mom3602_assign01_48591238

SURNAME AND INITIALS : MABENGO N.D. STUDENT NUMBER : 48591238 MODULE CODE : MOM3602 ASSIGNMENT NUMBER : 01

Page | 1 Mabengo N.D. 48591238 May 2015

QUESTION 1

1.1 Newtons second law gives:

dt

xdcF

xkF

dt

xdmF

damper

spring

2

2

resultant

Using vector superposition rule, we will have:

xcxkxm

dt

xdcxk

dt

xdm

FFF damperspring

2

2

resultant

Rearranging the above equation by diving both sides by m, we will have:

02

2

x

m

k

dt

dx

m

c

dt

xd

(Which is the equation for free vibration with viscous damping).

Let us now recall that there are two types of frequencies in vibration problems: n (natural

angular frequency) and d (the frequency of damped vibration). The relation between

angular frequency srad / and Hzv is given by:

Tv

v

1

2

Furthermore, the damped vibration d is expressed in the formula: nd 21 where

n is the natural angular frequency and is the damping ratio.

To find the frequency of vibration, we must work out n and .



Therefore,

n

c

c

c

nc

m

c

c

c

m

c

c

c

mkmm

kmc

22

222

So that:

267.0521.250

67

/521.25068.7204322

/309.1668.7

2043

c

c

n

c

c

mNskmc

sradm

k

Hence:

Hzv

srad

d

nd

501426573.22

71692669.15

2

/71692669.15309.16267.01122

We can conclude that the frequency of vibration when the mass is released is Hzv 501.2

1.2 The nature of damping is dependent on the magnitude of the damping factor .

If < 1: under damped

= 1: critically damped

> 1: over damped

From the evidence above, we can unassumingly conclude that the nature of the

damping for the system is UNDER DAMPED



1.3 An initial displacement of 48mm from the equilibrium is given. Thus:

smxsmtdt

dx

mxmtx

/0 is, that ,/00

048.0 is, that ,048.00

0

0

The general solution for a differential equation of the form 0 kxxcxm will give:

tsts eCeCtx 21 21

Where

n

m

k

m

c

m

c

m

mkccs

1

222

4

2

22

2,1

We will now substitute the known information into the above that will allow us to solve for the

roots s1,2 as follows:

i

s n

716.15354.4

928711.0309.16354.4

309.161267.0267.0

1

2

2

2,1

Hence:

is

is

716.15354.4

716.15354.4

2

1

To develop the expression, we must solve for the unknown constants C1 and C2 by replacing

the known initial conditions of 048.00 x and smx /0 . To do so we will use the expression

for the displacement that we can differentiate and obtain another expression for the velocity

x :



tsts

tsts

tsts

esCesC

eCeCdt

dtx

eCeCtx

21

21

21

2211

21

21

Substituting with the initial conditions mx 048.00 we obtain:

21

0

2

0

1

0

048.0

048.0

048.0

21

CC

eCeC

mx

ss

Also, with the initial condition smx /00 we will have:

212211

0

22

0

11

0

716.15354.4716.15354.40

0

0

0

21

CiCi

sCsC

esCesC

x

ss

We are now having a system of two equations with two unknowns:

0716.15354.4716.15354.4

048.0

21

21

CiCi

CC

i

iC

i

i

iC

iCii

iCiCi

iCiCCi

CiCi

CC

0066.0024.0

0066.0024.0048.0

0066.0024.0

432.31

7543.0208992.0

754368.0208992.0716.15716.15

754368.0208992.0716.15354.4716.15354.4

0716.15754368.0354.4208992.0716.15354.4

0048.0716.15354.4716.15354.4

048.0

2

1

1

11

111

11

12



Going back to the expression, let us substitute the values of C1 and C2 thereto:

ititttiti

tsts

eee

eiei

eCeCtx

716.15716.15354.4

716.15354.4716.15354.4

21

0066.0024.00066.0024.0

0066.0024.00066.0024.0

21

With 11 2 ii we will have:

tte

ittetx

itit

itt

titi

itit

itt

titi

t

t

it

it

716.15sin0132.0716.15cos048.0

716.15sin0132.0716.15cos048.0

716.15sin0066.0716.15cos0066.0

716.15sin0024.0716.15cos0024.0

716.15sin716.15cos0066.0024.00066.0024.0

716.15sin0066.0716.15cos0066.0

716.15sin0024.0716.15cos0024.0

716.15sin716.15cos0066.0024.00066.0024.0

354.4

2354.4

2

716.15

2

716.15



QUESTION 2

2.1 The displacement at any time )(tx will be given by the expression:

tt nn

eCeCtx

1

2

1

1

22

But with a different value of damping constant mNsc /318 and the same initial conditions,

that is, sradm

kn /309.16

68.7

2043 and mNskmcc /521.25068.7204322 ,

the damping ratio will give:

269.1521.250

318

The corollary is that the new system is over-damped and will have a different vibration

reaction.

tnt

n

tt

nn

nn

eCeC

eCeCdt

dx

1

2

21

1

2

1

2

1

1

22

22

11

Observing that e0 = 1, the above can be simplified as:

02212021

11 xCC

xCC

nn

Substituting the unknown values we thus have:

0149.0

0629.0

0437.33954.7

0309.161269.1269.1309.161269.1269.1

048.0048.0

2

1

21

2

2

1

2

1221

C

C

CC

CC

CCCC

Noting that 954.712 n and 437.3312 n it follows that:

437.33954.7 0149.00629.0 eetx t



Consequently, the expression for the displacement at any time

tt eetx 437.33954.7 0149.00629.0

1.2 The displacement after 0.026 seconds will give us:

m

eex

044902539.0

006246361.00511481.0

0149.00629.0026.0 026.0437.33026.0954.7

To find the velocity after the same time, we will implicitly differentiate the expression for the

displacement as follows:

tt

tt

tt

ee

ee

eedt

dtx

437.33954.7

437.33954.7

437.33954.7

4982113.05003066.0

437.330149.0954.70629.0

0149.00629.0

After 0.026 second the velocity will become:

sm

eex

/197979223.0

208859605.0406838829.0

4982113.05003066.0026.0 026.0437.33026.0954.7

In conclusion the displacement after 0.026s is 0.044902539m or 44.9mm and the velocity

after the same time is -0.197979223m/s or -197.9mm/s



QUESTION 3

Let us first calculate the damping ratio:

1029.1521.250

258

cc

c

Here the system is critically damped which means that that it is defined by the equation

0 kxxcxm and 21 ss . We know that:

t

xxtxetx n

n

n

n

tn

2

2

002

0 1sin1

1cos

The above cannot be directly applied in its current form since there is an indeterminate form

of 0

0present in the analytical solution because:

01lim01lim

2

1

2

1

tn

To solve this issue we will apply LHpital rule which will then give:

t

tt

ttt

n

nn

nnn

2

1

2

12

2

122

12

2

1

1coslim

212

1

212

11cos

lim1

1sinlim



Substituting the limits we will thus have:

txxxe

txxxe

txx

txetx

n

t

n

n

nt

n

n

n

n

t

n

n

n

000

2

2

1

00

0

2

2

002

01

1

1sinlim

1sin1

1coslim

With the known initial conditions, that is, mx 048.00 and smx /00 the above expression

will become:

te

tetx

t

t

782832.0048.0

048.0309.160048.0

309.16

309.16

The expression for displacement at any time is tetx t 782832.0048.0309.16

3.2 To find the expression of the velocity, we must first differentiate the expression of the

displacement as:

te

te

etee

ete

tedt

dtx

t

t

ttt

tt

t

7620709.12

782832.07620709.12782832.0

782832.076720709.12782832.0

782832.0782832.0048.0309.16

782832.0048.0

309.16

309.16

309.16309.16309.16

309.16309.16

309.16



After 0.027 seconds,

The displacement will give:

044505916.0

069128364.0

027.0782832.0048.0027.0

440343.0

027.0309.16

e

ex

The velocity will therefore be:

221843.0

344575914.0

027.07620709.12027.0

440343.0

027.0309.16

e

ex

The displacement after 0.027s is 0.044505916m or 44.505mm

The velocity after 0.027s is -0.221843m/s or -221.843mm/s



QUESTION 4

4.1 The length of the bar is , its mass is m , and the perpendicular distance between the axis passing through the bars centre of mass at point P and the axis about which the bar is

rotating passing through the point O is 4

1 so that:

2

2

2

2

2

48

7

48

7

412

1

4

1

12

1

mI

mmmI

d

mI

O

cm

From the free-body diagram, the angular acceleration will be expressed as:

2

2

dt

d

But we should recall that R

s where is the angle, s is the the arc length subtended

and R is the radius. Rearranging, we have Rs . In vibration problems, we only consider small displacement or rotations so this approximation is perfectly accurate. Applying this,

we will have:

222

111

Rs

Rs

The deflection of the spring is then 1s and the velocity of the damper is then 2s where:

222 Rs

Since there is only one bar which is rigid, it follows that the bar will rotate in equal angles,

thus 21 and we only need one variable to describe the motion of the bar.

Consequently, we observe that:

4

3

2

1

2

1

R

R



Knowing that the springs force is a retarding force opposite to the direction of displacement

and the damping force being the retarding force opposite to the direction of velocity, we

will have:

ccscF

kkkkskkF

damping

springs

2

1

2

1

4

3

4

3

1

2121221

Having the forces we can now calculate the corresponding torques taking into the account

the corresponding distances between the application of the respective forces and the pivot

are:

2

1

4

3

damper

spring

D

D

So that:

2

2

2121

4

1

2

1

2

1

16

9

4

3

4

3

ccDF

kkkkDF

damperdamperdamper

springspringspring

Applying vector superposition, we then have:

22214

1

16

9ckk

damperspringresultant

Newtons second law will allow us to obtain:

222

2148

7

4

1

16

9 mckk

Iresultant



The equation of the motion will be: 04

1

16

9

48

7 2221

2

ckkm

4.2 In the special case where 21 kk the above equation becomes:

04

12

16

9

48

7 222

ckm

mom3602_assign01_48591238

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