13
m
dd20
ÊËÁ
ˆ¯
=q
Natural frequency. wn
d g
d
d g
d2
12
13
13
13
2 2
1 5
1=
+( )+
=+( )
+.
( )
(a) To maximize the frequency, we need to take the derivative and set A equal to zero.
1 1 1 1 5 2
10
3 1 0
2 2
2 2
2
g
d
dt
d d d
d
d d
nw( )= + - +
+=
+ - =
( )( ) ( . )( )
( )
Solve for d. d = -0 3028 3 3028. .or
d = 0 303. m d = 303mm �
w
ww
n
n
n
22
2
1 5 0 3028 9 81
1 0 3028
16 200
4 0249
= ++
==
( . . )( . )
( . )
.
. rad/s wn = 4 02. rad/s �
(b) Period of oscillation. t pw
pn
n
= =2 2
4 0249. t n = 1 561. s �
72
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.52
A compound pendulum is defined as a rigid slab which oscillates about a fixed Point O called the center of suspension. Show that the period of oscillation of a compound pendulum is equal to the period of a simple pendulum of length OA, where the distance from A to the mass center G is GA k r= 2/ . Point A is defined as the center of oscillation and coincides with the center of percussion defined in Problem 17.66.
SOLUTION
+ S SM M W r I ma rt0 0= - = +( ) : sineff q a
- = +mgr mk mrsinq q q2 2&& &&
&&q q++
=gr
r k2 20sin (1)
For a simple pendulum of length OA l= ,
&&q q+ =g
l0 (2)
Comparing Equations (1) and (2), lr k
r= +2 2
GA l rk
r= - =
2
Q.E.D. �
73
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.53
A rigid slab oscillates about a fixed Point O. Show that the smallest period of oscillation occurs when the distance r from Point O to the mass center G is equal to k .
SOLUTION
See Solution to Problem 19.52 for derivation of
&&q q++
=gr
r k2 20sin
For small oscillations, sinq qª and
t pw
p pn
n
r k
gF gr
k
r= = + = +2
222 2 2
For smallest t n , we must have r kr+
2
as a minimum:
d r
dr
k
r
kr+( )
= - =
2
1 02
2
r k2 2= r k= Q.E.D. �
74
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.54
Show that if the compound pendulum of Problem 19.52 is suspended from A instead of O, the period of oscillation is the same as before and the new center of oscillation is located at O.
SOLUTION
Same derivation as in Problem 19.52 with r replaced by R.
Thus, &&q q++
=gR
R k20
Length of the equivalent simple pendulum is
LR k
RR
k
R
L l rk
lkr
= + = +
= - + =
2 2 2
2
2( )
Thus, the length of the equivalent simple pendulum is the same as in Problem 19.52. It follows that the period is the same and that the new center of oscillation is at O. Q.E.D.
75
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.55
The 8-kg uniform bar AB is hinged at C and is attached at A to a spring of constant k = 500 N/m. If end A is given a small displacement and released, determine (a) the frequency of small oscillations, (b) the smallest value of the spring constant k for which oscillations will occur.
SOLUTION
I ml
I
at
= = ÊËÁ
ˆ¯
= ◊
=
= =
1
12
1
128 0 250
0 04167
0 04
2 2( )( . )
.
.
kg m2
a q
a
&&00 04.
sin
&&qq qª
Equation of motion.
S SM MC C= ( )eff : - + = +( . ) . ( . )0 165 0 04 0 042 2k mg I mAq q q&& && ( . . ) ( . . )0 04167 0 01280 0 02722 0 32 0+ + - =&&q qk g (1)
(a) Frequency if N mk = ◊500 .
0 05447 10 47 0. ( . )&&q q+ =
fnn= =
( )wp p2 2
10 470 05447
..
fn = 2 21. Hz �
(b) For t n Æ or wn Æ 0, oscillations will not occur.
From Equation (1), wnk g2 0 02722 0 32
0 054470= - =. .
( . )
kg= =0 32
0 02722
0 32 9 81
0 02722
.
.
( . )( . )
( . ) k = 115 3. N/m �
76
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.56
A 20-kg uniform square plate is suspended from a pin located at the midpoint A of one of its 0.4 m edges and is attached to springs, each of constant k = 1 6. kN/m If corner B is given a small displacement and released, determine the frequency of the resulting vibration. Assume that each spring can act in either tension or compression.
SOLUTION
a q
a q
q q
=
= -
ª
&&&&a
b bt 2 2
sin
Equation of motion.
+
S SM M mgb
kb Ib
m0 02
2
22
2= ( ) - + = + Ê
ËÁˆ¯eff
: q q a a
I mb
mb mb
mb+ ÊËÁ
ˆ¯
= + =2
1
6 4
5
12
22
22
5
12 22 0
12
10
24
50
2 2mb mgb
kb
g
b
k
m
&&
&&
q q
q q
+ +ÊËÁ
ˆ¯
=
+ +ÊËÁ
ˆ¯
=
Data: b m
k
= == =
0 4 20
1 6 1600
.
.
m; kg
kN/m N/m
&&q q+ +È
ÎÍ
˘
˚˙ =( )( . )
( )( . )
( )( )
( )( )
12 9 81
10 0 4
24 1600
5 200
&&qw w
+ =
= =
413 43 0
413 43 20 3332
.
. .n n rad/s
Frequency. fnn= =
wp p2
20 333
2
. fn = 3 24. Hz �
77
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.57
Two uniform rods, each of mass m = 12 kg and length L = 800 mm, are welded together to form the assembly shown. Knowing that the constant of each spring is k = 500 N/m and that end A is given a small displacement and released, determine the frequency of the resulting motion.
SOLUTION
Equation of motion. S SM M0 0= ( )eff : + m gL
kL
I I mL
AC AC BD AC22
2 2
2 2
- ÊËÁ
ˆ¯
È
ÎÍÍ
˘
˚˙˙
= + + ÊËÁ
ˆ¯
q a a( )
m m m
I I I mL
BD AC
BD AC
= =
= = = 1
22
1
6
1
42
2 202
2
+ÊËÁ
ˆ¯
+ ÊËÁ
ˆ¯
-È
ÎÍÍ
˘
˚˙˙
=mL kL
mgL&&q q
10
24 2 20
6
52
22mL
kL mgL kL mg
mLn&&q q w+ -
È
ÎÍ
˘
˚˙ = fi = -( )
Data: L m k= = = =800 0 8 12 500 mm m, kg, N/m.
Frequency. wn2 6 500 0 8 12 9 81
5 12 0 835 285= - =[( )( . ) ( )( . )]
( )( )( . ).
ww
pn nnf= =5 9401
2. rad/s fn = 0 945. Hz �
78
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.58
The rod ABC of total mass m is bent as shown and is supported in a vertical plane by a pin at B and by a spring of constant k at C. If end C is given a small displacement and released, determine the frequency of the resulting motion in terms of m, L, and k.
SOLUTION
a q q
q q q
= =
= =
ª ª
&& &&aL
Im
LmL
t 2
1
12 2 241
22
sin cos
Equation of motion.
+ S SM Hmg L mg L
kL L Im
aB B= + - + = +( ) sin cos (eff ST: )cos2 2 2 2
2 22
q q q d q q&& ttL
2
mgL mgL
kL kLmL mL
4 4 12 42 2
2 2
q q d q q+ - - = +ST&& && (1)
But for equilibrium ( ),q = 0 SMm
gL
kLB = = -02 2
2dST (2)
79
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.58 (Continued)
Substituting Eq. (2) into Eq. (1), mgL
kLmL
kL mgL
mL
4 3
0
22
24
3
2
-ÊËÁ
ˆ¯
=
+-( )
=
q q
q q
&&
&&
Frequency. w
w
n
n
k
m
g
L
k
m
g
L
2 3 3
4
34
= -
= -
fnn=
wp2
fk
m
g
Ln = -3
2 4p �
80
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.59
A uniform disk of radius r = 250 mm is attached at A to a 650-mm rod AB of negligible mass which can rotate freely in a vertical plane about B. If the rod is displaced 2° from the position shown and released, determine the magnitude of the maximum velocity of Point A, assuming that the disk (a) is free to rotate in a bearing at A, (b) is riveted to the rod at A.
SOLUTION
I mr= 1
22
Kinematics: a q
a q
=
= =
&&&&a l lt
(a) The disk is free to rotate and is in curvilinear translation.
Thus, I a = 0
Equation of motion. S SM M mgl lmaB B t= - = ª( ) sin , sineff : q q q
ml mgl2 0&&q q+ =
Frequency. w
w
n
n
g
l2
9 81
0 65015 092
3 8849
=
=
==
.
..
. rad/s
t pwn
n
= =21 617. s t n = 1 617. s �
81
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.59 (Continued)
(b) When the disk is riveted at A, it rotates at an angular acceleration a.
Equation of motion. S SM M mgl I lma I mrB B t= - = + = ª( ) sin , , sineff : q a q q1
22
1
202 2mr ml mgl+Ê
ËÁˆ¯
+ =&&q q
Frequency. w
w
n r
n
gl
l
2
22
12
2 2
2
9 81 0 650
0 250 0 650
14 053
3
=+( )
=+
==
( . )( . )
( . ) ( . )
.
..7487 rad/s
t pwn
n
= =21 676. s t n = 1 676. s �
82
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.60
A 3-kg slender rod is suspended from a steel wire which is known to have a torsional spring constant K = ◊2 25. N m/rad. If the rod is rotated through 180° about the vertical and released, determine (a) the period of oscillation, (b) the maximum velocity of end A of the rod.
SOLUTION
Equation of motion. S SM M K IK
IK
I
G G
n n
= - = + =
+ = =
( )eff : q q q q
q w q w
&& &&
&&
0
02 2
Data: m
l
I ml
K
== =
= = ( )
= ◊= ◊
3
200 0 2
1
12
1
123 0 2
0 01
2 25
2 2
kg
mm m
kg m
N m/
2
.
( ) .
.
. rrad
rad/s
w
w
n
n
2 2 25
0 01225
15
= =
=
.
.
(a) Period of oscillation. t pw
pn
n
= =2 2
15 t n = 0 419. s �
Simple harmonic motion: q q w j
q w q w j
q w q
q w q
= +
= +
=
= =
m n
n m n
m n m
A m m A m
t
t
vl
l
sin ( )
cos( )
( )
&&
&2
1
2qq pm = ∞ =180 radians
(b) Maximum velocity at end A. ( ) ( . )( )( )vA m = ÊËÁ
ˆ¯
1
20 2 15 p ( ) .vA m = 4 71m/s �
83
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.61
A homogeneous wire bent to form the figure shown is attached to a pin support at A. Knowing that r = 220 mm and that Point B is pushed down 20 mm and released, determine the magnitude of the velocity of B, 8 s later.
SOLUTION
Determine location of the centroid G.
Let r = mass per unit length
Then total mass m r r r= + = +r p r p( ) ( )2 2
About C: mgc rr
g r g= + ÊËÁ
ˆ¯
=02
2 2p rp
r
yr
r c r cr
=
+ = =+
2
2 22
22
p
r p rp
for a semicircle
( ) ,( )
Equation of motion. S SM M a c c
mgc I mcat
n
0 0= = = =- = + ª
( ) :
sineff
sin
a q a qq a q q
&& &&
(( )I mc mgc I mgc+ + = + =200 0&& &&q q q q
Moment of inertia. I mc I
I I I m r mr
+ =
= + = +
20
0 0 02
22
1( ) ( )
( )semicirc. line semicirc. line 22
22
02
m r m rm
r
I r
semicirc. line = = =+
= ◊
rp r rp
r p
( )
rr rr mr+ ◊
È
ÎÍ
˘
˚˙ =
+( ) +ÈÎÍ
˘˚
23 2
2
3
2 2
pp
mrmg
r2
2
2
3
2
20
( ) ( )++È
Î͢˚
++
=p
p qp
q&&
+
84
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.61 (Continued)
Frequency. wp p
w w
n
n n
g
r2
23
23
2 2
2 2 9 81
0 220
23 418 4 8392
=+( ) =
+( )= =-
( )( . )
( . )
. .s rad/s
q q w f q= + =m n Bt y rsin( )
At t = 0, mm,
mm
y y
y y
y y
B B
B B m n
B B m
= =
= = + =
= =
20 0
0 02
20
&& ( ) cos( ),
( ) si
w f f p
nn ,
( sin .
02
20
202
4 8
+ÊËÁ
ˆ¯
=
= +ÊËÁ
ˆ¯
=
p
w p w
( ) mm
mm)
y
y t
B m
B n n& 3392
202
20
rad/s
mm&y t tB n n n= +ÊËÁ
ˆ¯
= -w w p w wcos ( ) sin
At t = 8 s, &yB = - = -( )( . )sin[( . )( )] ( . )( . )20 4 8392 4 8392 8 96 88 0 8492
= -82 2. mm/s vB = 82 2. mm/s � �
85
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.62
A homogeneous wire bent to form the figure shown is attached to a pin support at A. Knowing that r = 400 mm and that Point B is pushed down 40 mm and released, determine the magnitude of the acceleration of B, 10 s later.
SOLUTION
Determine location of the centroid G.
Let r = mass per unit length
Then total mass m r r r= + = +r p r p( ) ( )2 2
About C: mgc rr
g r g= + ÊËÁ
ˆ¯
=02
2 2p rp
r
yr
r c r cr
=
+ = =+
2
2 22
22
p
r p rp
for a semicircle
( )( )
Equation of motion. S SM M a c c
mgc I mcat
n
0 0= = = =- = + ª
( ) :
sineff
sin
a q a qq a q q
&& &&
(( )I mc mgc I mgc+ + = + =200 0&& &&q q q q
Moment of inertia. I mc I+ =20
I I I m r mr
m
0 0 02
22
12= + = +( ) ( )
( )semicirc. line semicirc. line
semicirrc. line= = =+
= ◊ + ◊È
ÎÍ
˘
˚˙ =
+
rp r rp
r pp
p
r m rm
r
I r rr r mr
22
2
3 202
2 2
( )
( )++È
Î͢˚
++È
Î͢˚
++
=
2
3
2
2
3
2
20
2mrmg
r
( ) ( )pp q
pq&&
+
86
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.62 (Continued)
Frequency. wp p
w w
n
n n
g
r2
23
23
2 2
2 2 9 81
0 4
12 8799 3
=+( ) =
+( )= =
( )( . )
( . )
. ) (rad/s ..
sin( )
sin( ) ( ) sin(
5889 rad/s
q q w f qq w f w
= + == + =
m n B
B m n B m
t y r
y r t y nnt + f)
At t = 0, mm
cos(0 )
mm
y y
t y y
y y
B B
B B m
B B m
= =
= = = + =
= =
40 0
0 02
40
&&( ): ( )
( ) sin
f f p
002
40
402
3 5889
+ÊËÁ
ˆ¯
=
= +ÊËÁ
ˆ¯
=
p
w p w
( )
( )sin .
y
y t
B m
B n n
mm
mm radd/s
&
&&
y t t
y
B n n n n
B n
= +ÊËÁ
ˆ¯
= -
= -
( )( )cos ( ) sin
( )
402
40
40 2
w w p w w
w(( ) +ÊËÁ
ˆ¯
=sin cosw p w wn n nt t2
40 2
At t = 10 s, ( ) ( )( . ) cos( . )( ) .
(
a y
vB t B
B
= = - = -&& 40 3 5889 3 5889 10 122 1242 mm/s2
)) ( )( . )sin ( . )( ) .= = - [ ] = -&yB 40 3 5889 3 5889 10 139 465 mm/s
a av
rB B tB= +
È
ÎÍ
˘
˚˙
È
ÎÍÍ
˘
˚˙˙
= + -È( ) ( . )
( . )22 2
12
22
122 124139 465
400ÎÎÍ
˘
˚˙
È
ÎÍÍ
˘
˚˙˙
=2
12
122 3. mm/s2 �
87
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.63
A uniform disk of radius r = 120 mm is welded at its center to two elastic rods of equal length with fixed ends at A and B. Knowing that the disk rotates through an 8∞ angle when a 500-mN m◊ couple is applied to the disk and that it oscillates with a period of 1.3 s when the couple is removed, determine (a) the mass of the disk, (b) the period of vibration if one of the rods is removed.
SOLUTION
Torsional spring constant.
kT
k
= = ◊
( )= ◊
q p0 5
3 581
.
.
N m
(8)
N m/rad180
Equation of motion.
S SM M K JK
J0 0 0= - = + =( ) :eff q q q q&& &&
Natural frequency and period. wnK
J2 =
Period. t pw
pnn
J
K= =2
2
Mass moment of inertia. JK
J mr
n= = ◊
= ◊ ◊ = =
tp p
2
2
2
2
2 2
2
1 35 3581
2
0 15331
2
( )
( . ) ( )
( )
.
N m/r
N m s11
20 120 2m( . ) m
(a) Mass of the disk. m =◊ ◊( . )( )
( . )
0 1533 2
0 120
2
2
N m s
m m = 21 3. kg �
(b) With one rod removed: ¢ = = = ◊KK
2
3 581
21 791
.. N m/rad
Period. t p p=¢
=◊ ◊
◊2 2
0 1533 2J
K
( . )N m s
1.791 N m/rad t = 1 838. s �
88
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.64
A 5 kg uniform rod CD of length l = 0 75. m is welded at C to two elastic rods, which have fixed ends at A and B and are known to have a combined torsional spring constant K = ◊30 N m/rad. Determine the period of small oscillation if the equilibrium position of CD is (a) vertical as shown, (b) horizontal.
SOLUTION
(a) Equation of motion.
+
a q a q
q q a
q
= = =
= - - = +
-
&& &&al l
M M K mgl
Il
ma
K
t
C C t
2 2
2 2S S( ) : ( ) sin ( )eff
-- = +1
2
1
42mgl I mlsinq q q&& &&
I ml mgl K
ml ml K
+ÊËÁ
ˆ¯
+ + =
+ÊËÁ
ˆ¯
+ +
1
4
1
20
1
12
1
4
1
2
2
2 2
&&
&&
q q q
q q
sin
mmgl
ml K mgl
K
ml
g
l
q
q q
q q
=
+ +ÊËÁ
ˆ¯
=
+ +ÊËÁ
ˆ¯
=
0
1
3
1
20
3 3
20
2
2
&&
&&
89
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.64 (Continued)
Data: K m l= ◊ = =
+ +
30 5 0 75
3 30
5 0 75
3 9 812
N m/rad kg m, , .
( )( )
( )( . )
( )( . )&&q(( )( . )
.
2 0 750
51 62 0
È
ÎÍ
˘
˚˙ =
+ =
q
q q&& Frequency. w wn n
2 51 62 7 1847= =. . rad/s
Period. t pw
p= =2 2
7 1847n . t n = 0 875. s �
(b) If the rod is horizontal, the gravity term is not present and the equation of motion is
&&q q
w
+ =
= = =
30
3 3 30
5 0 7532
2
22 2
K
mlK
mln
( )( )
( )( . )
w t pw
pn
n
= = =5 99382
5 65685.
. rad/s
2 t n = 1 111. s �
90
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.65
A 1.8-kg uniform plate in the shape of an equilateral triangle is suspended at its center of gravity from a steel wire which is known to have a torsional constant K = ◊35 mN m/rad. If the plate is rotated 360° about the vertical and then released, determine (a) the period of oscillation, (b) the maximum velocity of one of the vertices of the triangle.
SOLUTION
Mass moment of inertia of plate about a vertical axis:
h b A bh b= = =3
2
1
2
3
42
For area, I I bhb
I I I b
x y
z x y
= = =
= + =
1
36
3
96
3
48
34
4
For mass, Im
AI
bb mb
I
z=
= ÊËÁ
ˆ¯
Ê
ËÁˆ
¯=
=
( )
( . )( . )
area
m4
3
3
48
1
12
1
121 8 0 150
4 2
22 3 23 375 10= ¥ ◊-. kg m
Equation of motion. + = - =
+ =
S SM M K I
K
I
G G( ) :eff q q
q q
&&
&& 0
Frequency. w
w
n
n
K
I2
3
3
35 10
3 375 1010 37
3 2203
= = ¥¥
=
=
-
-..
. rad/s
91
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.65 (Continued)
(a) Period. t pw
p= =2 2
3 2203n . t = 1 951. s �
Maximum rotation. q pm = ∞ =360 2 rad
Maximum angular velocity. &q w q pm n m= = =( . )( ) .3 2203 2 20 234 rad/s
(b) Maximum velocity at a vertex.
v r h bm m m= = = = ÊËÁ
ˆ¯
Ê
ËÁˆ
¯& &q q2
3
2
3
3
2
2
3
3
20 150 20 234( . )( . )
vm = 1 752. m/s �
92
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.66
A horizontal platform P is held by several rigid bars which are connected to a vertical wire. The period of oscillation of the platform is found to be 2.2 s when the platform is empty and 3.8 s when an object A of unknown moment of inertia is placed on the platform with its mass center directly above the center of the plate. Knowing that the wire has a torsional constant K = ◊30 N m/rad, determine the centroidal moment of inertia of object A.
SOLUTION
Equation of motion. S SM M K I
K
I
K
I
G G
n
= - =
+ = =
( ) :eff q q
q q w
&&
&& & 0 2
Case 1. The platform is empty. w pt
p
w
n
n
IK
11
11
2 2
2 2
2 22 856
30
2 8563 6779
= = =
= = = ◊
..
( . ).
rad/s
kg m2
Case 2. Object A is on the platform. w pt
p
w
n
n
IK
22
22
2 2
2 2
3 81 653
30
1 65310 9793
= = =
= = = ◊
..
( . ).
rad/s
kg m22
Moment of inertia of object A. I I IA = -2 1 I A = ◊7 30. kg m2 �
93
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.67
A thin rectangular plate of sides a and b is suspended from four vertical wires of the same length l. Determine the period of small oscillations of the plate when (a) it is rotated through a small angle about a vertical axis through its mass center G, (b) it is given a small horizontal displacement in a direction perpendicular to AB, (c) it is given a small horizontal displacement in a direction perpendicular to BC.
SOLUTION
(a) Plate is rotated about vertical axis.
Similarly, for B, D, and C,
S SM M Tr
lr J T
mg
mgr
JlJ m a b
r
G G= - ◊ = =
= = = +
( ) :
( )
eff 44
01
12
22 2
q q
q q
&&
&&
22 2 21
4= +( )a b
t pw
p
t p p
nn
n
Jl
mgr
m a b l
mg a b
l
g
= =
=++
=
22
2 23
2
112
2 2
14
2 2
( )
( ) �
(b) Plate is moved horizontally. The plate is in curvilinear translation.
sin
cos
cos
q qq
q q q
ªª
ª =
1
l l a&& &&
94
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.67 (Continued)
+SF T mg
Tmg
y = = - =
=
0 4 0
4
( cos )q
+®S SF F T ma
l g
H H= - =
+ =
( ) : sineff 4
0
q
q q&&
t pnl
g= 2 �
(c) Since the oscillation about axes parallel to AB (and CD) is independent of the length of the sides of the plate, the period of vibration about axes parallel to BC (and AD) is the same.
t pnl
g= 2 �
95
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.68
A circular disk of radius r = 0 8. m is suspended at its center C from wires AB and BC soldered together at B. The torsional spring constants of the wires are K1 100= ◊ N m/rad for AB and K2 50= ◊ N m/rad for BC. If the period of oscillation is 0.5 s about the axis AC, determine the mass of the disk.
SOLUTION
Spring constant. Let T be the torque carried by the wires.
q q
q q q
B A C B
C A B A C B
T
K
T
K
K KT
T
K
K K
/ /
/ / /eq
eq
= =
= + = +ÊËÁ
ˆ¯
=
=
1 2
1 2
1
1 1
1 1 ++ =+
=+
= ◊1 100 50
100 5033 333
2
1 2
1 2KK
K K
K Keq N m/rad( )( )
.
Equation of motion. S SM M K I
K
IK
II
K
C C
nn
= - =
+ =
= =
( ) :eff eq
eq
eq eqor
q q
q q
ww
&&
&& 0
22
But t n = 0 5. s: w pt
pn
n
= = =2 2
0 512 566
.. rad/s
Then I = = ◊33 333
12 5660 21109
2
.
( . ). kg m2
For a circular disk, I mr= 1
22
mI
r= =2 2 0 21109
0 82 2
( )( . )
( . ) m = 0 660. kg �
96
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.69
Determine the period of small oscillations of a small particle which moves without friction inside a cylindrical surface of radius R.
SOLUTION
Datum at 2 :
Positionj
T
V WR m
1
1
0
1
== -( cos )q
Small oscillations:
( cos ) sin1 22 2
2
22
1
2
- = ª
=
qq q
q
mm m
mVWR
Positionk
v R T mv mR
V
m m m m= = =
=
& &q q22 2 2
2
1
2
1
20
Conservation of energy. T Y T V
WR mR
W mg
mgR mR
mm m n m
mn
1 1 2 2
22 2
22
02
1
20
2
1
2
+ = +
+ = + =
=
=
qq q w q
qw
& &
22 2qm
nWg
R=
Period of oscillations. t pw
pnn
R
g= =2
2 �
97
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.70
A 400 g sphere A and a 300 sphere C are attached to the ends of a rod AC of negligible weight which can rotate in a vertical plane about an axis at B. Determine the period of small oscillations of the rod.
SOLUTION
Datum at j:
Positionj
T
V W h W h
h BC
h BA
C C A A
C m
A m
1
1
0
1
1
== -= -= -
( cos )
( cos )
Small angles.
12
2
0 3 0 2
2
1
2
1
- =
= -
= -
cos
[( )( ) ( )( )]
( . )( . ) (
q
mm
C AmV m g BC m g BA
V kg m 00 4 0 1252
9 81
0 0981
2
1
. )( . ) ( . )
( . )
kg m m/s2[ ]ÊËÁ
ˆ
¯=
qm
V
Positionk
V T m v m v v
T m
C m A A m C m m
C m
2 2 02 2
22 2
01
2
1
20 2
1
20 2
= = + =
=
( ) ( ) , ( ) .
( . )
&
&
q
q ++ =
= +
1
20 125 0 125
1
20 3 0 2 0 4 0 1
2 2
2
m v
T
A m A m m( . ) ( ) .
( . )( . ) ( . )( .
& &q q
2 225
1
20 01825
2 2 2
22 2
)
( . )
2ÈÎ ˘ =
=
W
T
n m m n m
n m
q q w q
w q
&
98
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.70 (Continued)
Conservation of energy. T V T V mn m
n
1 1 2 2
22 2
2
0 0 09812
0 01825
20 0981
0 0182
+ = + + =
=
: ..
( . )
( .
qw q
w55
5 3753)
.=
Period of oscillations. t pw
pn
n
= =2 2
5 3753. t n = 2 71. s �
99
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.71
A 1.8-kg collar A is attached to a spring of constant 800 N/m and can slide without friction on a horizontal rod. If the collar is moved 70 mm to the left from its equilibrium position and released, determine the maximum velocity and maximum acceleration of the collar during the resulting motion.
SOLUTION
Datum at j:
Positionj
T V kxm1 120
1
2= =
Positionk
T mv V v x
x x
T V T V kx mx
m
m n m
m m
2 22
2 2
1 1 2 22 2
1
20
01
2
1
20
1
2
= = =
=
+ = + + = +
&&
&w
kkx m xk
mm n m n
n n
2 2 2
2 2
1
2
800
1 8
444 4 21 08
= = =
= =-
w w
w w
2 N/m
kg
s r
.
. . aad/s
&x xm n m= = =-w ( . )( . ) .21 08 0 070 1 4761 s m m/s �
&&x xm n m= = =-w2 221 08 0 070 31 1( . )( . ) . s m m/s2 �
100
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.72
A 1.5-kg collar A is attached to a spring of constant 1 kN/m and can slide without friction on a horizontal rod. The collar is at rest when it is struck with a mallet and given an initial velocity of 1 m/s. Determine the amplitude of the resulting motion and the maximum acceleration of the collar.
SOLUTION
m k= = =1 5 1 1000. kg kN/m N/m
Positionj Immediately after collar is struck:
x v
V T mv
= =
= = = = ◊
0 1
01
2
1
21 5 1 0 751 1
2 2
,
, ( . )( ) .
m/s
N m
Positionk v = 0, x is maximum, i.e., x x
V kx x x T
m
m m m
=
= = = =22 2 2
21
2
1
21000 500 0( )
Conservation of energy. T V T V
xm
1 1 2 2
20 75 0 0 500
+ = +
+ = +.
Amplitude of motion. xm = 0 03873. m xm = 38 7. mm �
Maximum force: F kxm m= = =( )( . ) .1000 0 03873 38 73 N
Maximum acceleration. aF
mmm= = 38 73
1 5
.
.
am = 25 82. m/s2 am = 25 8. m/s2 �
101
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.73
A uniform rod AB can rotate in a vertical plane about a horizontal axis at C located at a distance c above the mass center G of the rod. For small oscillations, determine the value of c for which the frequency of the motion will be maximum.
SOLUTION
Find wn as a function of c.
Datum at k:
Positionj T V mgh
V mgc
V mgc
m
mm m
m
1 1
1
22
1
2
0
1
1 22 2
2
= == -
- = ª
=
( cos )
cos sin
q
qq q
q
Positionk T I
I I mc ml mc
T ml
c
C m
C
m
22
2 2 2
2
22 2
1
21
12
1
2 12
=
= + = +
= +Ê
ËÁˆ
¯
&
&
q
q V
T V T V mgc ml
cm m
m n m
2
1 1 2 2
2 22
2
0
02 12 2
0
=
+ = + + = +Ê
ËÁˆ
¯+
=
q q
q w q
&
&
ggc ml
c
gc
c
n
n l
= +Ê
ËÁˆ
¯
=+( )
22 2
2
122
12
2
w
w
Maximum c, when d
dc
g c c g
c
nl
l
w212
2 2
122
02
0
2
2= =
+( ) -
+( ) =
l
c2
2
120- = c
l=12
�
102
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.74
A homogeneous wire of length 2 l is bent as shown and allowed to oscillate about a frictionless pin at B. Denoting by t0 the period of small oscillations when b = 0, determine the angle b for which the period of small oscillations is 2 0t .
SOLUTION
We denote by m the mass of half the wire.
Positionj Maximum deflections:
T V mgl
mgl
mgl
m m
m m
1 102 2
2
= = - - - +
= - +
, cos( ) cos( )
(cos cos sin
q b q b
q b q ssin cos cos sin sin )
cos cos
b q b q b
b q
+ -
= -
m m
mV mgl1
For small oscillations, cos
cos cos
q q
b b q
m m
mV mgl mgl
ª -
= - +
11
21
2
2
12
Positionk Maximum velocity:
T I I mlB m B22 21
22
1
3= = Ê
ËÁˆ¯
&q but
Thus, T ml
V mgl
mgl
m22 2
2
1
2
2
3
22
= ÊËÁ
ˆ¯
= - ÊËÁ
ˆ¯
= -
&q
b bcos cos
103
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.74 (Continued)
Conservation of energy. T V T V1 1 2 2+ = +
- + = -mgl mgl ml mglm mcos cos cosb b q q b1
2
1
32 2 2&
Setting &q q wm m n= , 1
2
1
32 2 2 2mgl mlm m ncos b q q w=
w b t pw
pbn
n
g
l
l
g2 3
2
22
2
3= = =cos
cos (1)
But for b = 0, t p0 22
3= l
g
For t t= 2 0 , 22
32 2
2
3p
bpl
g
l
gcos=
Ê
ËÁˆ
¯
Squaring and reducing,
1
41
4coscos
bb= = b = ∞75 5. �
104
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.75
The inner rim of an 40 kg flywheel is placed on a knife edge, and the period of its small oscillations is found to be 1.26 s. Determine the centroidal moment of inertia of the flywheel.
SOLUTION
Datum at j:
Positionj
T J Vm1 02
11
20= =&q
Positionk
T V mgh
h r r
r
V mgr
mm
m
m
2 2
2
2
2
2
0
1 22
2
2
= =
= - =
ª
=
( cos ) sinqq
q
q
Conservation of energy.
T V T V I mgrmm
1 1 2 2 02
21
20 0
2+ = + + = +: &q q
For simple harmonic motion, &q w q
w q q w t pw
pm n m
n m m n nn
I mgrmgr
J
I
mgr
=
= = = =02 2 2 2
0
22
2
204 4( )
Moment of inertia. I I mr I mrmgr
Imgr
mr
n
n
02 2
2
2
2
22
2
4
4
1 26 40
= + + =( )
=( )
- =
t
pt
p
( )
( ) ( . ) ( )(( . ) .( )
.9 81
4
0 35
240
0 35
22
2
pÊËÁ
ˆ¯
- ÊËÁ
ˆ¯
I = -2 7615 1 225. . I = ◊1 537. kg m2 �
105
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.76
A connecting rod is supported by a knife edge at Point A; the period of its small oscillations is observed to be 1.03 s. Knowing that the distance ra is 150 mm, determine the centroidal radius of gyration of the connecting rod.
SOLUTION
Positionj Displacement is maximum.
T V mgr mgra m a m1 120 1
1
2= = - ª, ( cos )q q
Positionk Velocity is maximum.
( )v r
T mv I mr mk
m r k
G m m
G m a m m
a
=
= + = +
= +
q q q
&
& & &2
2 2 2 2 2 2
2
1
2
1
2
1
2
1
21
222 2
2 0
( )=
&qm
V
For simple harmonic motion, &q w qm n m=
Conservation of energy. T V T V1 1 2 2+ = +
01
2
1
202 2 2 2 2
22 2
22
+ = +( ) +
=+
= -
mgr m r k
gr
r kk
grr
a m a n m
na
a
a
na
q w q
ww
or 22
Data: t w pt
pn n
n
ar g
= = = =
= = =
1 032 2
1 036 1002
150 0 15 9 81
..
.
. .
s rad/s
mm m m/s2
kk 22
29 81 0 15
6 10020 15 0 039543 0 0225 0 017043= - = - =( . )( . )
( . )( . ) . . . mm2
k = 0 13055. m k = 130 6. mm �
106
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.77
The rod ABC of total mass m is bent as shown and is supported in a vertical plane by a pin at B and a spring of constant k at C. If end C is given a small displacement and released, determine the frequency of the resulting motion in terms of m, L, and k.
SOLUTION
Positionj T I V kB m12
121
2
1
2= =&q d( )ST
Positionk T Vm
ghm
gL
k L
hL L
m m
m
2 22
2
02 2 2
1
2
21 2
2
= = - - + +
= - =
sin ( )
( cos ) sin
q q d
q
ST
q q q q d
q q
m
m m m
m m
hmL
Vm
gL m
gL
k L
2
4 2 4 2 2
1
2
2
22 2ª = - - + +
ª
( )
sin
ST
Conservation of energy.
T V T V
I km
gL mg L
k LB mm
m m
1 1 2 2
2 22
2 21
2
1
20
2 4 2 2
1
2
+ = +
+ = - - + +&q dq
q q dST STT ST2 2+È
Î͢˚L md q (1)
When the rod is in equilibrium,
+
SMm
gL
kLB = = -02 2
dST (2)
107
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.77 (Continued)
Substituting Eq. (2) into Eq. (1)
I kLmgL
I kLmgL
B m m m n m
B n m m
& &q q q w q
w q q
2 2 2
2 2 2 2
4
4
= -ÊËÁ
ˆ¯
=
= -ÊËÁ
ˆ¯
ww
w
n
mgL
B
B
n
mgL
mL
kL
I
Im
LmL
kL k
m
22
4
22
22
4
3
21
3 2 3
3 32
=-
= ÊËÁ
ˆ¯
=
=-
= - gg
L4
Frequency of oscillation.
&f k
m
g
Lnn= = -
wp p2
3
2 4 �
108
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.78
A 7.5 kg uniform cylinder can roll without sliding on an incline and is attached to a spring AB as shown. If the center of the cylinder is moved 10 mm down the incline and released, determine (a) the period of vibration, (b) the maximum velocity of the center of the cylinder.
SOLUTION
(a) Positionj
T V k r m1 120
1
2= = +( )d qST
Positionk
T J mv
V mgh k
m m22 2
22
1
2
1
21
2
= +
= +
&q
d( )ST
Conservation of energy.
T V T V k r J mv mgh k
k
m m m1 1 2 22 2 2 20
1
2
1
2
1
2
1
2+ = + + + = + + +: ( ) ( )d q q d
d
ST ST&
SST ST ST2 2 2 2 2 22 2+ + = + + +k r kr J mv mgh km m m md q q q d& (1)
When the disk is in equilibrium,
+SM mg r k rc = = -0 sin b dST
Also, h r m= sin b q
Thus,
mgh k r- =dST 0 (2)
109
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.78 (Continued)
Substituting Eq. (2) into Eq. (1)
kr J mv
v r r
kr J mr
m m m
m n m m m n m
m m n
2 2 2 2
2 2 2 2 2
q q
q w q q w q
q q w
= +
= = =
= +
&& &
( )
wwnkr
J mr2
2
2=
+ J mr= 1
22
wnkr
mr mr
k
m2
2
2 212
2
3=
+=
t pw
p pn
n
= = = =2 2
23
9007 5
2
800 702
( )( . )
.N/mkg
s �
(b) Maximum velocity.
v r
v r r
m m
m m n
m m n m
=
=
= = =
&&
q
q q w
q w q 10
10000 01. m
vm = =( . ) .0 01 80 0 0894m/s m/s �
110
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.79
Two uniform rods, each of weight m = 600 g and length l = 200 mm, are welded together to form the assembly shown. Knowing that the constant of each spring is k = 120 N/m and that end A is given a small displacement and released, determine the frequency of the resulting motion.
SOLUTION
Mass and moment of inertia of one rod. m = 600 g = 0.6 kg
I ml= = = ¥ ◊-1
12
1
120 6 0 2 2 102 2 3( . )( . ) kg m2
Approximation.
sin tan
cos sin
q q q
q
m m m
mm
m
ª ª
- = ª1 22
1
22 2
Spring constant: k = 120 N/m
Positionj T I ml
m m
m
12
2
3 2
21
2
1
2 2
21
22 10
= ÊËÁ
ˆ¯
+ ÊËÁ
ˆ¯
= ÊËÁ
ˆ¯
¥ +-
& &
&
q q
q( ) ( )11
20 6 0 1
5 10
0
2
3 2
1
ÊËÁ
ˆ¯ ( )
= ¥=
-
( . ) . &
&q
q
m
m
V
111
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.79 (Continued)
Positionk T
VWl
kl
m m
2
2
2
0
21 2
1
2 2
0 6 9 81 0
=
= - - + ÊËÁ
ˆ¯
ÊËÁ
ˆ¯
ª -
( cos )
( . )( . )( .
q q
22
2
1
22
1
2120
0 2
2
1 4943
22
2
)( ) ( )
.
.
q q
q
m m
m
ÊËÁ
ˆ¯
+ ÊËÁ
ˆ¯
ÊËÁ
ˆ¯
ª
Conservation of energy. T V T V1 1 2 2+ = +
5 10 0 0 1 4943
17 2876
3 2 2¥ + = +
=
- &&
q q
q qm m
m m
.
.
Simple harmonic motion. &q w qm n m=
wn = 17 2876. rad/s
Frequency. fnn=
wp2
fn = 2 75. Hz �
112
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.80
A slender 8-kg rod AB of length l = 600 mm is connected to two collars of negligible mass. Collar A is attached to a spring of constant k = 1.2 kN/m and can slide on a vertical rod, while collar B can slide freely on a horizontal rod. Knowing that the system is in equilibrium and that q = 40°, determine the period of vibration if collar B is given a small displacement and released.
SOLUTION
Vertical rod.
y l
y l
y l
x l
x l
x l
==
== -= -
= -
sin
cos
cos
cos
sin
sin
qd qdq
d qdqq
d qdq
d qd
&
& &&qy
yx
x= =2 2
Positionj (Maximum velocity, dq&m)
T J m x y
T ml
m m m
m
12 2 2
12
1
2
1
2
1
2
1
12
= + +ÈÎ ˘
= ÊËÁ
ˆ¯
( ) ( ) ( )
( )
dq d d
dq
& &
& 222 2
2
12
12 2
1
2
1
12
+ ÊËÁ
ˆ¯
+ ÊËÁ
ˆ¯
È
ÎÍÍ
˘
˚˙˙
=
ml l
T ml
msin cos ( )q q dq&
++ÈÎÍ
˘˚
=
= +
1
4
1
2
1
3
1
2
2 2 2
12
( ) ( )
( )
dq dq
d
& &m mml
V k mg yST
Positionk (Zero velocity, maximum dqm )
T
V k y mg y ym
2
22
0
1
2
=
= + + -( ) ( )d d dST
113
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.80 (Continued)
Conservation of energy. T V T V1 1 2 2+ = + :
1
2
1
3
1
20
1
21
3
2 2 2 2
2
ml k mgy y mg y y
ml
m m+ + + = + + + -( ) ( ) ( )
(
dq d d d d
d
&ST ST
&&q d d d d d dm xk mgy k y y mg y y) ( ) ( )2 2 2 22+ + = + + + -ST ST ST
(1)
But when the rod is in equilibrium,
+
SM mgx
k x mg kB = - = =2
0 2d dST ST (2)
Substituting Eq. (2) into Eq. (1),
ml k y y l
ml kl
m m m m
m m
2 2 2
2 2 2 2 2
1
31
3
( ) cos
( ) cos ( )
dq d d q dq
dq q dq
&
&
= =
=
For simple harmonic motion,
dq dq w f
dq dq w
dq w q dq
w
= +
=
=
=
m n
m m n
m n m
n
t
m k
k
m
sin( )
( ) cos ( )
&1
3
3
2 2 2 2
2 ccos( )
cos
.
2 2
2
3 1200
840
264 07
q
w
= ∞
=
N/m
kg
n
Period of vibration. t pw
pn
n
= =2 2
264 07. t n = 0 387. s �
114
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.81
A slender rod AB of length l = 600 mm and negligible mass is connected to two collars, each of mass 8 kg. Collar A is attached to a spring of constant k = 1 2. kN/m and can slide on a vertical rod, while collar B can slide freely on a horizontal rod. Knowing that the system is in equilibrium and that q = ∞40 , determine the period of vibration if collar A is given a small displacement and released.
SOLUTION
Horizontal rod.
Y l
l
l
x l
l
l
Y
g
x
x
==
=
== -
= -
sin
cos
cos
cos
sin
sin
qd qdq
d qdq
qd qdq
d qdq
&
&
Positionj (Maximum velocity, dq&m )
T m y m x
T m l l
T
m m
m
12 2
12 2 2
1
1
2
1
21
2
= +
= +
( ) ( )
[( cos ) ( sin ) ]( )
d d
q q dq
& &
&
==
=
1
20
2 2
1
ml
V
m( )dq&
Positionk (Zero velocity, maximum dq )
T
V k ym
2
22
0
1
2
=
= d
Conservation of energy. T V T V
ml k y
y l
ml kl
m m
m m
m
1 1 2 2
2 2 2
2 2
1
20
1
2
+ = +
+ =
=
=
( )
cos
( )
dq d
d qdq
dq
&
& 22 2 2cos ( )q dqm
115
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.81 (Continued)
Simple harmonic motion. dq dq w f
dq dq w
dq w q dq
w
= +
=
=
=
m n
m m n
m n m
n
t
ml kl
k
sin ( )
( ) cos ( )
&2 2 2 2 2 2
2
mm
n
cos
cos .
2
2 2 21200
840 88 02
q
w = = -N/m
kgs
Period of vibration. t pw
pn
n
= = =2 2
88 020 6697
.. s t n = 0 670. s �
116
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.82
A 3-kg slender rod AB is bolted to a 5-kg uniform disk. A spring of constant 280 N/m is attached to the disk and is unstretched in the position shown. If end B of the rod is given a small displacement and released, determine the period of vibration of the system.
SOLUTION
r
l
==
0 08
0 3
.
.
m
m
Positionj T J J
V
J m r
J m l
m A m
A AB
12 2
1
02
1
2
1
20
1
21
3
= +
=
=
=
disk rod
disk
rod
& &q q( )
( ) 22
Positionk T2 0=
V k r m gl
V kr
m AB m
mm m
22
22
22
1
2 21
1 22 2
1
2
= + -
- = ª
=
( ) ( cos )
cos sin
q q
qq q
qmmAB
l
m
m g2 2 2
2+
( )q
117
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.82 (Continued)
Conservation of energy.
T V T V m r m l kr m gAB m m AB1 1 2 2 02 2 2 2 21
2
1
2
1
30 0
1
2
1
2+ = + +Ê
ËÁˆ¯
+ = + +: &q q llm22q
For simple harmonic motion,
&q w q
w q q
w
m n m
AB n m AB m
n
m t m l kr m gl
=
+ÊËÁ
ˆ¯
= +ÊËÁ
ˆ¯
=
1
2
1
3 202 2 2 2 2 2
2 kkr m gl
m r m lAB
AB
n
2
12 0
2 13
2
22280 0 08 3 9 81
++
=+
w( )( . ) ( )( .N/m m kg m/s22 m
kg m kg m
)
( )( . ) ( )( . )
.
.
.0 32
12
2 13
2
2
5 0 08 3 0 300
6 207
0 1065
( )+
= =wn 88 55.
Period of vibration. t pw
pn
n
= =2 2
58 55. t n = 0 821. s �
118
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.83
A 400-g sphere A and a 300-g sphere C are attached to the ends of a 600-g rod AC, which can rotate in a vertical plane about an axis at B. Determine the period of small oscillations of the rod.
SOLUTION
Positionj
T m m m I
I
A C m AC m AC m12 2 2 21
20 125
1
20 2
1
20 0375
1
2= + + +( . ) ( . ) ( . )& & &q q q
AAC ACm
T
=
= + + +
1
120 325
1
20 4 0 125 0 3 0 2 0 6 0 0375
2
12 2 2
( . )
. ( . ) . ( . ) . ( . )11
120 6 0 325
1
20 024375
0
2 2
12
1
( . )( . )
( . ) ( )
ÈÎÍ
˘˚
= ◊
=
&
&
q
q
m
mT
V
N m
119
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.83 (Continued)
Positionk
T
V W W WA m C m AC m
2
2
0
0 125 1 0 2 1 0 0375 1
== - - + - + -. ( cos ) . ( cos ) . ( cos )q q q
11 2 22
0 4 9 81 0 125 0 3 9 81
22
2
- = ª
= - +
cos sin
( . )( . )( . ) ( . )( . )(
q qq
mmm
V
/
00 2 0 6 9 81 0 03752
0 318825
2
2
2
2
. ) ( . )( . )( . ) ( )
.
+[ ] ◊
=
q
q
m
mV
N m
Conservation of energy. T V T V m m1 1 2 22 21
20 024375 0 0
0 318825
2+ = + + = +: ( . )
.&q q
Simple harmonic motion.
&q w q
w
t pw
p
m n m
n
nn
=
= =
= =
2 0 318825
0 02437513 08
2 2
13 08
.
..
. t n = 1 737. s �
120
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.84
Three identical rods are connected as shown. If b l= 34 , determine the
frequency of small oscillations of the system.
SOLUTION
l = length of each rod
m = mass of each rod
Kinematics: vl
v b
m m
BE m m
=
=2&
&q
q( )
Positionj T
V mgl
mgb
V mg l b
m m
m
1
1
1
0
22
=
= - -
= - +
cos cos
( )cos
q q
q
Positionk V mgl
mgb
mg l b
T I mv m vm m BE m
2
22 2 2
22
21
2
1
2
1
2
= - -
= - +
= +ÈÎÍ
˘˚
+
=
( )
( )&q
11
12 2
1
2
1
3
1
2
2 22
2
22 2
ml ml
m b
T l b m
m m m
m
& & &
&
q q q
q
+ ÊËÁ
ˆ¯
+
= +ÊËÁ
ˆ¯
( )
22
121
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.84 (Continued)
Conservation of energy. T V T V mg l b l b m mg l b
mg
m m1 1 2 22 2 20
1
3
1
2+ = + - + = +Ê
ËÁˆ¯
- +: ( )cos ( )
(
q q&
ll b l b mm m+ - = +ÊËÁ
ˆ¯
)( cos )11
3
1
22 2 2q q&
For small oscillations, ( cos )
( )
11
21
2
1
3
1
2
2
2 2 2 2
- =
+ = +ÊËÁ
ˆ¯
q q
q q
m m
m mmg l b l b m &
But for simple harmonic motion, &q w q q w q
w
m n m m n m
n
mg l b l b m
gl b
l
= + = +ÊËÁ
ˆ¯
= +
: ( ) ( )1
2
1
3
1
2
1
2
2 2 2 2
213
22 12
2+ b
or wn gl b
l b2
2 23
2 3= +
+ (1)
For b l= 3
4, we have w
w
n
n
gl l
l l
gl
l
g
l
g
l
234
2 34
2
74
5916
2
32 3
3
1 4237
1 1932
=+
+ ( )=
=
=
.
.
f
g
l
nn=
=
wp
p
2
1 1932
2
. f
g
ln = 0 1899. �
122
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.85
An 800-g rod AB is bolted to a 1.2-kg disk. A spring of constant k = 12 N/m is attached to the center of the disk at A and to the wall at C. Knowing that the disk rolls without sliding, determine the period of small oscillations of the system.
SOLUTION
Positionj
T I ml
r I mG AB m AB m G m12
221
2
1
2 2
1
2
1
2= + -Ê
ËÁˆ¯
+ +( ) ( )& & &q q qdisk diskk
2 kg m
r
I ml
ml
r
m
G AB
AB
2 2
2 21
12
1
120 8 0 6 0 024
2
&q
( ) ( . )( . ) .= = = ◊
-ÊËÁ
ˆ¯
= - = ◊
= =
22
2
0 8 0 3 0 25 0 002
1
2
1
21
( . )( . . ) .
( ) ( .
kg m2
disk diskI m rG 22 0 25 0 0375
1 2 0 25 0 0750
1
2
2
2 2
1
)( . ) .
. ( . ) .
= ◊
= = ◊
=
kg m
kg m
2
disk2m r
T 00 024 0 002 0 0375 0 0750
1
20 1385
0
2
12
1
. . . .
[ . ]
+ + +[ ]
=
=
&
&
q
q
m
mT
V
Positionk T
V k r m gl
m AB m
mm m
2
22
22
0
1
2 21
1 22 2
=
= + -
- = ª
( ) ( cos )
cos sin
q q
qq q
(smalll angles)
N/m m kg m/s 2V m2
2 21
212 0 25 8 9 81
0 6= +( )( . ) ( )( . ).q mm
2ÊËÁ
ˆ¯
qm2
2
123
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.85 (Continued)
V
T V T V
m
m
m n
22
2
1 1 2 2
2
1
20 750 2 354
1
23 104
= +
= ◊
+ = +
=
[ . . ]
( . )
q
q
q w q
N m
2&mm
m n m
n
2
2 2
2
1
20 1385 0 0
1
23 104
0 1385
( . ) ( )
( . )
( .
q w q
w
+ = +
=◊
3.104
N m
2
kkg m
s
2◊
= -
)
.22 41 2 t pw
pn
n
= = =2 2
22 411 327
.. s �
124
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.86
Two uniform rods AB and CD, each of length l and mass m, are attached to gears as shown. Knowing that the mass of gear C is m and that the mass of gear A is 4 m, determine the period of small oscillations of the system.
SOLUTION
Kinematics: 2
2
2
r rA C
A C
A C
q qq q
q q
==
=& &
Let q qq q
q q
A m
m C m
m C m
==
=
2
2
( )
( )& &
Positionj T I I I I
ml
A m C m AB m CD m
AB m
12 2 2 21
2
1
22
1
2
1
22
1
2 2
= + + +
+ Ê
& & & &
&
q q q q
q
( ) ( )
ËËÁˆ¯
+ ÊËÁ
ˆ¯
2 21
2 22m
lCD m
&q
I m r mr
I m r mr
I ml I
A
C
AB CD
= =
= =
=
1
24 2 8
1
2
1
21
12
2 2
2 2
2
( )( )
( )( )
==
= +Ê
ËÁˆ
¯+ + + +
È
ÎÍÍ
˘
˚˙˙
=
1
12
1
28
24
12 3 4
1
2
2
12
2 2 2 22
1
ml
T m rr l l l
l
T m 1105
3
0
2 2 2
1
r l
V
m+ÈÎÍ
˘˚
=
&q
125
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.86 (Continued)
Positionk T
V mgl mgl
m m
1
1
0
21
21 4
=
= - + -( cos ) ( cos )q q
For small angles, 1 22 2
1 4 2 2 2
1
2 22
22
2 2
1
2
- = ª
- = =
= +
cos sin
cos sin
qq q
q q q
mm m
m m m
mV mgl mmm
m n m
mgl
T V T V
m r
22
1 1 2 22 2 2
2
1
2
5
2
1
210
Ê
ËÁˆ
¯=
+ = + =
+
q
q w q &55
30 0
1
2
5
22 2 2
2
l mgln mmÈ
Î͢˚
+ = +w qq
wn
gl
r l
gl
r l
2522 5
32
2 2
10
3
12 2
=+
=+
t pw
pnn
r l
gl= = +2
212 2
3
2 2
�
126
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.87
Two uniform rods AB and CD, each of length l and mass m, are attached to gears as shown. Knowing that the mass of gear C is m and that the mass of gear A is 4m, determine the period of small oscillations of the system.
SOLUTION
Kinematics: 2 2
2
r rA C A C
A C
q q q q
q q
= =
=
& &
Let q q q q
q qA m m C m
m C m
= =
=
2
2
( )
( )& &
Positionj T I I I I ml
A m C m AB m CD m AB m12 2 2 21
2
1
22
1
2
1
22
1
2 2= + + + + Ê& & & & &q q q q q( ) ( )
ËËÁˆ¯
+ ÊËÁ
ˆ¯
2 21
2 22m
lCD m
&q
I m r mr
I m r mr
I ml I
A
C
AB CD
= =
= =
=
1
24 2 8
1
2
1
21
12
2 2
2 2
2
( )( )
( )( )
==
= +Ê
ËÁˆ
¯+ + + +
È
ÎÍÍ
˘
˚˙˙
1
12
1
28
24
12 3 4
2
12
2 2 2 22 2
1
ml
T m rr l l l
l
T
m&q
== +ÈÎÍ
˘˚
=1
210
5
302 2 2
1m r l Vm&q
127
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.87 (Continued)
Positionk T
V mgl mgl
m m
2
2
0
21
21 2
=
= - - + -( cos ) ( cos )q q
For small angles, 1 22 2
1 2 2 2
2 2
22
2 2
2
2
- = ª
- = ª
= - +
cos sin
cos sin
qq q
q q q
q
mm m
m m m
mV mgl mgl
222
1
2
3
2
2
2
1 1 2 2
q
q
q w q
m
m
m n m
mgl
T V T V
=
+ = + = &
1
210
5
30 0
1
2
3
2
10
9
2 2 2 2 2
2322 5
32
m r l mgl
gl
r l
m n m
n
+ÈÎÍ
˘˚
+ = +
=+
=
q w q
w
ggl
r l60 102 2+ t pw
pnn
r l
gl= = +2
260 10
9
2 2
�
128
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.88
A 5-kg uniform rod CD is welded at C to a shaft of negligible mass which is welded to the centers of two 10-kg uniform disks A and B. Knowing that the disks roll without sliding, determine the period of small oscillations of the system.
SOLUTION
W W WA B= = disk
Positionj T I m r
I ml
r
A m m
CD m CD
12 2
2
1
22
1
22
1
2
1
2 2
= +
+ + -ÊË
( ) ( )( )disk disk& &
&
q q
q ÁÁˆ¯
= = = ◊
=
22
2 21
2
1
210 0 5 1 25
1
12
&qm
A
CD
I m r
I
( ) ( )( . ) .disk disk2kg m
mm l
T
CD2 2
1
1
125 1 5 0 9375
1
22 5 5 0 9375 0 3125
= = ◊
= + + +[ ]
( )( . ) .
. . .
kg m2
q
m
mT V
2
12
11
28 75 0= =( . ) &
Positionk T
V Wl
CD m
2
2
0
21
=
= -( cos )q
129
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.88 (Continued)
Small angles: 1 22 2
1
2 2
1
25 9 81 1 5
21
2
22
2
2
2
- = ª
=
=
=
cos sin
( )( . )( . )
q q q
q
q
mm
CDm
m
m
V W l
(( . )36 7875 2qm
Conservation of energy and simple harmonic motion.
T V T V
m n m
n m m
n
1 1 2 2
2 2 2
2
1
28 75 0
1
236 7875
36 7
+ = +
=
+ =
=
&q w q
w q q
w
( . ) ( . )
. 8875
8 754 2043
..=
Period of oscillations. t pw
pn
n
= =2 2
4 2043. t n = 3 06. s �
130
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.89
Four bars of the same mass m and equal length l are connected by pins at A, B, C, and D and can move in a horizontal plane. The bars are attached to four springs of the same constant k and are in equilibrium in the position shown ( ).q = ∞45 Determine the period of vibration if corners A and C are given small and equal displacements toward each other and released.
SOLUTION
Kinematics: BC xl
xl
xl
yl
yl
G G
G
G G
= = -
= -
= =
2 2
2
2 2
cos sin
sin
sin cos
q d q dq
d q dq
q d q
& &
ddq
d q dqyl
G =2
cos &
x l x l
x l
x l x l
x
C C
C
B B
B
= = -= -
= ==
cos sin
sin
sin cos
q d q dqd q dq
q d q dqd
& &
& ll cosq dq&y
y l y l
y l
C
B B
B m
== =
=
0
sin cos
cos
q d q dqd q dq& &
The kinetic energy is the same for all four bars.
131
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.89 (Continued)
Positionj T I m x y
I ml
T ml
m G m G m12 2 2
2
12
41
2
1
2
1
12
2
= + +ÈÎÍ
˘˚
=
=
( ) [( ) ( ) ]dq d d& & &
11
12
1
4
2
30
2 2 2
12
1
+ +ÈÎÍ
˘˚
=
=
(sin cos )q q d qm m m
T ml
V
&
Positionk T
V k x k y
V k l l
m m
m
2
22 2
22 2 2 2 2
0
21
22
1
2
=
= +
= +
=
( ) ( ) ( ) ( )
( sin cos )
d d
q q d q
kk l m2 2d q
Conservation of energy. T V T V
ml k lm m
1 1 2 2
2 2 2 22
30 0
+ = +
+ = +( ) ( )dq dq&
Simple harmonic motion. dq w dq
w
&m n m
nk
m
=
= ÊËÁ
ˆ¯
2 3
2
Period of vibration. t pwn
n
= 2 t pn
m
k= 2
2
3 �
132
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.90
The 10-kg rod AB is attached to two 4-kg disks as shown. Knowing that the disks roll without sliding, determine the frequency of small oscillations of the system.
SOLUTION
Positionk Positionj
r = 0 15. m
Masses and moments of inertia. m m
I I m r
m
A B
A B A A
AB
= =
= = =
= ◊=
4
1
2
1
24 0 15
0 045
10
2 2
2
kg
kg m
kg
( )( . )
.
Kinematics: v r
v
m A m m
AB m
= =
=
& &&
q q
q
0 15
0 05
.
.
Positionj (Maximum displacement) T
V WAB m m
m
1
1
0
0 1 10 9 81 0 1
9 81
== - = -= -
( . cos ) ( )( . )( . cos )
. cos
q qq
Positionk (Maximum speed) T m v I m v I m vA m A m B m B m AB AB22 2 2 2 21
2
1
2
1
2
1
2
1
2
21
24 0 15
= + + + +
=
& &q q
( )( . && & &
&q q q
q
m m m
m
V
) ( . ) ( )( . )
.
2 2 2
2
2
1
20 045
1
210 0 05
0 1475
+ÈÎÍ
˘˚
+
== -WWAB ( . ) ( )( . )( . ) .0 1 10 9 81 0 1 9 81= - = -
133
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.90 (Continued)
Conservation of energy. T V T V
m m
m m
1 1 2 2
2
2
9 81 0 1475 9 81
66 5085 1
+ = +
- = -
= -
. cos . .
. ( cos )
q q
q q
&&
ªª ÊËÁ
ˆ¯
==
66 50851
2
33 2542
5 7666
2
2
.
.
.
q
qq q
m
m
m m
Simple harmonic motion. &q w qm n m= wn = 5 7666. rad/s
Frequency. fnn= =
wp p2
5 7666
2
. fn = 0 918. Hz �
134
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.91
An inverted pendulum consisting of a sphere of mass m and a rigid bar ABC of length l and negligible weight is supported by a pin and bracket at C. A spring of constant k is attached to the bar at B and is undeformed when the bar is in the vertical position shown. Determine (a) the frequency of small oscillations, (b) the smallest value of a for which these oscillations will occur.
SOLUTION
(a) Positionj T m l ml
V
m m12 2 2
1
1
2
1
20
= =
=
( )& &q q
Positionk T
V k a mlm m
2
22
0
1
21
=
= - -( sin ) ( cos )q q
For small angles, sin
cos sin
[
q q
q
q q
m m
mm
m
m mV ka ml
ka ml
ª
- =
ª
= -
= -
1 22
2
1
2 2 21
2
2
2
22
2 2
2 ]]qm2
T V T V
ml ka mlm m
1 1 2 2
2 2 2 21
20 0
1
2
+ = +
+ = + -&q q[ ]
135
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.91 (Continued)
&q w q
w q
w
m n m
n m
n mg
mm
g
m
gl ka ml
ka ml
l
g
l
ka
ml
=
=
= -
= -
( )= -
È
ÎÍ
2 2 2 2
22
2
2
1˘
˚˙ f
g
l
ka
mln n= = -Ê
ËÁˆ
¯1
2
1
21
2
pw
p �
(b) fn = 0
ka
ml
2
1 0- � aml
k� �
136
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.92
For the inverted pendulum of Problem 19.91 and for given values of k, a, and l, it is observed that f = 1.5 Hz when m = 1 kg and that f = 0.8 Hz when m = 2 kg. Determine the largest value of W for which small oscillations will occur.
SOLUTION
See Solution to Problem 19.91 for the frequency in terms of W, k, a, and l.
fg
l
ka
Wln = -Ê
ËÁˆ
¯1
21
2
p
f Wg
l
ka
gln = = = -Ê
ËÁˆ
¯1 5 1
1
21
2
. ( ) ( )Hz g N 1.5p
(1)
f Wg
l
ka
gln = = = -Ê
ËÁˆ
¯0 8 2
1
2 21
2
. ( ) ( )Hz g N 0.8p
(2)
Dividing Eq. (1) by Eq. (2) and squaring, 1 5
0 8
1
1
2
2
2
2
2
2
2
2
.
.
( )
( )ÊËÁ
ˆ¯
=-( )-( ) = -
-
kagl
kagl
ka gl
ka gl
225
642 2
3 3196
1
2
3 31961
2 2
2
( ) ( )
.
.
ka gl ka gl
ka gl
fg
l
gl
Wln
- = -
=
= -ÊËp ÁÁ
ˆ¯
=
- =
f
g
W
n 0
3 31961 0
. W = 32 6. N �
137
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.93
A uniform rod of length L is supported by a ball-and-socket joint at A and by a vertical wire CD. Derive an expression for the period of oscillation of the rod if end B is given a small horizontal displacement and then released.
SOLUTION
Positionj (Maximum deflection)
Looking from above:
Horizontal displacement of C: x bC m= q
Looking from right: f qmC
mx
h
b
h= =
y h h
y hb
h
b
h
y yAG
ACy
C m m
C m m
m G
= - ª
= ÊËÁ
ˆ¯
=
= = -
( cos )11
2
1
2
1
2
2
2 22
f f
q q
CC m
m m
L
b
b
h
ybL
h
=Ê
ËÁˆ
¯
= ◊
12
22
2
1
2
1
4
q
q
We have T
V mgymgbL
hm m
1
12
0
1
4
=
= = q
Positionk (Maximum velocity)
Looking from above: T I mv
mL mL
T mL
m m
m
22 2
2 22
2
1
2
1
2
1
2
1
12
1
2 2
1
6
= +
= ÊËÁ
ˆ¯
+ ÊËÁ
ˆ¯
=
&
& &
q
q q
22 2
2 0
&qm
V =
Conservation of energy. T V T VmgbL
hmLm m1 1 2 2
2 2 21
4
1
6+ = + + =: 0 q q&
138
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.93 (Continued)
But for simple harmonic motion, &q w q
q w q
w
m n m
m n m
n
mgbL
hmL
bg
hL
=
=
=
1
4
1
63
2
2 2 2
2
( )
Period of vibration. t pwn
n
= 2 t pn
hL
bg= 2
2
3 �
139
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.94
A 2-kg uniform rod ABC is supported by a pin at B and is attached to a spring at C. It is connected at A to a 2-kg block DE, which is attached to a spring and can roll without friction. Knowing that each spring can act in tension or compression, determine the frequency of small oscillations of the system when the rod is rotated through a small angle and released.
SOLUTION
Positionj T I m C m x
I m l
R m R m E m
R R
12 2
12
2
1
2
1
2
1
21
121
122
= + +
=
=
( ) ( ) ( )
(
& & &q q
kg)(00.9 m)
kg m
kg(0.15 m) kg m
2
2
2 2
= ◊
= = ◊
=
0 135
2 0 0225
0 6
2
1
.
.
.
m C
x
l
q mm x m
x m x m
T
E m m
E m m
( ) ( )
. ( ) .
& && & &
12 2
1 12 2 2
1
2
0 6 0 72
=
= =
=
kg)(0.6q
q q11
20 135 0 0225 0 72
1
20 9
0
2
1
1
[ . . . ]
( . )
+ +
=
=
&
&
q
q
m
T
V
m
140
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.94 (Continued)
Positionk T
V k x k x m gc
x x
m m R m
m m
2
2 12
22
1 2
0
1
2
1
21
0 6 0 3
=
= + - -
= =
( ) ( ) ( cos )
( ) . .
q
q qmm
m m
m
m
V
1 22
1
250 50
2 2
22
- = ª
= +
cos sin
[( (
q q q
q N/m)(0.6 m) N/m)(02 ..3 m)
kg)(9.81 m/s
2
2
q
q
q
m
m
V
2
2
22
2 0 15
1
218 4 5 2 943
-
= + -
( )( . )
[ . . ] mm
V m221
219 55= ( . )q
Conservation of energy. T V T V m m1 1 2 22 20 9 0 0
1
219 55+ = + + = +:
1
2( . ) ( . )&q q
Simple harmonic motion. &q w q w q q
w
m n m n m m
n
= =
= = -
2 2 2 2 2
2
0 9 19 55
19 55
0 921 73
( . )( ) ( . )
.
.. s 2
Frequency of oscillations. fnn= =
wp p2
21 73
2
. fn = 0 742. Hz �
141
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.95
A 750-g uniform arm ABC is supported by a pin at B and is attached to a spring at A. It is connected at C to a 1.5-kg mass m which is attached to a spring. Knowing that each spring can act in tension or compression, determine the frequency of small oscillations of the system when the weight is given a small vertical displacement and released.
SOLUTION
Positionj k
kC
A
==
300
400
N/m
N/m
T I ml
I ml
BC m BCBC
m
BA m BAAB
12
22
2
1
2
1
2 2
1
2
1
2 2
= + ÊËÁ
ˆ¯
+ + ÊËÁ
( )& &
&
q q
q ˆ¯
+
+ ÊËÁ
ˆ¯
= + =
22 2
22 2
1
2
2
1
12
1
4
1
3
& &qm m
BC BBC
BC BC BC BC
my
I ml
m l m l mm l
I ml
m l m l m l
BC BC
BA BAAB
AB AB AB AB AB AB
2
22 2 2
2
1
12
1
4
1
3+ Ê
ËÁˆ¯
= + =
142
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.95 (Continued)
m m
m m
m l
BC ABC
BA ABC
BC BC
= =
= =
=
3
50 45
2
50 3
1
3
1
30 45 0 32
.
.
( . )( .
kg
kg
kg m)) kg m
) m kg m
2 2= ◊
= = ◊
=
0 0135
1
3
1
30 3 0 2 0 0042 2 2
.
( . ( . ) .m l
y l
BA BA
m B& CC m
m BC m m
m
my ml
T
&& & &
&
q
q q
q
2 2 2 2
2
1
1 5 0 3
0 135
1
20
= =
=
=
( . )( .
.
[ .
kg m)2
00135 0 004 0 135
0 15252
1
2
1
2
2
2
12
+ +
=
= +
. . ]
.
( ) ( )
&
&q
q
d d
m
C C A
m
V k kST ST AA2
Positionk T
V Wl W lW
k l
BC m BC BC mBA
l m
C BC m
AB
2
2
2
0
2 1
1
2
=
= + -Ê
ËÁ
ˆ
¯˜ -
+ +
q q q
q
( cos )
( (( ) ) ( ( ) )d q dST STC A AB m Ak l2 21
2+ +
when the system is in equilibrium ( ).q = 0
SM Wl Wl
k l k lB BC BCBC
C C BC A A AB= = + - -02
( ) ( )d dST ST (1)
1 22 2
2
22
2
- - =
= + ÊËÁ
ˆ¯
È
ÎÍÍ
˘
˚˙˙
-
cos sinqq q
q
mm m
BC BCBC
m BAV Wl Wl
Wll
k l
k l
AB mC BC m
C C BC m
2 2
1
2
1
2
22 2Ê
ËÁˆ¯
Ê
ËÁˆ
¯
È
ÎÍÍ
˘
˚˙˙
+
- +
d q( )ST kk
k l k l k
C C
A AB m A A AB m A A
( )
( ) ( )
d
q d q d
ST
ST ST
2
2 2 21
2
1
2+ - +
143
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.95 (Continued)
Taking Equation (1) into account,
V Wl
k l
k k l
BAAB m
C BC m
C C A AB
2
22 2
2 2
2 2
1
2
1
2
1
2
= - ÊËÁ
ˆ¯
+È
ÎÍÍ
+ +
d( )ST qq dm A Ak
V
2 2
22
1
2
1
20 3 9 81
0 2
2300 0 3
+ ˘˚
= - ÊËÁ
ˆ¯
+
( )
( . )( . ).
( )( . )
ST
++ÈÎÍ
˘˚
+ +
= - +
400 0 2
1
2
1
21
20 2943
2 2
2 2
2
( . )
( ) ( )
[ .
q
d d
m
C C A Ak k
V
ST ST
227 161
2
1
21
242 7057
1
2
2 2 2
22
+ + +
= +
] ( ) ( )
[ . ] (
q d d
q d
m C C A A
m C
k k
V k
ST ST
SST ST) ( )C A Ak2 21
2+ d
Conservation of energy.
T V T V k km C C A A1 1 2 22 2 21
20 1525
1
2
1
2
01
242
+ = + + +
= +
: ST ST( . ) ( ) ( )
(
&q d d
.. ) ( ) ( )70571
2
1
22 2 2q d dm C C A Ak k+ +ST ST
Simple harmonic motion. &q w q
w q q
w
m n m
n m m
n
=
=
= =
0 1525 42 7057
42 7057
0 1525280 037
2 2 2
2
. .
.
.. (raad/s)2
Frequency. fnn= =
wp p2
280 037
2
. fn = 2 66. Hz �
144
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.96*
Two uniform rods AB and BC, each of mass m and length l, are pinned together at A and are pin-connected to small rollers at B and C. A spring of constant k is attached to the pins at B and C, and the system is observed to be in equilibrium when each rod forms an angle b with the vertical. Determine the period of small oscillations when Point A is given a small downward deflection and released.
SOLUTION
x l
x l
xl
xl
yl
yl
C
C
G
G
G
G
==
=
= -
=
=
sin
cos
cos
sin
sin
bd bdb
b
d bdb
b
d
2
2
2
2
& &
ccos
cos
bdb
d bdb& &yl
G =2
Positionj T I m x y
mlml
m m C m G12 2 2
22
2
21
2
1
2
1
12 4
= + +ÈÎÍ
˘˚
= +
( ) ( ) ( )
(sin
db d d& & &
bb b db
db
+È
ÎÍ
˘
˚˙
=
=
cos )2 2
2
1
1
30
m
mml
V
&
Positionk T
V k x
l
l
m
m n m
2
22
2 2
2 2
0
1
22
1
24
2
=
=
=
=
=
( )
( cos )
cos
d
b
b
db w db&
145
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.96* (Continued)
T V T V1 1 2 2+ = +
Conservation of energy. 1
30 0 2
6
2 2 2 2 2 2
2 2
ml kl
k
m
n n n
n
w db bdb
w b
+ = +
=
cos
cos
Period of oscillations. t pw
p
bn
n km
= =( )
2 2
6 cos2 t p
bnm
k= 2
6cos �
146
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.97*
As a submerged body moves through a fluid, the particles of the fluid flow around the body and thus acquire kinetic energy. In the case of a sphere moving in an ideal fluid, the total kinetic energy acquired by the fluid is 1
42rVv , where r is the mass
density of the fluid, V is the volume of the sphere, and v is the velocity of the sphere. Consider a 500-g hollow spherical shell of radius 80 mm, which is held submerged in a tank of water by a spring of constant 500 N/m. (a) Neglecting fluid friction, determine the period of vibration of the shell when it is displaced vertically and then released. (b) Solve Part a, assuming that the tank is accelerated upward at the constant rate of 8 m/s2.
SOLUTION
This is not a damped vibration. However, the kinetic energy of the fluid must be included.
(a) Positionk T
V kxm
2
22
0
1
2
=
=
Positionj T T T m v Vv
V
s m m12 2
1
1
2
1
40
= + = +
=
spere fluid r
Conservation of energy and simple harmonic motion.
T V T V m v Vv kxs m m m1 1 2 22 2 21
40 0
1
2+ = + + + = +:
1
2r
v l x
m V x kx
k
m V
m m m n
s m n m
n
s
n
= =
+ÊËÁ
ˆ¯
=
=+
=
w
r w
wr
w
1
2
1
2
1
2
12
500
2 2 2
2
2 NN/m
kg)
kgm
( .
( . )
0 512
1
2
1
2
1000 4
30 08
33
+ ÊËÁ
ˆ¯
= ÊËÁ
ˆ¯
ÊË
r
r p
V
Vm
ÁÁˆ¯
=
= -
1 0723
3182
. kg
s 2wn
Period of vibration. t pw
pn
n
= =2 2
318
(b) Acceleration does not change mass. t n = 0 352. s �
147
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.98*
A thin plate of length l rests on a half cylinder of radius r. Derive an expression for the period of small oscillations of the plate.
SOLUTION
( sin )sin
( cos )
r r
r r
m m m
mm
q q q
ª
- ª
2
2
12
Positionj (Maximum deflection) T
V Wy
mgr
m
m
1
1
2
0
2
==
=q
Positionk ( ):q = 0 T I
ml
T ml
m
m
m n m
n
22
2 2
22 2
1
21
2
1
12
1
2
1
12
=
= ÊËÁ
ˆ¯
=
= ÊËÁ
ˆ¯
&
&
&
q
q
q w q
w qmm
T V T V
2
1 1 2 2+ = +
01
2
1
2
1
12
12
22
12
2 2 2 2
22
2
+ = ÊËÁ
ˆ¯
=
= =
mgr ml
gr
l
l
gr
m n m
n
nn
q w q
w
t pw
p t pn
l
gr=
3 �
148
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.99
A 50-kg block is attached to a spring of constant k = 20 kN/m and can move without friction in a vertical slot as shown. It is acted upon by a periodic force of magnitude P P tm f= sin ,w where w f = 18 rad/s. Knowing that the amplitude of the motion is 3 mm, determine the value of Pm .
SOLUTION
Equation of motion. mx kx P tm f&&+ = sinw
The steady state response is xm
Pkm
f
n
=( )
- ( )1ww
2
or P kxm mf
n
= -ÊËÁ
ˆ¯
È
ÎÍÍ
˘
˚˙˙
12w
w
Data: k
m
= ¥=
20 10
50
3 N/m
kg
w
ww
n
f
n
m
k
m
x
= = ¥ =
= =
= = ¥
20 10
5020
18
200 9
3
3
rad/s
rad/s
rad/s
mm 3
.
110 m3-
Pm = ¥ ¥ --( )( )[ ( . ) ]20 10 3 10 1 0 93 3 2 Pm = 11 40. N �
149
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.100
A 5 kg collar can slide on a frictionless horizontal rod and is attached to a spring of constant 500 N/m. It is acted upon by a periodic force of magnitude P P tm f= sin ,w where Pm = 15 N. Determine the amplitude of the motion of the collar if (a) w f = 5 rad/s, (b) w f = 10 rad/s.
SOLUTION
Eq. (19.33): xm
Pkm
f
n
=( )
- ( )12w
w
P k
m
k
mP
k
m
n
m
= ==
= = =
= =
15 500
5
500
510
50 0
N, N/m
kg
rad/s
N
500 N/m
w
. 33 m
Amplitude of vibration.
(a) w f = 5 rad/s xm =-
=0 03
1 100 04
2
.
( ). m | | 0.04 m(in phase)xm = �
(b) w f = 10 rad/s w wf n mx= fi xm �
150
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.101
A 5-kg collar can slide on a frictionless horizontal rod and is attached to a spring of constant k. It is acted upon by a periodic force of magnitude P P tm f= sin ,w where Pm = 10 N and w f = 5 rad/s. Determine the value of the spring constant k knowing that the motion of the collar has an amplitude of 150 mm and is (a) in phase with the applied force, (b) out of phase with the applied force.
SOLUTION
Eq. (19.33): xm
Pkm
f
n
=- ( )1
2ww
wnk
m2 =
xP
k m
kP
xm
mm
f
m
mf
=-
= +
w
w
2
2
Data: Pm = 10 N, m = 5 kg
w f
m
m
m
m
kP
x
P
x
=
= +
= +
5
5 5
125
2
rad/s
( )( )
(a) (In phase) xm = =150 0 15mm m.
k = +10
0 15125
. k = 191 7. N/m �
(b) (Out of phase) xm = - = -150 0 15mm m.
k =-
+10
0 15125
. k = 58 3. N/m �
151
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.102
A collar of mass m which slides on a frictionless horizontal rod is attached to a spring of constant k and is acted upon by a periodic force of magnitude P P tm f= sin .w Determine the range of values of w f for which the amplitude of the vibration exceeds two times the static deflection caused by a constant force of magnitude Pm .
SOLUTION
Circular natural frequency. wnk
m=
For forced vibration, the equation of motion is
mx kx p tm f&&+ = +sin ( )w jThe amplitude of vibration is
xm
pkm
f
n
f
n
=- ( )
=-( )1 1
2 2ww
ww
dST
For w wf n� and xm = 2 dST , we have
21
11
22
2
dd w
www
STST or=
- ( )-
ÊËÁ
ˆ¯
=f
n
f
n
w wf nk
m2 21
2
1
2= = w f
k
m=
2 (1)
For k
m f n2� �w w , | |xm exceeds ST2d
For w wf n� and xm = 2dST , we have
21
11
22
2
2d
dw w
w
wSTST or=
- -- =
( )f n
f
n
w wf nk
m2 23
2
3
2= = w f
k
m= 3
2 (2)
For w wn fk
m� �
3
2 | | exceeds STxm 2d
From Eqs. (1) and (2), Range: k
m
k
mf2
3
2� �w �
152
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.103
An 8-kg uniform disk of radius 200 mm is welded to a vertical shaft with a fixed end at B. The disk rotates through an angle of 3∞ when a static couple of magnitude 50 N m◊ is applied to it. If the disk is acted upon by a periodic torsional couple of magnitude T T tm f= sin ,w where Tm = ◊60 N m, determine the range of values of w f for which the amplitude of the vibration is less than the angle of rotation caused by a static couple of magnitude Tm .
SOLUTION
Mass moment of inertia: I mr= = = ◊1
2
1
28 0 200 0 162 2 2( )( . ) . kg m
Torsional spring constant: KT
T
K
=
= ◊= ∞ =
=
= ◊
q
q50
3 0 05236
0 50
0 05236954 93
N m
rad
N m/rad
.
.
..
Natural circular frequency: wnK
I= = =954 93
0 1677 254
.
.. rad/s
For forced vibration, qq
ww
ww
m
TKm
f
n
f
n
=- ( )
=- ( )1 1
2 2ST
For the amplitude | |qm to be less than qST, we must have w wf n� .
Then | | STSTq
ww
mf
n
=
( ) -2
1�
ww
f
n
ÊËÁ
ˆ¯
-2
1 1�
ww
w wf
nf n
ÊËÁ
ˆ¯
=2
2 2 2 77 254� � ( )( . )
w f �109 3. rad/s �
153
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.104
For the disk of Problem 19.103 determine the range of values of w f for which the amplitude of the vibration will be less than 3 5. .∞
SOLUTION
Mass moment of inertia: I mr= = = ◊1
2
1
28 0 200 0 162 2 2( )( . ) . kg m
Torsional spring constant: KT
T
K
=
= ◊ = ∞ =
= = ◊
qq50 3 0 05236
50
0 05236954 93
N m rad
N m/rad
, .
..
Natural circular frequency: wnK
I= = =954 93
0 1677 254
.
.. rad/s
For forced vibration, qq
q
ww
ww
m
TK
m
m
m
f
n
f
n
T
T
K
=- ( )
=- ( )
= ◊
= = =
1 1
60
60
954 930 062
2 2ST
ST
N m
.. 8832 rad
| | rad STq qm � �3 5 0 061087. .∞ =
For the amplitude | |qm to be less than qST , we must have w wf n� .
| | STqq
ww
ww
mf
n
f
n
=
( ) -=
( ) -2 2
1
0 062832
10 061087
..�
ww
f
n
ÊËÁ
ˆ¯
- =2
10 062832
0 0610871 02857�
.
..
ww
w wf
nf n
ÊËÁ
ˆ¯
2
2 02857 2 02857� �. . w f �110 0. rad/s �
154
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.105
An 8-kg block A slides in a vertical frictionless slot and is connected to a moving support B by means of a spring AB of constant k = 1.6 kN/m. Knowing that the displacement of the support is d d w= m f tsin , where dm = 150 mm, determine the range of values of w f for which the amplitude of the fluctuating force exerted by the spring on the block is less than 120 N.
SOLUTION
Natural circular frequency: wnk
m= = ¥ =1 6 10
814 1421
3.. rad/s
Eq. (19.33' ): xmm
f
n
=- ( )
dww1
2
Spring force: F K x km m m mf
n
= - - = - -- ( )
È
Î
ÍÍÍÍ
˘
˚
˙˙˙˙
( )d dww
11
12
=( )- ( )
k m
f
n
f
n
d
ww
ww
2
21
= ¥( )- ( )
=( )- ( )
( . )( . )1 6 10 0 1501
2401
3
2
2
2
2
ww
ww
ww
ww
f
n
f
n
f
n
f
n
Limit on spring force: | | NFm �120
2401
1201
1
2
2
2
2
2
ww
ww
ww
ww
f
n
f
n
f
n
f
n
( )- ( )
( )- ( )
� �or
155
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.105 (Continued)
In phase motion.
ww
ww
f
n
f
n
( )- ( )
2
21
1
2�
ww
ww
ww
ww
w w
f
n
f
n
f
n
f
n
f n
ÊËÁ
ˆ¯
-ÊËÁ
ˆ¯
ÊËÁ
ˆ¯
2 2
2
1
2
1
2
3
2
1
2
1
3
1
3
�
� �
� w f � 8 16. rad/s �
Out of phase motion.
ww
ww
ww
ww
f
n
f
n
f
n
f
n
( )( ) -
ÊËÁ
ˆ¯
ÊËÁ
ˆ¯
-
2
2
2 2
1
1
2
1
2
1
2
�
�
ww
f
n
ÊËÁ
ˆ¯
-2
1
2� No solution for w f .
156
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.106
Rod AB is rigidly attached to the frame of a motor running at a constant speed. When a collar of mass m is placed on the spring, it is observed to vibrate with an amplitude of 15 mm. When two collars, each of mass m, are placed on the spring, the amplitude is observed to be 18 mm. What amplitude of vibration should be expected when three collars, each of mass m, are placed on the spring? (Obtain two answers.)
SOLUTION
(a) One collar: ( ) ( )xk
mm n1 1215= =mm w
(b) Two collars: ( ) ( ) ( )xk
mm n n2 22
1218
2
1
2= = =mm w w
ww
wwn n
ÊËÁ
ˆ¯
=ÊËÁ
ˆ¯2 1
2
(c) Three collars:
( ) ( ) ( ) ,xk
mm n nn n
3 32
12
3 13
1
33= = =
ÊËÁ
ˆ¯
=ÊËÁ
ˆ¯
unknown, w w ww
ww
We also note that the amplitude dm of the displacement of the base remains constant.
Referring to Section 19.7, Figure 19.9, we note that, since ( ) ( ) ,( ) ( )x xm mn n
2 12 1
� � and ww
ww we must
have ww( )n 1
1� and ( ) .xm 1 0� However, ww( )n 2
may be either � �1 or 1, with ( )xm 2 being correspondingly
either � �0 0or .
1. Assuming ( ) :xm 2 0�
For one collar,
( )xmm
n
m
n
1
1
2
1
2
1
15
1
=
-ÊËÁ
ˆ¯
+ =
-ÊËÁ
ˆ¯
d
ww
d
ww
j mm (1)
For two collars,
( )xmm
n
m
n
2
2
2
1
2
1
18
1 2
=
-ÊËÁ
ˆ¯
+ =
-ÊËÁ
ˆ¯
d
ww
d
ww
j mm (2)
157
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.106 (Continued)
Dividing Eq. (2) by Eq. (1), member by member:
1 2
1
1 2
1
71
2
1
21
2
. ;=-
ÊËÁ
ˆ¯
-ÊËÁ
ˆ¯
ÊËÁ
ˆ¯
=
ww
ww
ww
n
n
n
we find
Substituting into Eq. (1), dm = -ÊËÁ
ˆ¯
=( )15 11
7
90
7mm mm
For three collars,
( )xmm
n
3
1
2
1 3
907
1 317
90
4=
-ÊËÁ
ˆ¯
=
ÊËÁ
ˆ¯
- ÊËÁ
ˆ¯
=d
ww
mmmm, ( ) .xm 3 22 5= mm �
2. Assuming ( ) :xm 2 0�
For two collars, we have - =
-ÊËÁ
ˆ¯
18
1 21
2mm
d
ww
m
n
(3)
Dividing Eq. (3) by Eq. (1), member by member:
- =-
ÊËÁ
ˆ¯
-ÊËÁ
ˆ¯
- +ÊËÁ
ˆ¯
= -Ê
1 2
1
1 2
1 2 2 4 1
1
2
1
2
1
2
.
. .
ww
ww
ww
ww
n
n
n nËËÁˆ¯
ÊËÁ
ˆ¯
= =
1
2
1
22 2
3 4
1 1
1 7
wwn
.
.
.
.
Substitute into Eq. (1), dm = -ÊËÁ
ˆ¯
=( ).
. .15 1
1 1
1 7
9
1 7mm mm
158
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PROBLEM 19.106 (Continued)
For three collars,
( ).
.
..
xmm
n
3
1
2
1 3
91 7
1 31 11 7
9
1 6=
-ÊËÁ
ˆ¯
=
ÊËÁ
ˆ¯
- ÊËÁ
ˆ¯
=-
d
ww
mm, ( ) .xm 3 5 63= - mm �
( )out of phase
Points corresponding to the two solutions are indicated below:
159
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.107
A cantilever beam AB supports a block which causes a static deflection of 50 mm at B. Assuming that the support at A undergoes a vertical periodic displacement d d w= m f tsin , where dm = 12 mm, determine the range of values of w f for which the amplitude of the motion of the block will be less than 25 mm. Neglect the weight of the beam and assume that the block does not leave the beam.
SOLUTION
For the static condition. mg k= dST
Natural circular frequency. wd
d
w
n
n
k
m
g
g
= =
= = =
= =
ST
2STm/s mm m
rad/s
9 81 50 0 05
9 81
0 0514 007
. , .
.
..
From Eqs. (19.31 and 19.33' ): ( )xm Bm
f
n
=- ( )
dww1
2
Conditions: | | m mxm B m� 0 025 0 012. .d =
In phase motion. 0 012
10 025
10 012
0 025
2
2
..
.
.
- ( )-
ÊËÁ
ˆ¯
ww
ww
f
n
f
n
�
�
0 52 0 522
. .� �ww
w wf
nf n
ÊËÁ
ˆ¯
w f �10 10. rad/s �
Out of phase motion. 0 012
10 025
2
..
ww
f
n( ) -�
ww
ww
f
n
f
n( ) - Ê
ËÁˆ¯
ÊËÁ
ˆ¯
-2 2
10 012
0 0251 0 48� �
.
..
ww
w wf
nf n
ÊËÁ
ˆ¯
2
1 48 1 48� �. . w f �17 04. rad/s �
160
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.108
A variable-speed motor is rigidly attached to a beam BC. When the speed of the motor is less than 600 rpm or more than 1200 rpm, a small object placed at A is observed to remain in contact with the beam. For speeds between 600 and 1200 rpm the object is observed to “dance” and actually to lose contact with the beam. Determine the speed at which resonance will occur.
SOLUTION
Let m be the unbalanced mass and r the eccentricity of the unbalanced mass. The vertical force exerted on the beam due to the rotating unbalanced mass is
P mr t P tf f m f= =w w w2 sin sin
Then from Eq. 19.33, xm
Pk
mr
km
f
n
f
f
n
=- ( )
=- ( )1 1
2 2
2
ww
w
ww
For simple harmonic motion, the acceleration is
a xm f m
mr
kf
f
n
= - =- ( )
ww
ww
22
4
1
When the object loses contact with the beam is | |.am Acceleration is greater than g. Let w1 600 62 832= =rpm rad/s..
| |aU
m
mrk
mr Uk
n
n
1 2 2
14
1
4 4
1 1=
- ( )=
-
w
ww
w
(1)
where Un
=ww
1 .
Let w w2 11200 125 664 2= = =rpm rad/s.
| |aU
m
mrk
mr Uk
n
n
2 2
2
2
24
2
4 4
1 4 1=
( ) -=
-
w
ww
w ( )
(2)
161
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PROBLEM 19.108 (Continued)
Dividing Eq. (1) by Eq. (2),
14 1
16 116 16 4 1
20 1717
20
17
20
20
1
2
22 2
2
1
= --
- = -
= =
= =
U
UU U
U U
nn
( )or
ww
w77
1 084651 1w w= .
wn = ( . )( )1 08465 600 rpm wn = 651 rpm �
162
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PROBLEM 19.109
An 8-kg block A slides in a vertical frictionless slot and is connected to a moving support B by means of a spring AB of constant k = 120 N/m. Knowing that the acceleration of the support is a a tm f= sin ,w where am = 1 5. m/s2
and w f = 5 rad/s, determine (a) the maximum displacement of block A, (b) the amplitude of the fluctuating force exerted by the spring on the block.
SOLUTION
(a) Support motion.
a a t
at
a
m f
m
ff
mm
f
= =
= -Ê
ËÁ
ˆ
¯˜
=-
= -
&&d w
dw
w
dw
sin
sin
.
(
2
2
21 5
5
m/s
rad/ssm
).
20 06= -
From Equation (19.31 and 19 33. ):¢
xk
mmm
nf
n
=-
= = =d
www
1
120152
2
2 2N/m
8 kgrad/s( )
xm = --
=0 06
10 09
2515
.
( ). m xm = 90 0. mm �
(b) x is out of phase with d for w f = 5 rad/s.
Thus,
F k xm m m= + = +( ) ( . . )d 120 0 09 0 06N/m m m
= 18 N Fm = 18 00. N �
163
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PROBLEM 19.110
A 20 g ball is connected to a paddle by means of an elastic cord AB of constant k = 7 5. N/m. Knowing that the paddle is moved vertically according to the relation d d w= m f tsin , where dm = 200 mm, determine the maximum allowable circular frequency w f if the cord is not to become slack.
SOLUTION
SF ma k x mx xk
mx= - = + Ê
ËÁˆ¯
= ( )d d&& &&
From Equation (19.31 and 19 33. ):¢ xmm
f
n
=-Ê
ˈ¯
dww
12
2
Data: m = =20 0 02kg kg.
k = 7 5. N/m dm = =200 0 2mm m.
wnk
m=
=
=
7 5
0 0219 3649
.
.. rad/s
The cord becomes slack if xm m- d exceeds dST, where
dST m= = ¥ =m
k
0 02 9 81
7 50 02616
. .
..
164
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PROBLEM 19.110 (Continued)
Then 0 2
10 2 0 02616
2
.. .
- ( )-
ww
f
n
�
0 2 0 2 0 2 0 02616 0 02616
0 22616
2 2
. . . . .
.
- +ÊËÁ
ˆ¯
-ÊËÁ
ˆ¯
ww
ww
w
f
n
f
n
f
�
ww
ww
n
f
n
ÊËÁ
ˆ¯
=
2
0 02616
0 02616
0 226160 3401
�
�
.
.
..
Maximum allowable circular frequency.
w f � ( . )( . )0 3401 19 3649 rad/s w f � 6 59. rad/s �
165
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PROBLEM 19.111
A simple pendulum of length l is suspended from a collar C, which is forced to move horizontally according to the relation x tC m f= d wsin . Determine the range of values of w f for which the amplitude of the motion of the bob is less than dm . (Assume that dm is small compared with the length l of the pendulum.)
SOLUTION
Geometry. x x l
x x
l
C
C
= +
=-
sin
sin
q
q
+ SF ma T mg T mgy y= ª - = ª0 0: cosq
+ SF max x= : - =T mxsinq &&
mxmg x x
l
xg
lx
g
lx
C
C
&&
&&
+-
=
+ =
( )0
Using the given motion of xC,
&&x g
lx
g
ltm f+ = d wsin
Circular natural frequency. wng
l=
&&x x tn n m f+ =w w d w2 2 sin
The steady state response is xmm
f
n
=- ( )
dww1
2
xmm
mf
n
22
2 22
1
=
- ( )ÈÎÍ
˘˚
dd
ww
�
Consider xm m2 2= d .
166
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PROBLEM 19.111 (Continued)
Then 1 1 2 12 2 2 4
-ÊËÁ
ˆ¯
È
ÎÍÍ
˘
˚˙˙
= -ÊËÁ
ˆ¯
+ÊËÁ
ˆ¯
=
ÊËÁ
ww
ww
ww
ww
f
n
f
n
f
n
f
n
ˆ¯
=ÊËÁ
ˆ¯
=
ÊËÁ
ˆ¯
= =
2 2
0 2
0 2
and
and
ww
ww
ww
f
n
f
n
f
n
For 0 2� � �ww
df
nm mx, | |
For ww
df
nm nx� �2, | |
Then w wf ng
l� 2
2= w fg
l�
2 �
167
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PROBLEM 19.112
The 1.2-kg bob of a simple pendulum of length l = 600 mm is suspended from a 1.4-kg collar C. The collar is forced to move according to the relation xC = d wm f tsin , with an amplitude dm = 10 mm and a frequency ff = 0.5 Hz. Determine (a) the amplitude of the motion of the bob, (b) the force that must be applied to collar C to maintain the motion.
SOLUTION
(a)
S
S
F ma
T mx
F T mg
x x
y
=- =
= - =sin
cos
&&0
For small angles cos .q ª 1 Acceleration in the y direction is second order and is neglected.
T mg
mx mg
x x
l
mxmg
lx
g
lx
mg
lm t
g
l
c
c f
n
== -
=-
+ = =
=
&&
&&
sin
sin
sin
q
q
d w
w2
&&&x x m tn n f+ =w w d w2 2 sin
From Equation ( . ):19 33¢
xm
mf
n
=-
dww
12
2
So w p p p
w
f n
n
f
g
l
2 2 2 2 2 2
2 2
2 4 0 5
9 81
0 60016 35
= = =
= = =
-
-
( ) ( . )
. /
..
s
m s
ms
2
xm =¥-
=-10 10
10 02523
3
16 35
2
mm
p.
. xm = 25 2. mm �
168
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PROBLEM 19.112 (Continued)
(b)
a x m tc c f f= = -&& d w w2 sin
+ SF m ax c c=
F T m ac c- =sinq
From Part (a): T mgx x
lc= =
-, sinq
Thus, F mgx x
lm x
m x m x m x
m x t
cc c
n n c c c
n m f
= --È
Î͢˚
+
= - + +
= -
&&
&&w w
w w
2 2
2 sin ++ -
= - +
m t m tn m f c f m fw d w w d w2 2
1 2 16 35 0 02523 1 2
sin sin
[ ( . )( . )( . ) ( . )(( . )( ) ( . ) ( )]sin
. sin
16 35 10 10 1 4 10 10
0 437
3 2 3¥ - ¥= -
- -p pp
t
t
F t= -0 437. sin ( )p N �
169
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.113
A motor of mass M is supported by springs with an equivalent spring constant k. The unbalance of its rotor is equivalent to a mass m located at a distance r from the axis of rotation. Show that when the angular velocity of the motor is w f , the amplitude xm of the motion of the motor is
xr
m
mM
f
n
f
n
=( )( )- ( )
ww
ww
2
21
where wnk
M= .
SOLUTION
Rotor Motor
+SF ma P t kx Mxm f= - =sinw &&Mx kx P t
xk
mx
P
Mt
k
M
m f
mf
n
&&
&&
+ =
+ =
=
sin
sin
w
w
w2
From Equation (19.33):
xm
Pkm
f
n
=- ( )1
2ww
But P
k
mr
kk M
P
kr
m
M
m fn
m f
n
= =
= ÊËÁ
ˆ¯
ÊËÁ
ˆ¯
ww
ww
22
2
Thus, xr
m
mM
f
n
f
n
=( )( )- ( )
ww
ww
2
21
Q.E.D. �
170
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.114
As the rotational speed of a spring-supported 100-kg motor is increased, the amplitude of the vibration due to the unbalance of its 15-kg rotor first increases and then decreases. It is observed that as very high speeds are reached, the amplitude of the vibration approaches 3.3 mm. Determine the distance between the mass center of the rotor and its axis of rotation. (Hint: Use the formula derived in Problem 19.113.)
SOLUTION
Use the equation derived in Problem 19.113 (above).
xr r
m
mM
mM
f
n
f
nf
n
=( )( )- ( )
=( )
-( )
ww
ww
ww
2
2 11
12
For very high speeds, 1
2ww
f
n( ) ® 0 and xm ®
rm
M,
thus, 3 315
100. mm = Ê
ËÁˆ¯
r r = 22 mm �
171
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.115
A motor of mass 200 kg is supported by springs having a total constant of 240 kN/m. The unbalance of the rotor is equivalent to a 30-g mass located 200 mm from the axis of rotation. Determine the range of allowable values of the motor speed if the amplitude of the vibration is not to exceed 1.5 mm.
SOLUTION
Let M m r= = =mass of motor, unbalance mass, eccentricity
M
m
r K
== =
= = = = ¥
200
30 0 03
200 0 2 240 240 103
kg
g kg
mm m kN/m N/m
.
.
Natural circular frequency: wnk
Mrm
M
= =¥
=
=
= ¥ -
240 10
20034 641
0 2 0 03
200
0 3 10
3
5
N/mrad/s.
( . )( . )
. mm mm= 0 03.
From the derivation given in Problem 19.113,
xm
rmM
f
n
f
n
f
n
f
n
=( )( )
- ( )=
( )- ( )
ww
ww
ww
ww
2
2
2
21
0 03
1
.mm
In phase motion with mm| | .xm �1 5
0 03
11 5
0 03 1 5 1 5
2
2
2 2
..
. . .
ww
ww
ww
ww
f
n
f
n
f
n
f
n
( )- ( )ÊËÁ
ˆ¯
-ÊËÁ
ˆ¯
�
�
11 53 1 5
1 5
1 530 99015
2
. .
.
..
ww
ww
f
n
f
n
ÊËÁ
ˆ¯
=
�
�
w f � ( . )( . ) .0 99015 34 641 34 30= rad/s w f � 328 rpm �
172
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.115 (Continued)
Out of phase motion with mm| | .xm = 1 5
0 03
11 5
0 03 1 5 1 5
2
2
2 2
..
. . .
ww
ww
ww
ww
f
n
f
n
f
n
f
n
( )( ) -
ÊËÁ
ˆ¯
ÊËÁ
ˆ¯
-
�
�
11 5 1 472
. .�ww
f
n
ÊËÁ
ˆ¯
ww
f
n
�1 5
1 471 01015
.
..=
w f � ( . )( . ) .1 01015 34 641 34 992= rad/s w f � 334 rpm �
173
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.116
As the rotational speed of a spring-supported motor is slowly increased from 300 to 500 rpm, the amplitude of the vibration due to the unbalance of its rotor is observed to increase continuously from 1.5 to 6 mm. Determine the speed at which resonance will occur.
SOLUTION
Let ¢m be the mass of rotor of the motor and m the total mass of the motor. Let e be the distance from the rotor axis to the mass center of the rotor. The magnitude of centrifugal force due to unbalance of the rotor is
P m em f= ¢ w2
and in rotation, the unbalanced force is
P t m e tm f f fsin sinw w w= ¢ 2
The equation of motion of the spring mounted motor is
mx kx m e t P tf f m f&&+ = ¢ =w w w2 sin sin
The steady state response is x
xm e
k
C
m
Pk
mf f
m
f
n
f
n
f
n
=- ( )
=¢
◊- ( )
=- ( )
1
1
1 1
2
2
2
2
2
ww
ww
ww
w w
Case 1. w w p pf
mx x
= = = =
= =
1
1
3002 300
6010
1 5
rpm rad/s
| | mm
( )
.
Case 2. w w p pf
mx x
1 2
2
5002 500
60
50
36
= = = =
= =
rpm rad/s
| | mm
( )
x
x
n
z
n
2
1
22
2
12
2
12 2
6
1 54
1
1
4 1
1
2
= = =+ ( )È
Î͢˚
+ ( )ÈÎÍ
˘˚
-
.
w
w
www
ww
ww
ÊÊËÁ
ˆ¯
È
ÎÍÍ
˘
˚˙˙
= -ÊËÁ
ˆ¯
È
ÎÍÍ
˘
˚˙˙
2
22 1
2
1 1wwwn
174
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.116 (Continued)
Multiplying by wn2 and transposing terms,
( )
( )( . )(
4 3 0
3
4
3 986 96 274
12
22 2
12
22
2 12
22
12
22
w w w w w
ww w
w w
- - =
=-
=
n
n11 56
3947 84 2741 566729 3
. )
. .. (
-= rad/s)2
wn = 82 032. rad/s wn = 783 rpm �
175
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.117
A100 kg motor is bolted to a light horizontal beam. The unbalance of its rotor is equivalent to a mass of 60 g located 100 mm from the axis of rotation. Knowing that resonance occurs at a motor speed of 400 rpm, determine the amplitude of the steady-state vibration at (a) 800 rpm, (b) 200 rpm, (c) 425 rpm.
SOLUTION
From Problem 19.113:
xr
m
mM
f
n
n
f
=( )( )- ( )
ww
ww
2
21
Resonance at 400 rpm means that wn = 400 rpm.
rm
MÊËÁ
ˆ¯
= ¥Ê
ËÁˆ
¯=
-( ) .100
60 10
1000 06
3
mm mm
(a) ww
f
n
ÊËÁ
ˆ¯
= ÊËÁ
ˆ¯
=2 2800
4004
xm =-
0 06 4
1 4
. ( ) xm = - 0 080. mm �
(b) ww
f
n
ÊËÁ
ˆ¯
= ÊËÁ
ˆ¯
=2 2200
400
1
4
xm =( )
-0 06
1
14
14
. xm = 0 020. mm �
(c) ww
f
n
ÊËÁ
ˆ¯
= ÊËÁ
ˆ¯
=2 2425
4001 1289.
xm =-
0 06 1 1289
1 1 1289
. ( . )
. xm = - 0 525. mm �
176
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.118
A 180-kg motor is bolted to a light horizontal beam. The unbalance of its rotor is equivalent to a 28-g mass located 150 mm from the axis of rotation, and the static deflection of the beam due to the weight of the motor is 12 mm. The amplitude of the vibration due to the unbalance can be decreased by adding a plate to the base of the motor. If the amplitude of vibration is to be less than 60 mm for motor speeds above 300 rpm, determine the required mass of the plate.
SOLUTION
Before the plate is added, M m
r1
3180 28 10
150 0 150
= = ¥= =
-kg, kg
mm m.
Equivalent spring constant: kW M g
k
= =
=¥
= ¥-
1 1
33180 9 81
12 10147 15 10
d dST ST
N/m( )( . )
.
Let M2 be the mass of motor plus the plate.
Natural circular frequency. wnk
M=
2
Forcing frequency: w
ww
w
f
f
n
f M
k
M
= =
ÊËÁ
ˆ¯
= =¥
300 31 416
31 416
147 15
22
22
rpm rad/s
2
.
( . )
. 1100 006707
3= . M2
From the derivation in Problem 19.113,
xm
rmM
f
n
f
n
=( )( )
- ( )2
2
21
ww
ww
For out of phase motion with xm = - ¥ -60 10 6 m,
- ¥ =ÈÎÍ
˘˚
--
¥ -
60 100 006707
1 0 0067076
0 150 28 102
2
3
2
( . )( ) ( . )
.
M M
M
- ¥ + ¥ = ¥
¥ =
- - -
-
60 10 60 10 0 006707 28 170 10
402 49 10 88
6 62
6
92
( )( . ) .
.
M
M ..
.
170 10
219 10
6
2
¥=
-
M kg
Added mass: DM M M= - = -2 1 219 10 180. DM = 39 1. kg �
177
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PROBLEM 19.119
The unbalance of the rotor of a 200-kg motor is equivalent to a 100-g mass located 150 mm from the axis of rotation. In order to limit to 1-N the amplitude of the fluctuating force exerted on the foundation when the motor is run at speeds of 100 rpm and above, a pad is to be placed between the motor and the foundation. Determine (a) the maximum allowable spring constant k of the pad, (b) the corresponding amplitude of the fluctuating force exerted on the foundation when the motor is run at 200 rpm.
SOLUTION
Mass of motor. M = 200 kg
Unbalance mass. m = =100 0 1g kg.
Eccentricity. r = =150 mm 0.15 m
Equation of motion: Mx kx P t mr tm f f f&&+ = =sin sinw w w2
( )- + =
=-
M k x mr
xmr
k M
f m f
mf
f
w w
w
w
2 2
2
2
Transmitted force. F kxkmr
k Mm m
f
f
= =-
w
w
2
2
For out of phase motion, | |Fkmr
M km
f
f
=-
w
w
2
2 (1)
(a) Required value of k.
Solve Eq. (1) for k. |F M k kmrm f f| ( )w w2 2- =
k mr F F Mf m m f( | |) | |w w2 2+ =
kF M
mr F
m f
f m
=+
| |
| |
w
w
2
2
Data: | |Fm = 1 N w f = =100 10 472rpm rad/s.
k =+
=( )( )( . )
( . )( . )( . ).
1 200 10 472
0 1 0 15 10 472 18292 3
2
2N/m k = 8 292. kN/m �
178
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PROBLEM 19.119 (Continued)
(b) Force amplitude at 200 rpm. w f = 20 944. rad/s
From Eq. (1), | |( . )( . )( . )( . )
( )( . ) .Fm =
-8292 3 0 1 0 15 20 944
200 20 944 8292 3
2
2
| | .Fm = 0 687 N �
179
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PROBLEM 19.120
A 180-kg motor is supported by springs of total constant 150 kN/m. The unbalance of the rotor is equivalent to a 28-g mass located 150 mm from the axis of rotation. Determine the range of speeds of the motor for which the amplitude of the fluctuating force exerted on the foundation is less than 20 N.
SOLUTION
Equation derived in Problem 19.113: xm
rmM
f
n
f
n
=( )( )
- ( )ww
ww
2
21
Force transmitted: F kxm m
kmrM
f
n
f
n
= =( )( )
- ( )ww
ww
2
21
Data: k
r
m g
M
kr
= = ¥= =
= = ¥=
-
150 150 10
150 0 150
28 28 10
180
3
3
kN/m N/m
mm m
kg
kg
.
mm
M
k
Mn
= ¥ ¥ =
= = ¥ =
-( )( . )( ).
.
150 10 0 150 28 10
1803 5
150 10
18028 8
3 3
3
N
w 668 rad/s
In phase motion with N| | .Fm � 20
3 5
120
3 5 20 20
23 5
2
2
2 2
.
.
.
ww
ww
ww
ww
w
f
n
f
n
f
n
f
n
( )- ( )ÊËÁ
ˆ¯
-ÊËÁ
ˆ¯
�
�
ff
nwÊËÁ
ˆ¯
2
20�
ww
w
f
n
f
�
�
20
23 5
20
23 528 868 26 63
.
.( . ) .= rad/s w f � 254 rpm �
180
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PROBLEM 19.120 (Continued)
Out of phase motion with N.| |Fm � 20
3 5
120
3 5 20 20
20 16
2
2
2 2
.
.
ww
ww
ww
ww
f
n
f
n
f
n
f
n
( )( ) -
ÊËÁ
ˆ¯
ÊËÁ
ˆ¯
-
�
�
� ..52w
wf
n
ÊËÁ
ˆ¯
ww
wf
nf� �
20
16 5
20
16 528 868 31 78
. .( . ) .= rad/s w f � 304 rpm �
181
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.121
A vibrometer used to measure the amplitude of vibrations consists essentially of a box containing a mass-spring system with a known natural frequency of 120 Hz. The box is rigidly attached to a surface, which is moving according to the equation y tm f= d wsin . If the amplitude zm of the motion of the mass relative to the box is used as a measure of the amplitude dm of the vibration of the surface, determine (a) the percent error when the frequency of the vibration is 600 Hz, (b) the frequency at which the error is zero.
SOLUTION
x t
y t
z
z x y
m
f
m f
f
n
=Ê
ËÁÁ
ˆ
¯˜˜
=
=
= =
dw
d w
ww
12
2-
-
sin
sin
relative motion
ddd w
dd
ww
ww
w
mm
f
m m
m
f
n
f
n
f
t
z
1
1
11
2
2
2
2
2
--
--
È
ÎÍÍ
˘
˚˙˙
=È
ÎÍÍ
˘
˚˙˙
=
sin
wwww
n
f
n
2
2
21-
(a) zm
m
f
n
f
n
d
ww
ww
= =( )
( )= =
2
2
2
21 1
25
241 0417
600120
2
600120
2- -
. �
Error = 4 17. %
(b) zm
m
f
n
f
n
d
ww
ww
= =11
2
2
2
2-
1 2
2
2
2
2120 84 853
2
2=
= = =
w
wf
n
f ff n ( ) . Hz fn = 84 9. Hz �
182
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.122
A certain accelerometer consists essentially of a box containing a mass-spring system with a known natural frequency of 2200 Hz. The box is rigidly attached to a surface, which is moving according to the equation y tm f= d wsin . If the amplitude zm of the motion of the mass relative to the box times a scale factorwn
2 is used as a measure of the maximum acceleration am m f= d w2 of the vibrating surface, determine the percent error when the frequency of the vibration is 600 Hz.
SOLUTION
x t
y t
z
z x y
m
f
m f
f
n
=Ê
ËÁÁ
ˆ
¯˜˜
=
=
= =
dw
d w
ww
12
2-
-
sin
sin
relative motion
ddd w
dd
ww
ww
w
mm
f
m m
m
f
n
f
n
f
t
z
1
1
11
2
2
2
2
2
--
--
È
ÎÍÍ
˘
˚˙˙
=È
ÎÍÍ
˘
˚˙˙
=
sin
wwww
n
f
n
2
2
21-
The actual acceleration is am f m= -w d2
The measurement is proportional to zm nw2.
Then z
a
zm n
m
m
m
n
f
f
n
wd
ww
ww
2 2
2
6002200
2
1
1
1
1
=Ê
ËÁ
ˆ
¯˜
=
( )=
( )
-
-
= 1 0804. Error = 8 04. % �
183
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PROBLEM 19.123
Figures (1) and (2) show how springs can be used to support a block in two different situations. In Figure (1), they help decrease the amplitude of the fluctuating force transmitted by the block to the foundation. In Figure (2), they help decrease the amplitude of the fluctuating displacement transmitted by the foundation to the block. The ratio of the transmitted force to the impressed force or the ratio of the transmitted displacement to the impressed displacement is called the transmissibility. Derive an equation for the transmissibility for each situation. Give your answer in terms of the ratio w wf n/ of the frequency wf of the impressed force or impressed displacement to the natural frequency wn of the spring-mass system. Show that in order to cause any reduction in transmissibility, the ratio w wf n/ must be greater than 2.
SOLUTION
(1) From Equation (19.33): xm
Pkm
f
n
=
( )12
- ww
Force transmitted: ( )P kx kT m m
Pkm
f
n
= = ( )È
Î
ÍÍÍ
˘
˚
˙˙˙1
2- w
w
Thus, Transmissibility = =
( )( )P
PT m
m f
n
1
12
- ww
�
(2) From Equation ( . ):19 33¢
Displacement transmitted: xmm
f
n
=
( )d
ww1
2-
Transmissibility = =
( )xm
m f
n
d ww
1
12
- �
For ( )PP
xT m
m
m
mor to be less than 1,d
1
1
12
- ww
f
n( )�
1 12
� - ww
f
n( )
ww
f
n
ÊËÁ
ˆ¯
2
2� ww
f
n
� 2 Q.E.D. �
184
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PROBLEM 19.124
A 30-kg disk is attached with an eccentricity e = 0 15. mm to the midpoint of a vertical shaft AB, which revolves at a constant angular velocity w f . Knowing that the spring constant k for horizontal movement of the disk is 650 kN/m, determine (a) the angular velocity w f at which resonance will occur, (b) the deflection r of the shaft when w f = 1200 rpm.
SOLUTION
G describes a circle about the axis AB of radius r e+ .
Thus, a r en f= +( )w2
Deflection of the shaft is F kr=
Thus, F ma
kr m r e
k
mm
k
krk
r e
re
n
f
nn
nf
f
n
f
=
= +
= =
= +
=
( )
( )
w
ww
ww
www
2
22
22
2
2
2
1-wwn
2
185
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PROBLEM 19.124 (Continued)
(a) Resonance occurs when w wf n r= , i.e., Æ
wnk
m=
=
==
( )
.
.
650 1000
30147 196
1405 6
¥
rad/s
rpm w wn f= = 1406 rpm �
(b) r =( )
( )( . ) .
.
0 15
1
12001405 6
2
12001405 6
2
mm
-
= 0 02964. mm r = 0 403. mm �
186
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.125
A small trailer and its load have a total mass of 250 kg. The trailer is supported by two springs, each of constant 10 kN/m, and is pulled over a road, the surface of which can be approximated by a sine curve with an amplitude of 40 mm and a wavelength of 5 m (i.e., the distance between successive crests is 5 m and the vertical distance from crest to trough is 80 mm). Determine (a) the speed at which resonance will occur, (b) the amplitude of the vibration of the trailer at a speed of 50 km/h.
SOLUTION
Total spring constant k = ¥
= ¥
2 10 10
20 10
3
3
( )N/m
N/m
(a) w
l
d
dl
n
m
m
k
m
yx
x
23
2
3
20 1080
5
40 40 10
= =¥
=
=
= = ¥
=
N/m
250 kg s
m
mm m
-
-
sin
==
= ÊËÁ
ˆ¯
= =
vt
yv
tvm fd
lt l w p
tsin , ,and
2
so y tv v
f f= ¥ = =40 102 2
53- sinw w p
lp
where
From Equation ( . ):19 33¢ xmm
f
n
=ÊË
ˆ¯
dww
12
2-
Resonance: w p wf nv= = =2
580 1s- ,
v = 7 1176. m/s v = 25 6. km/h �
(b) Amplitude at v
f
f
= =
= =
=
50 13 8889
2 13 8889
517 4533
3042
km/h m/s
rad/s
.
( . ).
.
w p
w 660 2 s-
xm = ¥ = ¥40 10
114 246 10
3
304 6280
3-
-
--
.. m xm = -14 25. mm �
187
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.126
Block A can move without friction in the slot as shown and is acted upon by a vertical periodic force of magnitude P P tm f= sin ,w where w f = 2 rad/s and Pm = 20 N. A spring of constant k is attached to the bottom of block A and to a 22-kg block B. Determine (a) the value of the constant k which will prevent a steady-state vibration of block A, (b) the corresponding amplitude of the vibration of block B.
SOLUTION
In steady state vibration, block A does not move and therefore, remains in its original equilibrium position.
Block A: +SF = 0
kx P tm f= - sinw (1)
Block B: SF m xB= &&
m x kx
x x t
k m
B
m n
n B
&&+ ==
=
0
2
sinw
w /
From Eq. (1): kx t P t
kx P
k
m
k m
m n m f
n f
m m
nB
B n
sin sinw w
w w
w
w
=
= =
=
=
=
-
-
2
2
rad/s
(a) Required spring constant. k = ( )( )22 2 2 k = 88 0. N/m �
(b) Corresponding amplitude of vibration of B.
k x P
xP
k
x
m m
mm
m
=
=
=
-
-
- 20
88
N
N/m xm = -0 227. N/m �
+
188
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.127
Show that in the case of heavy damping ( ),c cc� a body never passes through its position of equilibrium O (a) if it is released with no initial velocity from an arbitrary position or (b) if it is started from O with an arbitrary initial velocity.
SOLUTION
Since c cc� , we use Equation (19.42), where
l l1 20 0� �,
x c e c et t= +1 21 2l l (1)
vdx
dtc e c et t= = +1 1 2 2
1 2l ll l (2)
(a) t x v= = =0 00, , :x
From Eqs. (1) and (2): x c c
c c0 1 2
1 1 2 20
= += +l l
Solving for c1 and c2 , c x
cx
x
12
2 10
21
2 10
=
=
ll l
ll
--
-
Substituting for c1 and c2 in Eq. (1), xx
e et t= ÈÎ ˘2
2 12 1
1 2
l ll ll l
--
For x = 0: when t , we must have
l lll
l l l l1 2
2
1
2 1 2 10e e et t t- -= = ( ) (3)
Recall that
l l1 20 0� �, . Choosing l l1 and 2 so that l l1 2 0� � , we have
0 1 02
11� � �
ll
l land 2 -
Thus a positive solution for t � 0 for Equation (3) cannot exist, since it would require that e raised to a positive power be less than 1, which is impossible. Thus, x is never 0.
�
The x t- curve for this case is as shown.
189
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PROBLEM 19.127 (Continued)
(b) t x v v= = =0 0 0, , : Equations (1) and (2) yield
0 1 2
0 1 1 2 2
= += +
c c
v c cl l
Solving for c c1 2 and , cv
cv
10
2 1
22
2 1
=
=
--
-
l l
l l
Substituting into Eq. (1), xv
e et t= 0
2 1
2 1
l ll l
--[ ]
For x = 0, t π a
e et tl l2 1=
For c cc� , ;l l1 2- thus, no solution can exist for t, and x is never 0.
�
The x t- curve for thiscase is as shown.
190
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PROBLEM 19.128
Show that in the case of heavy damping ( ),c cc� a body released from an arbitrary position with an arbitrary initial velocity cannot pass more than once through its equilibrium position.
SOLUTION
Substitute the initial conditions, t x x v v= = =0 0 0, , in Equations (1) and (2) of Problem 19.127.
x c c v c c0 1 2 0 1 1 2 2= + = +l l
Solving for c c1 2 and , cv x
cv x
10 2 0
2 1
20 1 0
2 1
=
=
--
--
-
( )
( )
ll l
ll l
And substituting in Eq. (1) x v x e v x et t= ÈÎ ˘1
2 10 1 0 0 2 0
2 1
l ll ll l
-- - -( ) ( )
For x t= 0, : ( ) ( )
( )
( )( )
v x e v x e
ev x
v x
t
t t
t
0 1 0 0 2 0
0 2 0
0 1 0
2 1
2 1
1
- ---
-
l lll
l l
l l
=
=
=(( )
lnl l
ll2 1
0 2 0
0 1 0---
v x
v x
This defines one value of t only for x = 0, which will exist if the natural log is positive,
i.e., if v xv x
0 2 0
0 1 01-
-ll � . Assuming l l1 2 0� � ,
this occurs if v x0 1 0� l .
�
191
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.129
In the case of light damping, the displacements x1, x2, x3, shown in Figure 19.11 may be assumed equal to the maximum displacements. Show that the ratio of any two successive maximum displacements xn and xn+1 is constant and that the natural logarithm of this ratio, called the logarithmic decrement, is
ln( )
( )
x
x
c c
c c
n
n
c
c+=
12
2
1
p /
/-
SOLUTION
For light damping,
Equation (19.46): x x e tcm t= +( )
0 02-
sin( )w f
At given maximum displacement, t t x x
t
x x e
n n
n
ntc
m n
= =+ =
= ( )
,
sin( )w0
0
1f- 2
At next maximum displacement, t t x x
t
x x e
n n
n
ntc
m n
= =+ =
=
+ +
+
+( ) +
1 1
0 1
1 0
1
1
,
sin( )w f- 2
But w w pp
w
D n D n
n nD
t t
t t
+
+
=
=
1
1
2
2
-
-
Ratio of successive displacements: x
x
x e
x e
e e
n
n
t
t
t t
cm n
cm n
cm n n
cm D
+
( ) +
=
= =
+
+
1
0
0
2
2 1
2 1 22
-
-
- - pw
Thus, lnx
x
c
mn
n D+=
1
pw
(1)
From Equations (19.45) and (19.41): w w
w
D nc
Dc
c
c
c
m
c
c
=ÊËÁ
ˆ¯
= ÊËÁ
ˆ¯
1
21
2
2
-
-
Thus, lnx
x
c
m
m
c cc
n
n c
c
+=
ÊËÁ
ˆ¯
12
2 1
1
p
-
lnx
xn
n
cc
cc
c
c
+=
( )( )1
2
2
1
p
-Q.E.D. �
192
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PROBLEM 19.130
In practice, it is often difficult to determine the logarithmic decrement of a system with light damping defined in Problem 19.129 by measuring two successive maximum displacements. Show that the logarithmic decrement can also be expressed as ( ) ln ( ),1/ /k x xn n k+ where k is the number of cycles between readings of the maximum displacement.
SOLUTION
As in Problem 19.129, for maximum displacements xn and xn k+ at t tn n k n and t + + =, sin( )w0 1f
and sin( ) .wn n kt + + =f 1 x x e
x x e
nt
n kt
cm n
cm n k
=
=
( )
+( ) +
0
0
2
2
-
- ( )
Ratio of maximum displacements: x
x
x e
x een
n k
t
t
t tcm n
cm n k
cm n n k
+
( )( )
( )= =+
+0
0
2
2
2
-
-
- -
But w w pp
w
D n k D n
n n kD
t t k
t t k
+
+
=
=
-
-
( )2
2
Thus, x
x
c
m
kn
n k D+= +
ÊËÁ
ˆ¯2
2 pw
lnx
xk
c
mn
n k D+= p
w (2)
But from Problem 19.129, Equation (1):
log ln decrement = =+
x
x
c
mn
n D1
pw
Comparing with Equation (2), log ln decrement Q.E.D.=+
1
k
x
xn
n k
�
193
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PROBLEM 19.131
In a system with light damping ( ),c cc� the period of vibration is commonly defined as the time interval t p wd d= 2 / corresponding to two successive points where the displacement-time curve touches one of the limiting curves shown in Figure 19.11. Show that the interval of time (a) between a maximum positive displacement and the following maximum negative displacement is 1
2 t d , (b) between two successive zero displacements is 1
2 t d , (c) between a maximum positive displacement and the following zero displacement is greater than 1
4 t d .
SOLUTION
Equation (19.46): x x e tcm t
D= +( )0
2-sin( )w f
(a) Maxima (positive or negative) when &x = 0:
&x xc
me t x e t
cm
cmt
D Dt
D= ÊËÁ
ˆ¯
+ + +( ) ( )0 02
2 2- - -
sin( ) cos( )w w wf f
Thus, zero velocities occur at times when
&x tm
cDD= + =0
2, tan( )or w
wf (1)
The time to the first zero velocity, t1, is
t
mc
D
D
1
1 2
=( )È
΢˚tan- -w
w
f (2)
The time to the next zero velocity where the displacement is negative is
¢ =( ) +È
΢˚t
mc
D
D
1
1 2tan- -w p
w
f (3)
194
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PROBLEM 19.131 (Continued)
Subtracting Eq. (2) from Eq. (3),
¢ = =◊
=t tD
D D1 1 2 2
- pw
p tp
tQ.E.D.
(b) Zero displacements occur when
sin( )wq t + =f 0 or at intervals of
w p p pDt n+ =f , 2
Thus, ( ) ( )
( )( )t
tD D
1 01 0
2=¢ =
pw
pw
- -f fand
Time between ¢ = ¢ = = =02
2 21 0 1 0sD
D Dt t( ) ( )- -p pw
ptp
tQ.E.D.
Plot of Equation (1)
(c) The first maxima occurs at 1: ( )wDt1 + f
The first zero occurs at ( ( ) )w pD t1 0 + =f
From the above plot, ( ( ) ) ( )w w pD Dt t1 0 1 2
+ +f f- �
or ( ) ( )t t t tD
D1 0 1 1 0 12 4
- -� �pw
tQ.E.D.
Similar proofs can be made for subsequent maximum and minimum.
195
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PROBLEM 19.132
The block shown is depressed 30 mm from its equilibrium position and released. Knowing that after 10 cycles the maximum displacement of the block is 1.25 mm, determine (a) the damping factor c/c, (b) the value of the coefficient of viscous damping. (Hint: See Problems 19.129 and 19.130.)
SOLUTION
From Problems 19.130 and 19.129:
1 2
12k
x
x kn
n
cc
cc
c
c
ÊËÁ
ˆ¯ +
ÊËÁ
ˆ¯
=
( )ln
p
-
where number of cyclesk = = 10
(a) First maximum is x1 30= mm
Thus, n = 1 x
x
cc
cc
c
c
1
1
2
10
30
1 252 4
1
102 4 0 08755
2
1
+= =
=
=ÊËÁ
ˆ¯
..
ln . .
p
-
Damping factor. 12
0 08755
2 2 2
-c
c
c
cc c
ÊËÁ
ˆ¯
= ÊËÁ
ˆ¯
ÊËÁ
ˆ¯
p.
c
c
c
c
c
c
ÊËÁ
ˆ¯
ÊËÁ
ˆ¯
+È
ÎÍÍ
˘
˚˙˙
=
ÊËÁ
ˆ¯
=+( )
2 2
2
2
0 087551 1
1
5150 1
p.
== 0 0001941. c
cc
= 0 01393. �
196
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PROBLEM 19.132 (Continued)
(b) Critical damping coefficient. ck
mc = 2 m Eq 19.41)( .
or c km
c
c
c
c
c
=
== ◊
2
2 175
52 915
(
.
N/m)(4 kg)
N s/m
From Part (a), c
c
cc
=
=
0 01393
0 01393 52 915
.
( . )( . )
Coefficient of viscous damping. c = ◊0 737. N s/m �
197
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PROBLEM 19.133
A loaded railroad car of mass 15,000 kg is rolling at a constant velocity v0 when it couples with a spring and dashpot bumper system (Figure 1). The recorded displacement-time curve of the loaded railroad car after coupling is as shown (Figure 2). Determine (a) the damping constant, (b) the spring constant. (Hint: Use the definition of logarithmic decrement given in Problem 19.129.)
SOLUTION
Mass of railroad car: m = 15000 kg
The differential equation of motion for the system is
mx cx kx&& &+ + = 0
For light damping, the solution is given by Eq. (19.44):
x e c t c tcm t
d d= +( )- 21 2( sin cos )w w
From the displacement versus time curve,
t
w pt
pd
dd
=
= = =
0 41
2 2
0 4115 325
.
..
s
rad/s
At the first peak, x t t1 112 5= =. . mm and
At the second peak, x t t d2 13= = + mm and t .
Forming the ratio xx
2
1,
x
x
e
ee
x
xe
cm d
cm
cm
c dm
t
t
d2
1
1
2
2 1
2 1
2
2
= =
=
( ) +
( )( )
-
--
( )tt
t
(1)
198
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PROBLEM 19.133 (Continued)
(a) Damping constant.
From Eq. (1): c d
m
x
x
t2
1
2
=ÊËÁ
ˆ¯
ln
cm x
xd
=
=
= ◊
2
2 15000
0 41
12 5
3
104 423
1
2
3
tln
( )( )
.ln
.
. ¥10 N s/m c = ◊104 4. kN s/m �
(b) Spring constant.
Equation for wd : wdk
m
c
m2
2
2= Ê
ËÁˆ¯
-
k mc
md= +
= +
=
w22
23 2
4
15000 15 325104 423 10
4 15000
3 70
( )( . )( . )
( )( )
.
¥
446 106¥ N/m k = 3 70 106. ¥ N/m �
199
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PROBLEM 19.134
A 4-kg block A is dropped from a height of 800 mm onto a 9-kg block B which is at rest. Block B is supported by a spring of constant k = 1500 N/m and is attached to a dashpot of damping coefficient c = ◊230 N s/m. Knowing that there is no rebound, determine the maximum distance the blocks will move after the impact.
SOLUTION
Velocity of Block A just before impact.
v ghA =
==
2
2 9 81 0 8
3 962
( . )( . )
. m/s
Velocity of Blocks A and B immediately after impact.
Conservation of momentum.
m v m v m m v
v
v
A A B B A B+ = + ¢+ = + ¢
¢ =
( )
( )( . ) ( )
.
4 3 962 0 4 9
1 219 m/s &x0 1 219= + . m/s Ø = &x0
Static deflection (Block A): xm g
kA
0
4 9 82
15000 02619
=
=
=
-
-
-
( )( . )
. mx = 0, Equivalent position for both blocks:
c kmc =
== ◊
2
2 1500 13
279 3
( )( )
. N s/m
Since c cc< , Equation (19.44): x e c t c t
c
m
cm t
D D= +
=
=
( )-
-
21 2
1
2
230
2 13
8 846
[ sin cos ]
( )( )
.
w w
s
200
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PROBLEM 19.134 (Continued)
Expression forwD : w
w
D
D
k
m
c
m
x
22
2
2
1500
13
230
2 13
6 094
= ÊËÁ
ˆ¯
=ÊËÁ
ˆ¯
=
=
-
-( )( )
. rad/s
ee c t c tt-8 8461 26 094 6 094. ( sin . cos . )+
Initial conditions: x
t x
x e c c
0
0
00
1 2
0 02619
0 1 219
0 02619 0 1
== = +
= = +
-
-
.
( ) .
. [ ( ) (
m
m/s&))]
.
( ) . [ ( ) ( . )( )]( . )
c
x e c
2
8 846 01
0 02619
0 8 846 0 0 02619 1
=
= +
+
-
- --&ee c c( . )( )[ . ( ) ( )] .
. ( . )( .
-
- -
8 846 01 26 094 1 0 1 219
1 219 8 846 0 026
+ == 119 6 094
0 16202
0 16202 6 094 0 02619
1
1
8 846
) .
.
( . sin . . c.
+=
=
c
c
x e tt- - oos . )6 094t
’
Maximum deflection occurs when &x = 0
&x e t ttm m
m= =
+
0 8 846 0 16202 6 094 0 02619 6 0948 846- --. ( . sin . . cos . ).
ee t ttm m
m-
-
8 846 6 094 0 1620 6 094 0 02619 6 094
0
. [ . ][ . cos . . sin . ]
[(
+= 88 846 0 16202 6 094 0 02619 6 094
8 846 0
. )( . ) ( . )( . )]sin .
[( . )( .
++
tm
- - 002619 6 094 0 1620 6 094
0 1 274 6 094 1 219
) ( . )( . )]cos .
. sin . . c
+= +
t
tm
- oos .
tan ..
..
6 094
6 0941 219
1 2740 957
t
t = =
Time at maximum deflection s= = =
=
t
x e
m
m
tan .
..
-1 0 957
6 0940 1253
--
-
( . )( . )[ . sin( . )( . )
. cos( .
8 846 0 1253 0 1620 6 094 0 1253
0 02619 6 0944 0 1253
0 3301 0 1120 0 0189 0 307
)( . )]
( . )( . . ) .xm = =- m
Blocks move, static deflection + xm Total distance m mm= + = =0 02619 0 307 0 0569 56 9. . . . �
201
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.135
Solve Problem 19.134, assuming that the damping coefficient of the dashpot is c = ◊300 N s/m.
SOLUTION
Velocity of Block A just before impact.
v ghA =
==
2
2 9 81 0 8
3 962
( . )( . )
. m/s
Velocity of Blocks A and B immediately after impact.
Conservation of momentum.
m v m v m m v
v
v
A A B B A B+ = + ¢+ = + ¢
¢ =
( )
( )( . ) ( )
.
4 3 962 0 4 9
1 219 m/s
&x0 1 219= + . m/s Ø = &x0
Static deflection (Block A). xm g
kA
0
4 9 82
15000 02619
=
=
=
-
-
-
( )( . )
. m
x = 0, Equivalent position for both blocks:
c kmc =
== ◊
2
2 1500 13
279 3
( )( )
. N s/m
Since c cc� , the system is heavily damped.
Eq. 19.42: x c e c et t= +1 21 2l l
202
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.135 (Continued)
Eq. 19.40: l = ± ÊËÁ
ˆ¯
= ± ÊËÁ
ˆ¯
= ±
- -
- -
-
c
m
c
m
k
m2 2
300
26
300
26
1500
13
11 538 4
2
2
. .2213
15 751
7 325
1
11
21
115 751
27 325
l
l
=
=
= +
=
-
-
-
-
-
- -
.
.. .
s
s
x c e c e
x
t t
& 55 751 7 325111 751
27 325. .. .c e c et t- --
Initial conditions: x
x0
0
0 02619
1 219
==
- .
.
m
m/s&Then -
- -0 02619
1 219 15 751 7 3251 2
1 2
.
. . .
= +=
c c
c c
Solving the simultaneous equations, c
c1 0 1219
0 09571
==
- .
.
m
m2
Then x e e
x
t t= +
=
-
- -
- -0 1219 0 09571
15 751 0 1219
15 751 7 325. . ( )
( . )( . )
. . m
& ee e
e e
t t
t
- -
-
-
-
15 751 7 325
15 751
7 325 0 09571
1 92 0 70108
. .
.
( . )( . )
. .= --7 325. t
The maximum value of x occurs when &x = 0
0 1 92 0 70108
1 92 0 70108
15 751 7 325
15 751 7
=
=
. .
. .
. .
.
e e
e e
t t
t
m m
m
- -
- -
-..
( . . )
.
.
.
.
.
325
15 751 0 7325
8 426
1 92
0 70108
2 7386
8 426
t
t
t
m
m
m
e
e
t
=
=
-
mm
m
m
t
x e
==
= +
ln( . )
.
. .( . )( . )
2 7386
0 11956
0 1219 0 015 751 0 11956
s
- - 99571
0 018542 0 039867
0 02132
7 325 0 11956e-
-
( . )( . )
. .
.
= += m
Total deflection static deflection + == +=
xm
0 02619 0 02132
0
. .
.004751 m Total deflection = 47.5 mm �
203
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.136
The barrel of a field gun weighs 750 kg and is returned into firing position after recoil by a recuperator of constant c = ◊18 kN s/m Determine (a) the constant k which should be used for the recuperator to return the barrel into firing position in the shortest possible time without any oscillation, (b) the time needed for the barrel to move back two-thirds of the way from its maximum-recoil position to its firing position.
SOLUTION
(a) A critically damped system regains its equilibrium position in the shortest time.
c c
mk
m
c==
=
18000
2
Then km
cc
=( )
=( )
=2
218000
2
2
750108 kN/m �
(b) For a critically damped system, Equation (19.43):
x c c t e nt= +( )1 2-w
We take t = 0 at maximum deflection x0 .
Thus, &xx x
( )
( )
0 0
0 0
==
Using the initial conditions, x x c e c x
x x c t e nt
( ) ( ) ,
( )
0 00 10
1 0
0 2
= = + =
= +
so -w
and &&
x x c t e c e
x x c c xn
t t
n n
n n= + += = + =
--
- -ww w
w w( )
( ) ,0 2 2
0 2 2 00 0 so
Thus, x x t entn= +0 1( )w w-
For xx
t ek
mnt
nn= = + =0
3
1
31, ( ) ,w ww- with (1)
We obtain wn = =108
75012
31¥ -10
s( )
rad /
From solution of (1) wnt = 2 289. ,
t = =2 289
120 19075
.. s t = 0 1908. s �
204
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.137
A uniform rod of mass m is supported by a pin at A and a spring of constant k at B and is connected at D to a dashpot of damping coefficient c. Determine in terms of m, k, and c, for small oscillations, (a) the differential equation of motion, (b) the critical damping coefficient cc.
SOLUTION
In equilibrium, the force in the spring is mg.
For small angles, sin cosq q q
d q
d q
ª ª 1
2y
l
y l
B
C
=
=
(a) Newton’s Law: S SM MA A= ( )eff
+mgl
kl
mgl
cl l I mal
t2 2 2 2- -q q a+Ê
ËÁˆ¯
= +&
Kinematics: a q
a q
=
= =
&&&&a
l lt 2 2
I ml
cl kl
I ml
+ ÊËÁ
ˆ¯
È
ÎÍÍ
˘
˚˙˙
+ + ÊËÁ
ˆ¯
=
+ ÊËÁ
ˆ¯
=
2 20
2
1
3
22
2
2
&& &q q q
mml2 && &q q q+ ÊËÁ
ˆ¯
+ ÊËÁ
ˆ¯
=3 3
40
c
m
k
m �
(b) Substituting q l= e t into the differential equation obtained in (a), we obtain the characteristic equation,
l l2 3 3
40+ Ê
ËÁˆ¯
+ =c
m
k
m
and obtain the roots l =( ) ( )- -3 3 2 3
2
cm
cm
kmm
The critical damping coefficient, cc , is the value of c, for which the radicand is zero.
Thus, 3 32c
m
k
mcÊ
ËÁˆ¯
= ckm
c =3
�
205
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.138
A 2-kg uniform rod is supported by a pin at O and a spring at A, and is connected to a dashpot at B. Determine (a) the differential equation of motion for small oscillations, (b) the angle that the rod will form with the horizontal 5 s after end B has been pushed 25 mm down and released.
SOLUTION
Small angles: sin ,
. .
. .
.
q q qd q q
d q q
d q
ª ª cos 1
y
y
y
A
C
B
= ( ) =
= ( ) =
= ( ) =
0 2 0 2
0 2 0 2
0 6
m
m
m 00 6. q
(a) Newton’s Law: S SM M0 0= ( )eff
+ - -0 2 0 2 2 9 81 0 6. . ( )( . ) .m m m( ) + ( ) ( )F Fs D
= ( )I mata + 0 2. m (1)
F k y k
F c y c
I ml
s A A A
D B
= + = +( )= =
= =
( ( ) ) . ( )
.
d d q d
d qST ST0 2
0 6
1
12
1
122
& &
mm( . )0 84
752 =
m
Kinematics: a q a q= = ( ) =&& &&, . . mat 0 2 0 2
Thus, from Eq. (1), 4
750 2 0 6 0 2 0 2 0 2 22 2m
m c k A+ÈÎÍ
˘˚
+ + ( ) -( . ) ( . ) . ( . ) ( . )(&& &q q q + (dST ))( . )9 81 0= (2)
But in equilibrium, SM0 0=
+ k kA A( ) ( )( . )( . ) , . ( ) ( . )( )( . )d dST ST(0.2) - 2 9 81 0 2 0 0 2 0 2 2 9 81= =
206
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PROBLEM 19.138 (Continued)
Equation (2) becomes
7
75
9
25
1
250
7
75
7
752 0
ÊËÁ
ˆ¯
+ ÊËÁ
ˆ¯
ÊËÁ
ˆ¯
=
= ÊËÁ
ˆ¯
=
m c k
m
&& &q q q
( ) ..
( ) .
18667
9
25
9
258 2 88
25
50
252
c
k
= ÊËÁ
ˆ¯
=
= = 0 18667 2 88 2 0. .&& &q q q+ + = �
(b) Substituting e tl into the above differential equation,
0 18667 2 88 2 0
0 7289 14 699
2. .
. , .
l ll
+ + == -- i
Since the roots are negative and real, the system is heavily damped. (Eq. 19.46):
q
q
= +
= - --
c e c e
c e c
t t
t
10 7289
114 699
10 7289
20 7289 14 699
- -. .
.( . ) ( .& )) .e t-14 699 (3)
Initial conditions. q
q
01
0
1 2
1
25
6000 04168
0
0 04168
0 0 07289 14 6
= =
== += - -
-sin .
.
. .
rad
c c
c 999 2c
Solving for c1 and c2: c c1 230 04385 2 1747 10= = - -. , . ¥
At t = 5, q = -=
-0 04385 2 1747 10
0 1146
0 7289 5 3. .
.
( . )( )e e- ¥= 0.657∞rad
207
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PROBLEM 19.139
A 500-kg machine element is supported by two springs, each of constant 50 kN/m. A periodic force of 150 N amplitude is applied to the element with a frequency of 2.8 Hz. Knowing that the coefficient of damping is 1 8. kN s/m,◊ determine the amplitude of the steady-state vibration of the element.
SOLUTION
Eq. (19.52): xP
k m cm
m
f f
=( ) +- w w2 2 2( )
Total spring constant: k =
=
( )(2 503
kN/m)
100 N/m¥10
w p p pf f
m
m
f
m
P
c
x
= = =
=== ◊
=
2 2 2 8 5 6
500
150
1800
150
( . ) .
[
rad/s
N
N s/m
kg
1100 000 500 5 6 1800 5 6
150
2 9982 10 1 0028
2 2 2
9
, ( )( . ) ] [ ( . ) ]
. .
- p p+
=¥ + ¥¥
= ¥ -10
2 3714 10
9
3. m 2.3714 mm= xm = 2 37. mm �
208
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.140
In Problem 19.139, determine the required value of the constant of each spring if the amplitude of the steady-state vibration is to be 1.25 mm.
SOLUTION
w p p pf f
m
f
P
mW
g
c
= = =
=
= =
= ◊
2 2 2 8 5 6
150
500
1800
( )( . ) . rad/s
N
N s/m
kg
xxm = ¥ -1 25 3. mm 1.25 10 m¥
Eq. (19.52): xP
k m cm
m
f f
=+( ) ( )- w w2 2 2
Solve for k. ( ) ( )k m cP
xf fm
m- w w2 2 2
2
+ = ÊËÁ
ˆ¯
k mP
xcf
m
mf= +
ÊËÁ
ˆ¯
= + ÊËÁ
ˆ¯-
w w
p
22
2
23
2
500 5 6150
1 25 10
-
¥
( )
( )( . ).
--
-
[( )( . )]
. . .
.
1800 5 6
154 755 10 1 44 10 1 00281 10
154 75
2
3 10 9
p
= ¥ + ¥ ¥
= 55 10 115 746 10
270 501 10
3 3
3
¥ + ¥
= ¥
.
. N/m
The above calculated value is the equivalent spring constant for two springs.
For each spring, the spring constant is k
2.
k
2135 3= . kN/m �
209
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PROBLEM 19.141
In the case of the forced vibration of a system, determine the range of values of the damping factor c cc/ for which the magnification factor will always decrease as the frequency ratio w wf n/ increases.
SOLUTION
From Eq. ( . ) :19 53 ¢
Magnification factor: xmPk c
c
m
f
n c
f
n
=
- ( )ÈÎÍ
˘˚
+ ( )( )ÈÎÍ
˘˚
1
1 22 2 2w
www
Find value of c
cc
for which there is no maximum for xmPkm
as ww
f
n
increases.
d
d
x cc
mPmk
f
n
f
n c
f
n
( )( )
=- - ( )Ê
ËÁˆ¯
-( ) +È
ÎÍ
˘
˚˙
-
2
2
22 1 1 4
1
2
2
ww
ww
ww(( )È
Î͢˚
+ ( )( )ÈÎÍ
˘˚
ÏÌÔ
ÓÔ
¸˝Ô
Ô
=
- + ÊËÁ
ˆ¯
2 2 22
2
0
2 2
ccc
f
n
f
n
ww
ww
22 2
2 2
2
4 0
1 2
2+ =
ÊËÁ
ˆ¯ = -
c
c
c
c
c
f
nc
ww
For c
cc
2
2
1
2� , there is no maximum for
xmPkm( ) and the magnification factor will decrease as
ww
f
n
increases.
c
cc
�1
2
c
cc
� 0 707. �
210
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.142
Show that for a small value of the damping factor c cc/ , the maximum amplitude of a forced vibration occurs when w wf nª and that the corresponding value of the magnification factor is 1
2 ( ).c cc/
SOLUTION
From Eq. (19.53' ):
Magnification factor = =
- ( )ÈÎÍ
˘˚
+ ( )( )È
xmPk c
c
m
f
n c
f
n
1
1 22 2
ww
wwÎÎÍ
˘˚
2
Find value of ww
f
n
for which xmPkm
is a maximum.
0
2 1 1 4
1
2
2
2 2
2
=( )( )
= -- ( )È
Î͢˚
-( ) + ( )È
ÎÍ
˘
˚˙
-
d
d
x cc
mPmk
f
n
f
n c
ww
ww
www
ww
ww
f
n c
f
n
cc
f
n
( )ÈÎÍ
˘˚
+ ( )( )ÈÎÍ
˘˚
ÏÌÔ
ÓÔ
¸˝Ô
Ô
- +ÊËÁ
ˆ
2 2 22
2
2 2¯
+ÊËÁ
ˆ¯
=2 2
4 0c
cc
For small c
cc
, ww
w wf
nf nª ª1
For ww
f
n
= 1
xmPk c
c
m
c
=-[ ] + ( )È
΢˚
1
1 1 2 12 2
x c
cm
Pk
c
m( ) = 1
2 �
211
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PROBLEM 19.143
A 50-kg motor is directly supported by a light horizontal beam, which has a static deflection of 6 mm due to the mass of the motor. The unbalance of the rotor is equivalent to a mass of 100 g located 75 mm from the axis of rotation. Knowing that the amplitude of the vibration of the motor is 0.8 mm at a speed of 400 rpm, determine (a) the damping factor c cc/ , (b) the coefficient of damping c.
SOLUTION
Spring constant: km mg= = =
¥= ¥ ◊-d dST ST
N m( )( . )
.50 9 81
6 1081 75 10
33
Natural undamped circular frequency: wnk
m= = ¥ =81 75 10
5040 435
3.. rad/s
Unbalance: ¢ = == =
m g
r
100 0 100
75 0 075
.
.
kg
mm m
Forcing frequency: w f = =400 41 888rpm rad/s.
Unbalance force: P m rm f= ¢ = =w2 20 100 0 075 41 888 13 1595( . )( . )( . ) . N
Static deflection: dST m= =¥
= ¥ -P
km 13 1595
81 75 100 16097 10
33.
..
Amplitude: xm = = ¥ -0 8 0 8 10 3. .mm m
Frequency ratio: ww
f
n
= =41 888
40 4351 0359
.
..
Eq. (19.53): xm
cc
f
n c
f
n
=
- ( )ÈÎÍ
˘˚
+ ( )( )ÈÎÍ
˘˚
d
ww
ww
ST
1 22 2 2
1 22 2 2
-ÊËÁ
ˆ¯
È
ÎÍÍ
˘
˚˙˙
+ÊËÁ
ˆ¯
ÊËÁ
ˆ¯
È
ÎÍÍ
˘
˚˙˙
=ww
ww
df
n c
f
n m
c
c xSTÊÊ
ËÁˆ¯
- +ÊËÁ
ˆ¯
È
ÎÍÍ
˘
˚˙˙
=
2
2 22
1 1 0359 2 1 03590 16097
[ ( . ) ] ( . ).c
cc
¥¥¥
È
ÎÍ
˘
˚˙
+ÊËÁ
ˆ¯
=
-
-10
0 8 10
0 0053523 4 2924 0 040486
3
3
2
2
.
. . .c
c
c
c
c
cc
ÊËÁ
ˆ¯
=2
0 008185.
212
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PROBLEM 19.143 (Continued)
(a) Damping factor. c
cc
= 0 090472. c
cc
= 0 0905. �
Critical damping factor. c kmc =
= ¥
= ¥ ◊
2
2 81 75 10 50
4 0435 10
3
3
( . )( )
. N s/m
(b) Coefficient of damping. cc
cc
cc=
ÊËÁ
ˆ¯
= ¥( . )( . )0 090472 4 0435 103 c = ◊366 N s/m �
213
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PROBLEM 19.144
A 15-kg motor is supported by four springs, each of constant 45 kN/m. The unbalance of the motor is equivalent to a mass of 20 g located 125 mm from the axis of rotation. Knowing that the motor is constrained to move vertically, determine the amplitude of the steady-state vibration of the motor at a speed of 1500 rpm, assuming (a) that no damping is present, (b) that the damping factor c cc/ is equal to 1.3.
SOLUTION
Mass of motor: m = 15 kg
Equivalent spring constant: k = ¥
= ¥
( )( )4 45 10
180 10
3
3 lb/ft
Undamped natural circular frequency: wnk
m= = ¥
=
180 10
15109 545
3
. rad/s
Forcing frequency: w f = =1500 157 08rpm rad/s.
Frequency ratio: ww
f
n
= =157 08
109 5451 43394
.
..
Unbalance: ¢ = = ¥= =
-m g
r
20 20 10
125 0 125
3 kg
mm m.
Unbalance force: P m rm f= ¢ = ¥ =-w2 3 220 10 0 125 157 08 61 685( )( . )( . ) . N
Static deflection: dST
m
= =¥
= ¥ -
P
km 61 685
180 10
0 3427 10
3
3
.
.
Eq. (19.53): xm
cc
f
n c
f
n
=
- ( )ÈÎÍ
˘˚
+ ( )( )ÈÎÍ
˘˚
d
ww
ww
ST
1 22 2 2
(a) No damping. 1 1 1 43394 1 115222 2
2 2- ( )ÈÎÍ
˘˚
= - =ww
f
n
[ ( . ) ] .
xm =
= ¥
= ¥
-
-
dST
m
1 11522
0 3427 10
1 05618
0 324 10
3
3
.
.
.
. xm = 0 324. mm �
214
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.144 (Continued)
(b) c
cc
cc
f
n c
f
n= - ( )È
Î͢˚
+ ( )( )ÈÎÍ
˘˚
1 3 1 22 2 2
. ww
ww
= + =
= = ¥
1 11522 2 1 3 1 43394 15 015
15 015
0 3427 10
2. [( )( . )( . )] .
.
.xm
dST--
-= ¥
3
3
3 8749
0 0884 10
.
. m xm = 0 0884. mm �
215
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.145
A 100-kg motor is supported by four springs, each of constant 90 kN/m, and is connected to the ground by a dashpot having a coefficient of damping c = ◊6500 N s/m. The motor is constrained to move vertically, and the amplitude of its motion is observed to be 2.1 mm at a speed of 1200 rpm. Knowing that the mass of the rotor is 15 kg, determine the distance between the mass center of the rotor and the axis of the shaft.
SOLUTION
Equation (19.52): Rotor
xP
k m cm
m
f f
f
f
=-( ) +
= ÈÎÍ
˘˚
= -
w w
w p
w
2 2 2
22
2 2
1200 2
60
15 791
( )
( )( )
, s
kk
P m e
e
m f
= ¥
= ¥
= ¢
==
-
4 90 10
360 10
15 15 791
236 8
3
3
2
2
( )
( ) ( , )
,
N/m
kg s
w
665e
2 1 10236 865
360 10 100 15 791 6500 15 79
3
3 2 2.
,
[( ) ( )( , )] ( ) ( ,¥ =
¥ - +- m
e
11
0 16141
13 01 10 3
)
.
.
=
= ¥ -
e
e m e = 13 01. mm �
216
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.146
A counter-rotating eccentric mass exciter consisting of two rotating 400-g masses describing circles of 150-mm radius at the same speed, but in opposite senses, is placed on a machine element to induce a steady-state vibration of the element and to determine some of the dynamic characteristics of the element. At a speed of 1200 rpm, a stroboscope shows the eccentric masses to be exactly under their respective axes of rotation and the element to be passing through its position of static equilibrium. Knowing that the amplitude of the motion of the element at that speed is 15 mm, and that the total mass of the system is 140 kg, determine (a) the combined spring constant k, (b) the damping factor c cc/ .
SOLUTION
Forcing frequency: w f = =1200 125 664rpm rad/s.
Unbalance of one mass: m g
r
= == =
400 0 4
150 0 15
.
.
kg
mm m
Shaking force: P mr t
t
f f
f
=
=
= ¥
2
2 0 4 0 15 125 664
1 89497 10
2
2
3
w w
w
sin
( )( . )( . )( . ) sin
. siin
.
w f
m
t
P = ¥1 89497 103 N
Total mass: M = 140 kg
By Eqs. (19.48) and (19.52), the vibratory response of the system is
x x tm f= -sin( )w j
where xP
k M cm
m
f f
=-( ) +w w2 2 2( )
(1)
and tanjw
w=
-
c
k M
f
f2
(2)
Since j p= ∞ =902
, tan .j w- =and k M f2 0
(a) Combined spring constant.
k M f=
=
= ¥
w2
2
6
140 125 664
2 2108 10
( )( . )
. N/m k = 2 21. Mn/m �
217
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.146 (Continued)
The observed amplitude is xm = =15 0 015mm m.
From Eq. (1): cP
xk M
P
xf
m
mf
m
f m
=ÊËÁ
ˆ¯
- - =
= ¥
1
1 89497 10
125 664 0 01
22
3
ww
w( )
.
( . )( . 55
1 00531 103
)
.= ¥ ◊N s/m
Critical damping coefficient: c kMc =
= ¥
= ¥ ◊
2
2 2 2108 10 140
35 186 10
6
3
( . )( )
. N s/m
(b) Damping factor. c
cc
= ¥¥
1 00531 10
35 186 10
3
3
.
.
c
cc
= 0 0286. �
218
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.147
A simplified model of a washing machine is shown. A bundle of wet clothes forms a mass mb of 10 kg in the machine and causes a rotating unbalance. The rotating mass is 20 kg (including mb) and the radius of the washer basket e is 25 cm. Knowing the washer has an equivalent spring constant k = 1000 N/m and damping ratio z = c/ cc = 0.05 and during the spin cycle the drum rotates at 250 rpm, determine the amplitude of the motion and the magnitude of the force transmitted to the sides of the washing machine.
SOLUTION
Forced circular frequency: w pf = =( )( )
.2 250
6026 18 rad/s
System mass: m = 20 kg
Spring constant: k = 1000 N/m
Natural circular frequency: wnk
m= = =1000
207 0711. rad/s
Critical damping constant: c kmc = = = ◊2 2 1000 20 282 84( )( ) . N s/m
Damping constant: cc
cc
c
=ÊËÁ
ˆ¯
= = ◊( . )( . ) .0 05 141 42 14 1421 N s/m
Unbalance force: P m e
P
m b f
m
=
= =
w2
210 0 25 26 18 1713 48( )( . )( . ) .kg m rad/s N
The differential equation of motion is
mx cx kx P tm f&& &+ + = sinwThe steady state response is
x x tm f= -sin( )w j &x x tf m f= -w w jcos( )
where xP
k m cm
m
f f
=-( ) +
=- +
w w2 2 2
2 2
1713 48
1000 20 26 18 141421
( )
.
[ ( )( . ) ] [( ))( . )]
.
( , . ) ( . )
.
, .
26 18
1713 48
12 707 8 370 24
1713 48
12 713 2
2
2 2=
- += = 00 13 478. , m
xm = 134 8. mm �
219
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.147 (Continued)
(a) Amplitude of vibration. x t
x t
f
f
= -
= -
=
0 13478
26 18 0 13478
3 5285
. sin( )
( . )( . )cos( )
. cos
w j
w j&(( )w jf t -
Spring force: kx t
t
f
f
= -
= -
( )( . )sin( )
. sin( )
1000 0 13478
134 78
w j
w j
Damping force: cx t
t
f
f
& = -
= -
( . )( . )cos( )
. cos( )
14 1421 3 5285
49 901
w j
w j
(b) Total force: F t tf f= - + -134 78 49 901. sin( ) . cos( )w j w j
Let F F t F t
F t
m f m f
m f
= - + -
= - +
cos sin( ) sin sin( )
sin( )
y w j j w j
w j y
Maximum force. F F Fm m m2 2 2 2 2= +cos siny y
= +=
( . ) ( . )
,
134 78 49 901
20 656
2 2
Fm = 143 7. N �
220
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.148
A machine element is supported by springs and is connected to a dashpot as shown. Show that if a periodic force of magnitude P = Pm sin wf t is applied to the element, the amplitude of the fluctuating force transmitted to the foundation is
F Pm m
cc
cc
c
f
n
f
n c
f
n
=+ ( )( )È
Î͢˚
- ( )ÈÎÍ
˘˚
+ ( )( )ÈÎ
1 2
1 2
2
2 2 2
ww
ww
wwÍÍ
˘˚
2
SOLUTION
From Equation (19.48), the motion of the machine is x x tm f= -sin( )w f
The force transmitted to the foundation is
Springs: F kx kx ts m f= = -sin( )w f
Dashpot: F cx cx t
F x k c t
D m f f
T m f f f
= = -
= - + -
& w w f
w f w w f
cos( )
[ sin( ) cos( )]
or recalling the identity,
A y B y A B y
B
A BA
A B
F x k cT m
sin cos sin( )
sin
cos
(
+ = + +
=+
=+
= +
2 2
2 2
2 2
2
f
y
y
w ff f t) sin( )2ÈÎ
˘˚ - +w f y
Thus, the amplitude of FT is F x k cm m f= +2 2( )w (1)
From Equation (19.53): xm
Pk
cc
m
f
n c
f
n
=
- ( )ÈÎÍ
˘˚
+ ( )ÈÎÍ
˘˚1 2
2 2 2ww
ww
Substituting for xm in Equation (1), FP
k
m
m
mc
k
cc
n
f
f
n c
f
n
=+ ( )
- ( )ÈÎÍ
˘˚
+ ( )( )ÈÎÍ
˘˚
=
1
1 2
2
2 2
2
2
w
ww
ww
w
(2)
221
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.148 (Continued)
and Equation (19.41), c m
mc
c
k
c
m
c
c
c n
n
f f
n c
f
n
=
=
= =ÊËÁ
ˆ¯
ÊËÁ
ˆ¯
2
2
22
ww
w w
w
ww
Substituting in Eq. (2), FP
m
mcc
cc
c
f
n
f
n c
f
n
=+ ( )( )È
Î͢˚
- ( )ÈÎÍ
˘˚
+ ( )( )ÈÎÍ
1 2
1 2
2
2 2
ww
ww
ww
˘˚
2 Q.E.D. �
222
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PROBLEM 19.149
A 100-kg machine element supported by four springs, each of constant k = 200 N/m, is subjected to a periodic force of frequency 0.8 Hz and amplitude 100 N. Determine the amplitude of the fluctuating force transmitted to the foundation if (a) a dashpot with a coefficient of damping c = 420 N · s/m is connected to the machine element and to the ground, (b) the dashpot is removed.
SOLUTION
Forcing frequency: w p p pf ff= = =2 2 0 8 1 6( )( . ) . rad/s
Exciting force amplitude: Pm = 100 N
Mass: m = 100 kg
Equivalent spring constant: k = =( )( )4 200 800N/m N/m
Natural frequency: wnk
m=
=
800
1002 8284.
Frequency ratio: ww
pf
n
=
=
1 6
2 8284
1 77715
.
.
.
Critical damping coefficient: c kmc =
== ◊
2
2 800 100
565 685
( )( )
. N s/m
From the derivation given in Problem 19.148, the amplitude of the force transmitted to the foundation is
FP
m
mcc
cc
c
f
n
f
n c
f
n
=+ ( )( )È
Î͢˚
- ( )ÈÎÍ
˘˚
+ ( )( )ÈÎÍ
1 2
1 2
2
2 2
ww
ww
ww
˘˚
-ÊËÁ
ˆ¯
= - = -
2
221 1 1 77715 2 1583
ww
f
n
( . ) .
(1)
223
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PROBLEM 19.149 (Continued)
(a) Fm when c = ◊420 N s/m: c
c
c
c
c
c
f
n
=
=
ÊËÁ
ˆ¯
ÊËÁ
ˆ¯
=
420
565 685
0 74263
2 2 0 74263 1
.
.
( )( . )( .ww
777715
2 63894
)
.=
From Eq. (1): Fm =+
- +
100 1 2 63894
2 1583 2 63894
2
2 2
( . )
( . ) ( . )
= 100 7 964
11 6223
.
. Fm = 82 8. N �
(b) Fm when c = 0: FP
mm
f
n
=- ( )
=1
100
2 15832ww
. Fm = 46 3. N �
224
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.150*
For a steady-state vibration with damping under a harmonic force, show that the mechanical energy dissipated per cycle by the dashpot is E cxm f= p w2 , where c is the coefficient of damping, xm is the amplitude of the motion, and w f is the circular frequency of the harmonic force.
SOLUTION
Energy is dissipated by the dashpot.
From Equation (19.48), the deflection of the system is
x x tm f= -sin( )w j
The force on the dashpot. F cx
F cx tD
D m f f
== -&
w w jcos( )
The work done in a complete cycle with
c
E F dx
dx x t
ff
d
m f f
f
=
= ¥
= -
Ú
2
0
2
pw
w w
p w/ i.e., force distance( )
cos( jj
w w j
w jw
p w
)
cos ( )
cos ( )[ cos(
dt
E cx t dt
tt
m f f
Df
f= -
- =- -
Ú 2 2 2
0
2
2 1 2
/
ff
ww f
w w
p w
)]
cos( )
sin(
21 2
2
2
2
2 2
0
2
2 2
E cxt
dt
Ecx
tt
m ff
m f f
f=- -
= --
Ú/
ffw
p w)
f
fÈ
ÎÍÍ
˘
˚˙˙0
2 /
Ecxm f
f f
= - - -È
ÎÍÍ
˘
˚˙˙
2 2
2
2 22
w pw w
p j j(sin( ) sin )0
E cxm f= p w2 Q.E.D. �
225
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.151*
The suspension of an automobile can be approximated by the simplified spring-and-dashpot system shown. (a) Write the differential equation defining the vertical displacement of the mass m when the system moves at a speed v over a road with a sinusoidal cross section of amplitude dm and wavelength L. (b) Derive an expression for the amplitude of the vertical displacement of the mass m.
SOLUTION
(a)
+ SF ma W k x cdx
dt
d
dtm
d x
dt= - + - - -Ê
ËÁˆ¯
=: ( )d d dST
2
2
Recalling that W k= dST, we write
md x
dtc
dx
dtkx k c
d
dt
2
2+ + = +d d
(1)
Motion of wheel is a sine curve, d d w= m f tsin . The interval of time needed to travel a distance L at a
speed v is tL
v= , which is the period of the road surface.
Thus, w pt
p pf
fLv
v
L= = =2 2 2
and d d w d d pw= =m f
mLv
ftd
dttsin cos
2
Thus, Equation (1) is
md x
dtc
dx
dtkx k t c tf f f m
2
+ + = +( sin cos )w w w d �
226
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PROBLEM 19.151* (Continued)
(b) From the identity A y B y A B y
B
A BA
A B
sin cos sin( )
sin
cos
+ = + +
=+
=+
2 2
2 2
2 2
y
j
j
We can write the differential equation
md x
dtc
dx
dtkx k c t
c
k
m f f
f
2
22 2
1
+ + = + +
= -
d w w y
yw
( ) sin( )
tan
The solution to this equation is analogous to Equations 19.47 and 19.48, with
P k cm m f= +d w2 2( )
x x tm f= - +sin( )w j y (where analogous to Equations (19.52)) �
xk c
k m cm
f
f f
=+
-( ) +
d w
w w
2 2
2 2 2
( )
( ) �
tanjw
w=
-
c
k m
f
f2
�
tanyw
=c
kf
�
227
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PROBLEM 19.152*
Two blocks A and B, each of mass m, are supported as shown by three springs of the same constant k. Blocks A and B are connected by a dashpot, and block B is connected to the ground by two dashpots, each dashpot having the same coefficient of damping c. Block A is subjected to a force of magnitude P = P tm fsin .w Write the differential equations defining the displacements xA and xB of the two blocks from their equilibrium positions.
SOLUTION
Since the origins of coordinates are chosen from the equilibrium position, we may omit the initial spring compressions and the effect of gravity
For load A,
+ SF ma P t k x x c x x mxA m f B A B A A= + - + - =: sin ( ) ( )w 2 & & && (1)
For load B,
+ SF ma k x x c x x kx cx mxB B A B A B B B= - - - - - - =: 2 2( ) ( )& & & && (2)
Rearranging Equations (1) and (2), we find: mx c x x k x x P tA A B A B m f&& & &+ - + - =( ) ( ) sin2 w �
mx cx cx kx kxB B A B A&& & &+ - + - =3 3 2 0 �
228
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PROBLEM 19.153
Express in terms of L, C, and E the range of values of the resistance R for which oscillations will take place in the circuit shown when switch S is closed.
SOLUTION
For a mechanical system, oscillations take place if c cc� (lightly damped).
But from Equation (19.41), c mk
mkmc = =2 2
Therefore, c km� 2 (1)
From Table 19.2: c ® R
m ® L
k ® 1
C (2)
Substituting in Eq. (1) the analogous electrical values in Eq. (2), we find that oscillations will take place if
RC
L� 21Ê
ËÁˆ¯
( ) RL
C� 2 �
229
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.154
Consider the circuit of Problem 19.153 when the capacitor C is removed. If switch S is closed at time t = 0, determine (a) the final value of the current in the circuit, (b) the time t at which the current will have reached ( )1 1- /e times its final value. (The desired value of t is known as the time constant of the circuit.)
SOLUTION
Electrical system Mechanical system
The mechanical analogue of closing a switch S is the sudden application of a constant force of magnitude P to the mass.
(a) Final value of the current corresponds to the final velocity of the mass, and since the capacitance is zero, the spring constant is also zero
+ SF ma P Cdx
dtm
d x
dt= - =:
2
2 (1)
Final velocity occurs when d x
dt
P Cdx
dt
dx
dtv
2
20
0
=
- = =final final
final
vP
Cfinal =
From Table 19.2: v ® L, P ® E, C ® R
Thus, LE
Rfinal = �
230
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.154 (Continued)
(b) Rearranging Equation (1), we have
md x
dtC
dx
dtP
2
2+ =
Substitute dx
dtAe
P
C
d x
dtA et t= + = -- -l ll;
2
m A e C AeP
CPt t-ÈÎ ˘ + +È
Î͢˚
=- -l l l
- + = =m Cc
ml l0
Thus, dx
dtAe
p
cc m t= +-( )/
At t = 0, dx
dtA
P
CA
P
C= = + = -0 0
vdx
dt
p
ce c m t= = -ÈÎ ˘-1 ( )/
From Table 19.2: v ® c, p ® E, c ® R, m ® L
LE
Re R L t= -ÈÎ ˘-1 ( )/
For LE
R e= Ê
ËÁˆ¯
-ÊËÁ
ˆ¯
11
, R
LtÊ
ËÁˆ¯
= 1 tL
R= �
231
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PROBLEM 19.155
Draw the electrical analogue of the mechanical system shown. (Hint: Draw the loops corresponding to the free bodies m and A.)
SOLUTION
We note that both the spring and the dashpot affect the motion of Point A. Thus, one loop in the electrical circuit should consist of a capacitor k cfi( )1 and a resistance ( ).c Rfi
The other loop consists of ( sinP tm fw ®E tm fsin ),w an inductor (m®L) and the resistor (c ®R).
Since the resistor is common to both loops, the circuit is
�
232
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PROBLEM 19.156
Draw the electrical analogue of the mechanical system shown. (Hint: Draw the loops corresponding to the free bodies m and A.)
SOLUTION
Loop 1 (Point A): k1 ® 1
12c
k, ® 1
21c
c, ® R1
Loop 2 (Mass m): k2 ® 1
2cm, ® L c, 2 ® R2
With k2 ® 12c common to both loops,
�
233
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.157
Write the differential equations defining (a) the displacements of the mass m and of the Point A, (b) the charges on the capacitors of the electrical analogue.
SOLUTION
(a) Mechanical system.
Point A: + SF = 0: cd
dtx x kxA m A( )- + = 0 �
Mass m:
+ SF ma cd
dtx x P t m
d x
dtm A m f
m= - - = -: ( ) sinw2
2
md x
dtc
d
dtx x P tm
m A m f
2
2+ - =( ) sinw �
(b) Electrical analogue.
From Table 19.2:
m ® L
c ® R
k ® 1
Cx ® q
P ® E
234
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.157 (Continued)
Substituting into the results from Part (a), the analogous electrical characteristics,
Rd
dtq q
CqA m n( )- + Ê
ËÁˆ¯
=10 �
Ld q
dtR
d
dtq q E tm
m A m f
2
2+ - =( ) sinw �
Note: These equations can also be obtained by summing the voltage drops around the loops in the circuit of Problem 19.155.
235
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.158
Write the differential equations defining (a) the displacements of the mass m and of the Point A, (b) the charges on the capacitors of the electrical analogue.
SOLUTION
(a) Mechanical system.
+ SF = 0
Point A: k x cdx
dtk x xA
AA m1 1 2 0+ + - =( ) c
dx
dtk k x k xA
A m1 1 2 2 0+ + - =( ) �
Mass m: + SF ma k x x cdx
dtm
d x
dtA m
m m= - - =: ( )2 2
2
2 m
d x
dtc
dx
dtk x xm m
m A
2
2 2 2 0+ + - =( ) �
(b) Electrical analogue.
Substituting into the results from Part (a) using the analogous electrical characteristics from Table 19.2 (see left),
Rdq
dt C Cq
CqA
A m11 2 2
1 1 10+ +
ÊËÁ
ˆ¯
- = �
Ld q
dtR
dq
dt Cq qm m
m A
2
2 22
10+ + - =( ) �
236
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.159
A thin square plate of side a can oscillate about an axis AB located at a distance b from its mass center G. (a) Determine the period of small oscillations if b a= 1
2 . (b) Determine a second value of b for which the period of small oscillations is the same as that found in Part a.
SOLUTION
Let the plate be rotated through angle q about the axis as shown.
+ S SM M mgb I ma bAB AB t= = - -( ) sin ( )( )eff : q a
Kinematics: a q= &&
a b bt = =ª
a qq
&&sin 0
Moment of inertia: I ma= 1
122
Then ( )I mb mgb
a b gb
gb
a b
+ + =
+ÊËÁ
ˆ¯
+ =
++
=
2
2 2
2 2
0
1
120
12
120
&&
&&
&&
q q
q q
q q
Natural circular frequency: wngb
a b=
+12
122 2
(a) b a= 1
2: wn
ga
a a
g
a=
+=16
3
3
22 2
Period of vibration: t pwn
n
= 2 t pn
a
g= 2
2
3 �
237
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.159 (Continued)
(b) Another value of b giving the same period:
wngb
a b
g
a
b
a b a
ab a b
=+
=
+=
= +
12
12
3
2
12
12
3
2
24 3 36
2 2
2 2
2 2
36 24 3 0
0 5 0 16667
2 2b ab a
b a b a
- + == =. .and b a= 0 1667. �
238
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.160
A 150-kg electromagnet is at rest and is holding 100 kg of scrap steel when the current is turned off and the steel is dropped. Knowing that the cable and the supporting crane have a total stiffness equivalent to a spring of constant 200 kN/m, determine (a) the frequency, the amplitude, and the maximum velocity of the resulting motion, (b) the minimum tension which will occur in the cable during the motion, (c) the velocity of the magnet 0.03 s after the current is turned off.
SOLUTION
Data: m m k1 23150 100 200 10= = = ¥ kg kg N/m
From the first two sketches, T kx m m gm0 1 2+ = +( ) (1)
T m g0 1= (2)
Subtracting Eq. (2) from Eq. (1), kx m gm = 2
xm g
km = =¥
= ¥ =-23
3100 9 81
200 104 905 10
( )( . ). m 4.91 mm
Natural circular frequency: wnk
m= = ¥ =
1
3200 10
15036 515. rad/s
Natural frequency: f fnn
n= = =w
p p2
36 515
25 81
.. Hz
Maximum velocity: v xm n m= = ¥ =-w ( . )( . ) .36 515 4 905 10 0 17913 m/s
(a) Resulting motion: amplitude mmxm = 4 91. �
frequency Hzfn = 5 81. �
maximum velocity m/svm = 0 1791. �
239
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.160 (Continued)
(b) Minimum value of tension occurs when x xm= - .
T T kx
m g m g
m m g
mmin
( )
( )( . )
= -= -= -=
0
1 2
1 2
50 9 81 Tmin = 491 N �
The motion is given by x x t
x x tm n
n m n
= += +
sin( )
cos( )
w jw w j&
Initally, x x
x
x x t
m
n m n
0
0
1
0
2
2
= - = -= =
= -
= -ÊËÁ
ˆ¯
or sin
or cos 0
jj
j p
w w p
&
& cos
(c) Velocity at s.t = 0 03.
ww j
w j
n
n
n
t
t
t
= =- = -
-
( . )( . ) .
.
)
36 515 0 03 1 09545
0 47535
rad
rad
cos( == 0 88913.
&x = ¥ -( . )( . )( . )36 515 4 905 10 0 889133 &x = 0 1592. m/s� �
240
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.161
Disks A and B weigh 15 kg and 6 kg, respectively, and a small 2.5-kg block C is attached to the rim of disk B. Assuming that no slipping occurs between the disks, determine the period of small oscillations of the system.
SOLUTION
Small oscillations: h rr
B mB B= - ª( cos )12
2
Positionj r rB B A A& &q q=
T m r I Ir
r
Im r
Im r
C B m B m AB
Am
BB B
AA
12 2
2
2
1
2
1
2
1
2
2
= + +ÊËÁ
ˆ¯
=
=
( )& & &q q q
AA
C BB B A A B
AmT m r
m r m r r
r
T
2
12
2 2 22
2
1
2 2 2= + + Ê
ËÁˆ¯
ÊËÁ
ˆ¯
È
ÎÍÍ
˘
˚˙˙&q
112 2
1
1
2 2 20
= + +ÊËÁ
ˆ¯
ÈÎÍ
˘˚
=
mm m
r
V
CB A
B m&q
241
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.161 (Continued)
Positionk T e
V m gh
m r
C
C m
2
2
2
0
2
==
=&q
Conservation of energy and simple harmonic motion.
T V T V
mm m
rm
m n m
CB A
B n m
1 1 2 2
2 2 21
2 2 20 0
+ = +
=
+ +ÊËÁ
ˆ¯
ÈÎÍ
˘˚
+ = +
&q w q
w q CC B m
nC
CB A B
n
gr
m
mm m
g
r
q
w
w
2
2
2
2
2
2 5 9 81
2 515
=+ +È
Î͢˚
=+ +
( )
( . )( . )
.( 66
20 15
12 57692 2
)( . )
.
ÈÎÍ
˘˚
= -wn s
Period of small oscillations. t pw
pn
n
= =2 2
12 5769. t n = 1 772. s �
242
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.162
A period of 6.00 s is observed for the angular oscillations of a 120-g gyroscope rotor suspended from a wire as shown. Knowing that a period of 3.80 s is obtained when a 30 mm-diameter steel sphere is suspended in the same fashion, determine the centroidal radius of gyration of the rotor. (Density of steel = 7800 kg/m3.)
SOLUTION
+ S SM M K I= - =( ) :eff q q&&
&&q q
w
t p
+ =
=
=
K
IK
I
I
K
n
0
2
2
(1)
KI= 4 2
2
pt
(2)
IK= t
p
2
24 (3)
For the sphere, rd= = =2
30
215 mm = 0.015 m
Volume: V rs = =
= ¥ -
4
3
4
30 015
1 41372 10
3 3
5
p p ( . )
. m3
Mass: m SVs s=
= ¥=
-( )( . )
.
7800 1 41372 10
0 11027
5kg/m
kg
3
Moment of inertia: I m rs= =
= ¥ -
2
5
2
50 11027 0 015
9 9243 10
2 2
6
( . )( . )
. kgm2
Period: t s = 3 80. s
243
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.162 (Continued)
From Eq. (2): K =¥
= ¥ ◊
-
-
4 9 9243 10
3 80
2 7133 10
2 6
2
5
p ( . )
( . )
. N m/rad
For the rotor, m = ==
120 0 12
6 00
g kg
s
.
.t
From Eq. (3): I = ¥
= ¥ ◊
-
-
( . )( . )
.
2 7133 106 00
4
2 4742 10
52
2
5
pkg m2
Radius of gyration. I mk= 2
kI
m=
= ¥
=
-2 4742 10
0 120 014359
5.
.. m k = 14 36. mm �
244
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.163
A 1.5-kg block B is connected by a cord to a 2-kg block A, which is suspended from a spring of constant 3 kN/m. Knowing that the system is at rest when the cord is cut, determine (a) the frequency, the amplitude, and the maximum velocity of the resulting motion, (b) the minimum tension that will occur in the spring during the motion, (c) the velocity of block A 0.3 s after the cord has been cut.
SOLUTION
Before the cord is cut, the tension in the spring is
T m m gA B0
2 1 5 9 81
34 335
= += +=
( )
( . )( . )
. N
The elongation of the spring is eT
k00
3
3
34 335
3 10
11 445 10
=¢
=¥
= ¥ -
.
. m
After the cord is cut, the tension in the equilibrium position is
¢ ===
T m gA0
2 0 9 81
19 62
( . )( . )
. N
The corresponding elongation is ¢ =¢
=¥
= ¥ -
eT
k00
3
3
19 62
3 10
6 54 10
.
. m
Let x be measured downward from the equilibrium position.
T T kx= ¢ +0
(a) Newton’s Second Law after the cord is cut.mx m g T
m x m g kx T kx
m x kx
k
m
A
A A
A
nA
&&&&
&&
= -+ - - = -
+ =
= = ¥ =
0
3
0
3 10
238 72w . 998
2
38 7298
2
rad/s
fnn= =
wp p
. fn = 6 16. Hz �
245
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.163 (Continued)
The resulting motion is x x t
x xm n
n m t
= += +
sin ( )
cos( )
w jw w j&
Initial condition: x e e
x
x
xm
n m
0 0 0
3
0
3
4 905 10
0
0 75375 10
0
= - ¢
= ¥=
¥ ==
-
-
.
. sin
cos
m
&j
w j
j p=
=24 905xm . m xm = 4 91. mm �
x t
x t
n
n
= ¥ +ÊËÁ
ˆ¯
= +ÊËÁ
ˆ¯
-4 905 102
0 189972
3. sin ( )
. cos
w p
w p
m
(m/& ss) &xm = 0 1900. m/s �
(b) Minimum tension occurs when x is minimum.
T T kxmmin
. ( )( . )
= ¢ -
= - ¥ ¥ -0
3 319 62 3 10 4 905 10 Tmin .= 4 91 N �
(c) Velocity when t = 0.3 s.
w j pnt
x
- = +
==
( . )( . )
.
( . )cos( .
38 7298 0 32
13 1897
0 18997 13 1
radians
& 8897
0 18997 0 8119
)
( . )( . )= &x = 0 1542. m/s ¯ �
246
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PROBLEM 19.164
Two rods, each of mass m and length L, are welded together to form the assembly shown. Determine (a) the distance b for which the frequency of small oscillations of the assembly is maximum, (b) the corresponding maximum frequency.
SOLUTION
Positionj V1 0=
T mv mv I I
v b
vL
CD AB CD m AB m
CD m
AB
12 2 2 21
2
2
= + + +ÈÎ ˘
=
= ÊËÁ
ˆ¯
& &
&
&
q q
q
qmm
CD AB
m
I I
mL
T m bL
L L
=
=
= + ÊËÁ
ˆ¯ + +
È
ÎÍÍ
˘
˚˙˙
1
12
1
2 2
1
12
1
12
2
12
22 2 2&q
== +Ê
ËÁˆ
¯m
bL
m2
5
122
22&q
247
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PROBLEM 19.164 (Continued)
Positionk V mgb mgL
m m2 12
1= - + -( cos ) ( cos )q q
Small angles: 1 22 2
2 20
22
2
2
2
- = ÊËÁ
ˆ¯ ª
= +ÊËÁ
ˆ¯
=
cos sinq q aq
q
mm m
mV mg bL
T
Conservation of energy and simple harmonic motion.
T V T V
m b Lmg
bL
m m
m
1 1 2 2
2 2 2 21
2
5
120 0
2 2
+ = +
+ÈÎÍ
˘˚
+ = + +ÈÎÍ
˘˚
=
&
&q q
q wnn m
n
Lg b
b L
q
w2 22 5
122
=+( )
+( ) (1)
(a) Distance b for maximum frequency.
Maximum whenww
nnd
db2
2
0=
d
db
b L g g b b
b L
b Lb
nLw2 2 5
122
2
2 512
2 2
2
20
5
12
=+( ) - +( )
+( )=
- - + ÊËÁ
ˆ¯
( )
˜ =
=- + ( )
=
L
bL L L
L L
2
2 2012
2
0
20 316 1 317
m. , . b L= 0 316. �
(b) Corresponding maximum frequency.
From Eq. (1) and the answer to Part (a):
w
wp p
n
nn
g
L
g
L
fg
L
22 5
12
0 316 0 5
0 316
1 580
2
1 580
2
= ++ÈÎ ˘
=
= =
[ . . ]
( . )
.
.
== 0 200.g
LHz �
248
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.165
As the rotating speed of a spring-supported motor is slowly increased from 200 to 500 rpm, the amplitude of the vibration due to the unbalance of the rotor is observed to decrease steadily from 8 mm to 2.5 mm. Determine (a) the speed at which resonance would occur, (b) the amplitude of the steady-state vibration at a speed of 100 rpm.
SOLUTION
Since the amplitude decreases with increasing speed over the range w1 200= rpm to w2 500= rpm, the motion is out of phase with the force.
xm
Pk
mr
k
mrM
m
f
n
f
f
n
f
n
f
n
=- ( )
=- ( )
=( )( )
- ( )1 1 12 2
2
2
2
ww
w
ww
ww
ww
| |xm
mrM
f
n
f
n
=( )( )( ) -
ww
ww
2
21
Let un
=ww
1 , where w1 200 20 944= =rpm rad/s. .
| |xu
um
mrM
1
2
2 1=
( )-
(1)
At w w
ww
f
f
n
u u
= =
= =
2 500
500
2002 5
rpm
.
| |( . )
.x
u
um
mrM
2
2
2
6 25
6 25 1=
( )-
(2)
Dividing Eq. (1) by Eq. (2),
| |
| |
( . )
( )( . ).
.
x
x
u u
u u
u u
m
m
1
2
2 2
2 2
4
6 25 1
1 6 25
83 2
6 25
= --
= =
-
mm
2.5 mm22 4 2
4 2
20 20
13 75 19 019
13 751 17551
= -
- = = =
u u
u u u..
.
(a) Natural frequency. ww
n u= = =1 20 944
1 1755117 817
.
.. rad/s wn = 170 1. rpm �
249
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 19.165 (Continued)
At 200 rpm, | | .x um 138 8 10 1 17551= = ¥ =-mm m,
From Eq. (1): 8 101 17551
1 17551 1
19
5 25
2 2105 10
32
2¥ =
( )-
=
= ¥
-
-
mrM
mr
Mmr
M
( . )
( . )
.
. 33 m
(b) Amplitude at 100 rpm. w w
ww
f n
f
n
=
= =
10 472
10 472
17 8170 58775
.
.
..
rad/s �
The motion is in phase with the force.
x
x
m
mrM
m
f
n
f
n
=( )( )
- ( )= ¥
-
-
ww
ww
2
2
3 2
1
2 2105 10 0 58775
1 0 587
( . )( . )
( . 775
1 167 10
2
3
)
.= ¥ - m xm = 1 167. mm �
250
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PROBLEM 19.166
The compressor shown has a mass of 250 kg and operates at 2000 rpm. At this operating condition, undesirable vibration occurs when the compressor is attached directly to the ground. To reduce the vibration of the concrete floor that is resting on clay soil, it is proposed to isolate the compressor by mounting it on a square concrete block separated from the rest of the floor as shown. The density of concrete is 2400 kg/m3 and the spring constant for the soil is found to be 80 106¥ N/m. The geometry of the compressor leads to choosing a block that is 1.5 m by 1.5 m. Determine the depth h that will reduce the force transmitted to the ground by 75%.
SOLUTION
Forced circular frequency corresponding to 2000 rpm:
w pf = =( )( )
.2 2000
60209 44 rad/s
Natural circular frequency for compressor attached directly to the ground:
w11
680 10565 68= =
¥=k
m
N/m
250 kgrad/s.
Let P be the force due to the compressor unbalance and F1 be the force transmitted to the ground.
F kx kP
F
P
Pk
f
n
f
n
f
n
1 1 2 2
12 209 44
565
1 1
1
1
1
1
= =- ( )
=- ( )
=- ( )
=-
ww
ww
ww
.
.668
21 15951
( )= .
251
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PROBLEM 19.166 (Continued)
Modify the system by making the mass much larger so that the natural frequency of modified system is much less than the forcing frequency. Let w2 be the new natural circular frequency. For the ground force to be reduced by 75%,
F
P
F
P f
n
f
2 12
2
2
0 25 0 25 1 15951 0 289881
1= = = =
( ) -
ÊËÁ
ˆ¯
=
. ( . )( . ) .ww
ww
111
0 289884 4497 2 109
4 4497
209 44
4 4497
2
22
2 2
+ = =
= = =
.. .
.
( . )
.
ww
ww
f
f 99 8579 10
80 10
3 2
22
2
222
6
. ( )
(
¥
=
= =¥¥
rad/s
N/m
9.8579 10 rad/3
w
w
k
m
mk
ss)kg
2= 8115
Required properties of attached concrete block:
mass kg
volumemass
density
kg
kg/
= - = - =
= =
m m2 1 8115 250 7865
7865
2400 mmm
area m m m
depthvolume
area
m
33
2
3
=
= ¥ =
= =
3 277
1 5 1 5 2 25
3 277
2
.
. . .
.
.2251 456
mm
2= . h = 1 456. m �
252
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PROBLEM 19.167
If either a simple or a compound pendulum is used to determine experimentally the acceleration of gravity g, difficulties are encountered. In the case of the simple pendulum, the string is not truly weightless, while in the case of the compound pendulum, the exact location of the mass center is difficult to establish. In the case of a compound pendulum, the difficulty can be eliminated by using a reversible, or Kater, pendulum. Two knife edges A and B are placed so that they are obviously not at the same distance from the mass center G, and the distance l is measured with great precision. The position of a counterweight D is then adjusted so that the period of oscillation t is the same when either knife edge is used. Show that the period t obtained is equal to that of a true simple pendulum of length l and that g = 4p 2l/t 2.
SOLUTION
From Problem 19.52, the length of an equivalent simple pendulum is:
l rk
rA = +2
and l Rk
RB = +2
But t t
p p
A B
A Bl
g
l
g
=
=2 2
Thus, l lA B=
For l l
rk
rR
k
R
r R k R r R k r
r R r R k r R
r R
A B=
+ = +
+ = +
- = -- =
2 2
2 2 2 2
2
0
[ ] [ ]
( )
Thus, r R k= 2
or rk
R
Rk
r
=
=
2
2
253
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PROBLEM 19.167 (Continued)
Thus, AG GA BG GB= ¢ = ¢and
That is, A A B B= ¢ = ¢and
Noting that l l l
l
g
A B= =
=t p2
or gl= 4 2
2
pt
254
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PROBLEM 19.168
A 400-kg motor supported by four springs, each of constant 150 kN/m, is constrained to move vertically. Knowing that the unbalance of the rotor is equivalent to a 23-g mass located at a distance of 100 mm from the axis of rotation, determine for a speed of 800 rpm (a) the amplitude of the fluctuating force transmitted to the foundation, (b) the amplitude of the vertical motion of the motor.
SOLUTION
Total mass: M = 400 kg
Unbalance: ¢ = == =
m
r
23 0 023
100 0 100
g kg
mm m
.
.
Forcing frequency: w f =
=
800
83 776
rpm
rad/s.
Spring constant: k = ¥
= ¥
( )( )4 150 10
600 10
3
3
N/m
N/m
Natural frequency: wnk
m=
= ¥
=
600 10
40038 730
3
. rad/s
Frequency ratio: ww
f
n
= 2 1631.
Unbalance force: P m rm f= ¢
==
w2
20 023 0 100 83 776
16 1424
( . )( . )( . )
. N
Static deflection: P
km =
¥= ¥ -
16 1424
600 10
26 904 10
3
6
.
. m
255
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PROBLEM 19.168 (Continued)
Amplitude of vibration. From Eq. (19.33):
xm
Pkm
f
n
=- ( )
= ¥-
= - ¥
-
-
1
26 904 10
1 2 1631
7 313 10
2
6
2
6
ww
.
( . )
. m
(a) Transmitted force. F kxm m= - = - ¥ - ¥ -( )( . )600 10 7 313 103 6 Fm = 4 39. N �
(b) Amplitude of motion. | | .xm = ¥ -7 313 10 6 m | | .xm = 0 00731 mm �
256
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PROBLEM 19.169
Solve Problem 19.168, assuming that a dashpot of constant c = ◊6500 N s/m is introduced between the motor and the ground.
PROBLEM 19.168 A 400-kg motor supported by four springs, each of constant 150 kN/m, is constrained to move vertically. Knowing that the unbalance of the rotor is equivalent to a 23-g mass located at a distance of 100 mm from the axis of rotation, determine for a speed of 800 rpm (a) the amplitude of the fluctuating force transmitted to the foundation, (b) the amplitude of the vertical motion of the motor.
SOLUTION
Total mass: M = 400 kg
Unbalance: m
r
= == =
23 0 023
100 0 100
g kg
mm m
.
.
Forcing frequency: w f =
=
800
83 776
rpm
rad/s.
Spring constant: ( )( )4 150 10 600 103 3¥ = ¥N/m N/m
Natural frequency: wnk
m= = ¥
=
600 10
40038 730
3
. rad/s
Frequency ratio: ww
f
n
= 2 1631.
Viscous damping coefficient: c = ◊6500 N s/m
Critical damping coefficient: c kMc = = ¥= ◊
2 2 600 10 400
30 984
3( )( )
, N s/m
Damping factor: c
cc
= 0 20978.
Unbalance force: P mrm f= =
=
w2 20 023 0 100 83 776
16 1424
( . )( . )( . )
. N
Static deflection: dST
m
= =¥
= ¥ -
P
km 16 1424
600 10
26 904 10
3
6
.
.
257
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PROBLEM 19.169 (Continued)
Amplitude of vibration. Use Eq. (19.53).
xm
Pk
cc
m
f
n c
f
n
=
- ( )ÈÎÍ
˘˚
+ ( )( )ÈÎÍ
˘˚
1 1 22 2 2
ww
ww
where 1 1 2 1631 3 6792
2-ÊËÁ
ˆ¯
= - = -ww
f
n
( . ) .
and 2 2 0 20978 2 1631 0 90755
26 904
c
c
x
c
f
n
m
ÊËÁ
ˆ¯
ÊËÁ
ˆ¯
= =
=
ww
( )( . )( . ) .
. ¥¥
- +
= ¥
-
-
10
3 679 0 90755
7 1000 10
3
2 2
6
( . ) ( . )
. m
Resulting motion: x x t
x x t
m f
f m f
= -
= -
sin ( )
cos( )
w j
w w j&Spring force: F kx kx t ts m f f= = - = -sin ( ) .w j w j4 26 sin ( )
Damping force: F cx c x t td f m f f= = - = -& w w j w jcos( ) . cos( )3 8663
Let F F t F F ts m f d m f= - = -cos sin ( ) sin cos( )y w j y w jand
Total force: F F t F t
F t
m f m f
m f
= - + -
= - +
cos sin ( ) sin cos( )
sin ( )
y w j y w j
w j y
(a) Force amplitude. F F F
F kx c x
m m m
m m f m
= +
= +
( cos ) ( sin )
( ) ( )
y y
w
2 2
2 2
= +( . ) ( . )4 26 3 86632 2 Fm = 5 75. N �
(b) Amplitude of vibration. xm = 0 00710. mm �
258
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PROBLEM 19.170
A small ball of mass m attached at the midpoint of a tightly stretched elastic cord of length l can slide on a horizontal plane. The ball is given a small displacement in a direction perpendicular to the cord and released. Assuming the tension T in the cord to remain constant, (a) write the differential equation of motion of the ball, (b) determine the period of vibration.
SOLUTION
(a) Differential equation of motion.
+ SF ma T mx= = -: sin2 q &&
For small x, sin tanq qª = ( ) =x x
ll2
2
mx Tx
l&&+ Ê
ËÁˆ¯ =( )2
20 �
mxT
lx&&+ Ê
ËÁˆ¯
=40
Natural circular frequency. w
w
n
n
T
ml
T
ml
2 4
2
=
=
(b) Period of vibration. t pwn
n
= 2 t pn
ml
T= �