chapter 1 & 2 - introduction to vibrations and single dof systems (1)

7/30/2019 Chapter 1 & 2 - Introduction to Vibrations and Single DOF Systems (1)

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There are two general classes of vibrations - free

and forced.

Free vibration

takes place when a system oscillates under theaction of forces inherent in the system itself or

after an initial disturbance.

Example : oscillation of a simple pendulum

Forced vibration

Vibration that takes place under the excitation of

external forces.

Often a repeating type of force, e.g. the oscillation

arises in machines such as diesel engine.


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Natural frequency

Property of the dynamic system established

by its mass(m) and stiffness(k).

Resonance

A condition whereby the frequency of the

external force coincides with one of the

natural frequencies of the system. The system undergoes a dangerously large

oscillation.

Causing failure of structures such as bridges,

buildings, turbines and airplane wing etc.


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Degree of freedom (DOF)

The number of independent coordinates

required to describe the motion of the system.

Free particle undergoing general motion in

space will have 3 DOF.

Rigid body, will have six degrees of freedom,

i.e. 3 components of position and angles

defining its orientation.

Continuous elastic body will require an infinite

number of coordinates to describe motion.


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Period motion the motion is repeated in equal

intervals of time, T.

Periodthe repetition time, tof the oscillation and

its reciprocal. Frequency, f

If the motion is designated by the time function x(t),

1f

txtx


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Harmonic motion is often represented as the

projection on a straight line of a point that is

moving on a circle at constant speed. With the angular speed of the line O-P designated

by , the displacementxcan be written as

)1(....sin tAx


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is generally measured in radians per second

and referred as angular frequency. The

relationship is

Where T= period(s), f= frequency(cycles/s) of theharmonic motion.

Differentiating the x in eq. (1) twice, the velocity

and acceleration of system

)2(....22

f


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Differentiating the x in eq. (1) twice, the velocityand acceleration of the system turned out as

)3(....)sin(cos 2

tAtAx

)4(....)sin(sin22

tAtAx


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Basic vibration model of a simple oscillatory

system consists ofa mass, a spring and a damper

or dashpot.

Spring force-deflection relationship is consideredto be linear following the Hooke's law,

The viscous damping generally representeddashpot, is described by force proportional to the

velocity,

NkxF

NxcF


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Spring-mass system and FBD


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a simple undamped spring-mass system.

One degree of freedom (DOF) as its described by a

single coordinatex.

Applying Newton's second law,

)5(....mgwk


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With x chosen to be positive downward, together with

all quantities, i.e. force, velocity, and acceleration.

Once again applying Newtons second law of motion;

Since , therefore

)( xkwFxm wk

)6(....kxxm


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Defining the circular frequency by the equation;

Eq. (6) can be written as

n

)7(....

m

kn

)8(....02

xx n


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Eq.(8), a homogeneous second order linear differential

equation has the following solution

Where A and B are the necessary constants andevaluated from initial conditions and

And eq.(9) is reduced to

)9(....cossin tBtAxnn

)0(x )0(x

)10(....cos)0(sin)0(

txtx

x nnn


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The naturalperiod of the oscillation is established from

And the natural frequency is

)11(....2 k

m

2n

)12(....211

mkfn


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These quantities can be expressed in terms of the static

deflection, by observing eq.(5), thus eq.(12)

can be expressed as

Note: depend only on the mass and stiffnessof the system.

)13(....2

1

gfn

mgk

nnf and,


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Viscous damping force is expressed by the equation

c is a constant of proportionality.

By adding the damper, the equation of motion

becomes

)14(....xcFd

)15(....)(tFkxxcxm


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Solution of this equation divided into two parts.

If the homogenous solution correspondsphysically to that offree-damped vibration.

If the solution of the system is added with

the particular solution that due to the excitation

irrespective of the homogenous solution.

0)( tF

0)( tF


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Particular solutionxp(t) orSteady-state Responsexss.

i. If F(t) = F0 = constant(step input), the we have

The solution is assumed in the same form as the input that

is

From which we obtain,

Applying the initial conditions

Finally, the total solution is

0Fkxxcxm

constant)( Ctxp

0)( KFCtxp

00 )0(and)0( xxxx

textex

tx dt

d

t

d

n nn

sinsin)( 00


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i. If F(t) = F0 sin(t) (sinusoidal input), then we have

We have

The solution is assumed in the form as the input but with

phase angle, that is

)sin(0 tFkxxcxm

)sin()( 0 tXtxp

)sin(0 tFkxxcxm


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Homogenous solution:

Then, assume a solution of the form;

Wheres is a constant. Substituting into the differentialequation eq.(16),

)16(....0 kxxcxm

)17(....stex

0)(2 stekcsms


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Satisfying all values oft;

Eq.(18) is a characteristic equation and has two roots,

solved by using the quadratic formula;

)18(....0

2

m

k

sm

c

s

)19(....22

2

2,1m

k

m

c

m

cs


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Hence the general solution is given by the equation:

WhereA andB are constants to be evaluated from the

initial condition and

Eq.(19) substitute into (20) gives:

)20(....21tsts

BeAex

)21(....))2/(())2/(()2(

22

tmkmctmkmctmc BeAeex

)0(x )0(x


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The first term , is simply an exponentially

decaying function of time.

As the damping term is larger than , the

exponents in the previous equation are real numbersand no oscillations are possible. This condition is

referred as overdamped.

when the damping term is less than , the

exponent becomes an imaginary number,

tmce

2

22mc mk

22mc mk

tmcmki 22


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Thus,

The terms of eq.(21) within the parentheses are

oscillatory. Hence, this is referred as underdamped.

In limiting case between the oscillatory and non-oscillatory motion where , the resultant

of the exponent is zero. Then, the corresponding case

is said as critical damping, cc.

tm

c

m

k

itm

c

m

k

e

tmcmki22

2

2sin2cos

2

mkmc 22


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Any damping can then be expressed in terms of the

critical damping by a non dimensional number zetha,called the damping ratio.

And

)22(....222 kmmm

kmc nc

)23(....

cc

c

)24(....22

nc

m

c

m

c


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i. Oscillatory motion Underdamped Case:

The frequency for damped oscillation is equal to:

)25(....22

11

titi

t nnn BeAeex

)26(....12 2 nd

d

)0.1(


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Damped oscillation )1(


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ii. Non oscillatory Motion Overdamped Case:

The motion is exponentially decreased with time as shown

below:

A periodic motion

)27(....11 22 tt nn

BeAex

)0.1(

)0.1(


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ii. Critically Damped Motion:

Three types of response with initial displacement is

shown in the figure below:

Critically Damped Motion

)28(....)( tneBtAx

)0.1(

)0(x

)0.1(


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Commonly created by the unbalance in rotating

machinery. It is less likely to occur compared to periodic

and other types of excitation.

Basically required in order to understand how the systemwill respond to more general types of excitation.

Considering a single DOF system with viscous damping

excited by a harmonic force . The equation of

motion becomes

)29(....sin tFkxxcxm o

tFo sin


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Viscously Damped System with Harmonic Excitation

Solution to this system, eq.(29) consists of two parts

Complementary function:- solution of thehomogeneous equation, which for this case is a

damped free vibration.

Particular integral particular solution of a steady-

state oscillation of the excitation.


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Particular solution:

WhereX- the amplitude of oscillation

- the phase of the displacement with respect to

the excitation force

By substituting eq.(30) into eq.(29), the amplitude and thephase are solved.

Recall that in the Harmonic Motion the phases of velocity

and acceleration are ahead of the displacement by 90 and

180 respectively.

)30(....)sin( tXx


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Vector Relationship for Forced Vibration with Damping

From the above diagram, it is clearly seen that

and

)31(....

222 cmk

FX o

)32(....tan2

1

mk

c


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Expressing eq.(31) and eq.(32) in non-dimensional term,

divide the numerator and denominator of eq.(31) and

eq.(32) by k, ones obtain:

And

)33(....

1

222

k

c

k

m

kF

X

o

)34(....

1

tan2

k

m

k

c


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Further expressing in terms of the following quantities:

oscilltionundampedoffrequencynaturalm

kn

dampingcriticalmc nc 2

factordampingc

c

c

n

c

c k

c

c

c

k

c

2


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The non-dimensional expression for phase and amplitude

then become:

and

)35(.....

21

1

222

nn

oF

Xk

2

1

2

tan

n

n


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Eq.(35) and eq.(36) indicate that the non-dimensional

amplitude , and the phase angle, are functions of

the frequency ratio, and the damping factor, can be

plotted as below:

oFXk

n


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For mechanical system as shown, derive its mathematical

model and determine its frequency of vibration in hertz if

k=100N/m and m=2kg. Obtain the displacementx if the initial

conditions are x(0) =0.05m and (0)=0 m/s. What is themaximum velocity and acceleration attained by the mass?

mk

x


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FBD:

Applying newtons second law, yields

or (mathematical model or equation of motion)

The frequency of the vibration is given by

which is the undamped natural frequency of the system and in the unit of hertz,

mkx

x

m m

x

: xmF xmkx )1(....0 kxxm

sradm

kn 07.7

2

100


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For the undamped free vibration, the displacement is given by

which is the complimentary orhomogeneous solution of the forced vibration.

Differentiating eq.(1) yields,

Substituting the initial conditions into eq.(1) and (2), we

obtain

Hzf nn 125.12

07.7

2

)1(....)sin( tCtx n

)2(....)cos( tCtx nn00 )0(and)0( xxxx

0sin0 xCx 0cos0 xCx n


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Solving for

Substituting the given initial conditions ,we have

and

Differentiating eq.(2), yields

2

0

2

0 )( nxxC

yieldsandC

001tan xx n smxmx /0)0(and05.0)0(

m05.0)07.70(05.0 22 A rad571.1007.705.0tan1

(2)....m)rad571.107.7sin(05.0)( ttx

(4)....m/s)rad571.107.7sin()07.7(05.0)()( 22 ttxta

(3)....m/s)rad571.107.7cos()07.7(05.0)()( ttxtv


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Notice that sine and cosine function oscillates between 1 to -1, therefore, the

maximum values occur when this function is equal 1 or -1. Hence, we

m/s3535.0)1)(07.7(05.0max v

22

max m/s5.2)1()07.7(05.0

a


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The logarithmic decrement , is introduced to determine the reduction trend of

two successive amplitudes. It is applied for underdamped system. An example

of decrement curve is as shown in the figure below:


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The ratio of two successive amplitudes ofa full cycle distant such as A1andA2are measured and computed as

The logarithmic decrement is defined as the natural logarithm of this ratio

For many systems, the first amplitude A0

is somehow not very clear and

because of that the general form is written as

)37(....)ln(2

1

A

A

)36(....)()(

2

1

1

1

2

1

dn

dn

n

ee

e

Ce

Ce

A

ATt

t

t

t

)38(....)ln(...)ln()ln(14

3

2

1

k

k

A

A

A

A

A

A


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From eq.(36), we get

which then rearrange into

if the value of is very small, then and eq.(40) simplifies to

)40(....4

22

)39(....1

2

1

22

22

n

n

d

ndn

)41(...2

112


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When value of is too small, the decrement of two successive amplitude is not

so significant and the amplitude ratio close to 1 where .

thus, in order to overcome this condition, we have to consider n number of

cycles for measuring the logarithmic decrement, the modified formula is given

as follow;

Where

)42(....1

2ln

2

n

A

An

nk

k

01ln

knk

k

AnAA

aftercyclesamplitudeanisamplitudeknownanyis


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When value of is too small, the decrement of two successive amplitude is not

so significant and the amplitude ratio close to 1 where .

thus, in order to overcome this condition, we have to consider n number of

cycles for measuring the logarithmic decrement, the modified formula is given

as follow;

Where

)42(....1

2ln

2

n

A

An

nk

k

01ln

knk

k

AnAA

aftercyclesamplitudeanisamplitudeknownanyis


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Given that the amplitude of free vibration is reduced by 65%

from its maximum value after 4 complete cycles, determine

for this system

a) The damping ratio, and

b) The corresponding value of the viscous damping

coefficient form = 5 kgand k = 0.5 kN/m.


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Let the maximum amplitude be A0, then after 4 complete

cycles (n = 4),A4 = 0.35A0 (since it reduces by 65%).

By using eq.(42);

With n = 4 and k = 0

nk

k

A

A

nln

1

262.0

35.0

1ln4

1

35.0ln4

1ln4

1

0

0

4

0

A

A

A

A


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Using eq.(40)

Since the value of is quite small, eq.(41) could also be

applied, from which we get

(b) Applying

0417.0262.04

262.0

22

0417.0)142.3(2

262.0

msNkmc /.17.4))5(5002)(0417.0(2


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Consider the mechanical system as shown in figure. If the

10-kg block is moved 0.16m to the right of the equilibrium

position released from rest ort = 0 s, determine

(a) The frequency of the resulting motion, and

(b) The displacement of the block at t = 0.3 s.

Given k = 40 N/m and that the force exerted by the dashpot

is 1.8 N when the speed of the block is 0.12 m/s

m

kx

c


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FBD: Kinetic Diagram:

Applying newtons second law, yields

or (mathematical model or equation of motion)

The frequency of the vibration is given by

mkx

x

m m

: xmF xmxckx )1(....0 kxxcxm

srad210

40

m

kn

c


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And the viscous damping coefficient is obtained from the relation

For the underdamped free vibration, the displacement is given by

where

ed)(underdamp375.0)10(402

15

2

km

c

N.s/m1512.0/8.1/ xFc

)1(....)sin()( tCetx d

tn

xcF

rad/s854.1375.0121 22 nd


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By differentiating eq.(1), we get

Substituting the initial condition into eq.(1) and

(2), we obtain

Solving for

)3(...sin)0( 0xCx

sin0xC

)2(....)cos()sin()( tCetCetx d

t

dd

t

nnn

00 )0()0( xxandxx

)4(...cossin)0( 0xCCx dn

yields,and C

nd xxx 0001tan

n

d

dn

dn

xx

x

xx

x

xxx

00

0

00

0

000

tan

tan

cossinsinsin)4(eq.intoCSubs.


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Substituting the given initial conditions

And the displacement in meters is

Hence, at time t = 0.3 s gives

m1358.0)186.1)3.0(854.1sin(1726.0)3.0()3.0(75.0 ex

1726.0)186.1sin(16.0 C

m/s0)0(m16.0)0( xandx

m)186.1854.1sin(1726.0)( 75.0 tetx t

rad186.1)2375.016.00(854.116.0tan 1


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Determine the natural frequency of the spring-mass system in bothrad/s and Hz. Given k = 2.5 kN/m and m = 20 kg.

m

k x


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For the spring-mass-damper system shown in figure below, determine

(a) The natural frequency in rad/s

(b) Damping ratio

Classify the system as underdamped, critically damped, oroverdamped.

Given k = 2.5 kN/m, m = 20 kg and c = 120 N/m/s

m

k x

c


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For the spring-mass-damper system shown in figure below, determine

(a) The natural frequency in rad/s

(b) The Damping ratio

Classify the system as underdamped, critically damped, oroverdamped.

Given k = 3 kN/m, m = 10 kg and c = 0.5 N/m/s

c

m

k

y


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An industrial press is mounted on a rubber pad to isolate it from itsfoundation. If the rubber pad is compressed 5 mm by the self-weight

of the press, find the natural frequency of the system.


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Consider a rod of mass m, length l, pivoted at O, and is connected toa spring of stiffness kat a distance a from O and also carries a loadM

at a free endB which actuated by the forcef(t).

M

kO

a

B

f(t)

l


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Firstly, equate FDB and kinetic diagram of the system as in figurebelow,

Knowing that, the spring displacement is ,

the acceleration of the mass centre of the rod is ,

acceleration of end B is ,

And the moment of inertia of the rod is

Mg

O

B

f(t)k(y+st)

mg

A

G

FBD

MaB

O

B

maG

A

G

KD

I

ay

2laG

IaB 2

12

1 mlI


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Then, applying Newtons 2nd law, yields

At static equilibrium, we have

From eq.(1) we obtain

effectiveOexternalO MM+

)1(...)(22))((2)()( lMl

ll

mIaak

l

mglMgltf st

)2(...0)(2

)(

ak

lmglMg st

)3(...)(2212

1))(()(

2lMl

llmmlaakltf


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Alternatively, eq.(4) can be derived by considering only moment ofexternal forces without the effect of the weight since its moment will

cancel out with the moment of static equilibrium.

where simplifying eq.(5)

becomes

which is identical to eq.(4).

)5(...)()()( OIaakltf

)6(...)(3

1 222 ltfkaMlml

:)( OO IM+

2

,

2

,,,3

1MlIandmlIwithIII loadOrodOloadOrodOO


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Some may consider the vertical moment arm when writing eq.(5) as

However, since we assumed that the angular displacement is too

small, . Therefore eq.(7) becomes

Which eventually yields eq.(4) or (6).

1cosandsin

)7(....)cos)(sin(cos)( OIaakltf

)8(....))(()( OIaakltf


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A rod OA hinged at O and connected to spring at end A which wasinitially in equilibrium is released abruptly. If the stiffness of the spring is

given as k = 100N/m, as well as the mass and length of the rod are

given as m = 2.0kg and l = 250mm respectively. Determine the

mathematical model of the system and its natural frequency.

O

lA

k


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FBD of rod OA,

Equate to KD,

Therefore, its equation of motion is,

O A

kl

mg

:IM

O

2

3

1where)( mlIIlkl OO

)1(....02 klIO


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Take a look at eq.(1), the weight of the rod is not considered in thissolution as it will cancel out with the static equilibrium equation of the

system.

From eq.(1), the undamped natural frequency of the system is

rad/s25.122

)100(33

m

kn


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For the torsional mechanical system as shown, derive itsmathematical model and determine its frequency of vibration if kt =

100Nm/rad, r = 120mm and m = 3kg. Obtain the displacement if

the initial conditions are .rad/s0)0(andrad05.0)0(

Jkt

r


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FBD:

Applying Newton's 2nd law,

Mathematical model is yielded as

The frequency is represented as

)1(....0 tkJor

J

tk

JT22

kg.m0144.03

1where mrJJkt

rad/s3.49

0144.0

35

J

ktn


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The maximum velocity attained by the mass of a simple harmonicoscillator is 10 cm/s and the period of oscillation is 2 s. If the mass is

released with an initial displacement of 2 cm, find

(a) The amplitude

(b) The initial velocity

(c) The maximum acceleration and

(d) The phase angle


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Three springs and a mass are attached to rigid, weightless barPQ asshown in figure below. Find the natural frequency of vibration of the

system.


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Find the natural frequency of vibration of a spring-mass systemarranged on an inclined plane as shown below.


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For plane motion, we have for

(a) A particle, the potential energy and kinetic energy given,

respectively as

(elastic potential energy, where x=0 in the

equilibrium position)

(gravitational potential energy, where x is

measured from the equilibrium position)

(b) A rigid body, the kinetic energy given as

2

2

1kxVe

mgxVg

2

2

1xmT

22

2

1

2

1ImvT G


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Using the principle of conservation of energy, we write the totalenergy as

and that

constanVT ge VVV where

0VTdt

d


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If the given system is a conservative one, the total kinetic energy ofthe system is zero at the maximum displacement, but is maximum at

the static equilibrium point.

On the other hand, the potential energy is vice versa of the above

statement.Hence, the relationship is given as

This is known as Rayleighs Method. Result of the above equation

will yield the natural frequency of the system.

systemtheofenergytotal).().( maxmax EPEK


78/86

Using the energy method, determine the undamped natural frequencyof the block having m shown in the figure below and also its natural

period of vibration. Assume the block does not slip on the surface of

contact as it oscillates.

m

kx


79/86

Applying energy method, we can write

The equation of motion is

From which

constantVT 222

1

2

1kxxmVT

:0

VTdt

d0

2

2

2

2 xkxxxm

0kxxm

thatand

m

kn s284.6

2

k

mT

n


80/86

Find the natural frequency of the pulley system show in figure belowby neglecting the friction and masses of the pulleys. Solve by using

Rayleighs method.


81/86

Use energy method to find the natural frequency of the

system shown in the figure below.


82/86

The system shown in figure below has a natural frequency

of 5 Hz for the following data: m = 10kg, J0=5kg-m2, r1=10cm,

r2=25cm. When the system is disturbed by giving it an initial

displacement, the amplitude of free vibration is reduced by

80% in 10 cycles. Determine the values ofkand c.


83/86

A boy riding a bicycle can be modeled as a spring-mass-damper

system with an equivalent weight, stiffness, and damping constant of

800N, 50,000N/m and 1,000N-s/m, respectively. The differential setting of

the concrete blocks on the road caused the level surface to decrease

suddenly as indicated in the figure. If the speed of the bicycle is 5m/s(18km/hr), determine the displacement of the boy in the vertical

direction. Assume that the bicycle is free of vertical vibration before

encountering the step change in the vertical displacement.


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A weight of 50N is suspended from a spring of stiffness 4000N/m and is

subjected to a harmonic force of amplitude 60N and frequency of 6Hz.

Find

(a) The extension of the spring due to the suspended weight

(b) The static displacement of the spring due to the maximum applied

force, and

(c) The amplitude of the forced motion of the weight.


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Consider a spring-mass-damper system with k = 4000N/m, m = 10kg,

and c = 40N-s/m.

Find the steady-state and total responses of the system under the

harmonic force F(t) = 200 cos10t N and the initial condition

.0andm1.0 00 xx


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chapter 1 & 2 - introduction to vibrations and single dof systems (1)

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