Paper-II

Chapter- Damped vibration

Free vibrations:

When a body continues to oscillate with its own characteristics frequency. Such

oscillations are known as free or natural vibrations of the body. Ideally, the body

vibrates with its natural frequency for an indefinite time with a constant amplitude.

Damped vibrations:

In practice, the amplitude of vibrations decays with time and finally the body

comes to rest at its mean position. So, we may conclude that there is a damping

force on the vibrating body and this may be due to viscosity of the medium or other

frictional forces. Such vibrations of decaying amplitude are referred to as resisted or

damped vibrations.

The motion of a simple pendulum is an example of a damped simple harmonic

motion, because the amplitude gradually decreases with time and the bob finally

comes to rest.

Analytical treatment of damped vibrations:

Let x be the displacement of a particle of mass m performing a damped motion

from its initial position of rest. The following forces act on the body.

(i) Restoring force (F1):

The restoring force (F1) is proportional to the displacement x and tends to bring

the particle back to its initial position. i.e.

F1 ∝ x

1

F1 = −sx (1)

where s is called the stiffness constant.

(ii) Retarding or resisting force (F2):

This force is caused by friction. This force is proportional to the instantaneous

velocity of the particle.

F2 ∝ v

F1 = −kv = −kdxdt

(2)

where k is the restoring force per unit velocity. The -ve sign indicates that the force

opposes the motion. The system is subjected to viscous friction.

So, the net force acting on the bob is

F = F1 + F2

F = −sx− kdxdt

(3)

From Newton’s 2nd law of motion, we get

F = md2x

dt2(4)

So, from (3) we get

md2x

dt2= −sx− kdx

dt

d2x

dt2= − s

mx− k

m

dx

dt

d2x

dt2+k

m

dx

dt+

s

mx = 0

d2x

dt2+ 2b

dx

dt+ ω2

0x = 0 (5)

where 2b = km

and ω2 = sm

.

This is the differential equation of damped harmonic motion. Let auxiliary solution

of the equation (5) is

x = Aemt

2

So we get from (5)

m2 + 2bm+ ω2 = 0

m = −b±√b2 − ω2

So, we can write

m1 = −b+√b2 − ω2

m2 = −b−√b2 − ω2

So, the solution of (4) is

x = A1em1t + A2e

m2t

x = A1e(−b+

√b2−ω2)t + A2e

(−b−√b2−ω2)t

x = e−bt(A1e

√b2−ω2t + A2e

−√b2−ω2t

)

x = e−bt(A1e

√b2−ω2t + A2e

−√b2−ω2t

)

x = e−bt(A1e

√b2−ω2t + A2e

−√b2−ω2t

)(6)

Where A1 and A2 are constant to be determined from the initial condition. Let

b′ =√b2 − ω2. From (6) we get

q = e−bt(A1e

b′t + A2e−b′t

)(7)

Case I: Over damped motion (b > ω0):

When damping is very large, i.e. b > ω0 Now, at t = 0, x = x0, from (7) we get

x0 = A1 + A2

A1 + A2 = x0 (8)

At t = 0, dxdt

= v0, from (7) we get

dx

dt= be−bt

(A1e

b′t + A2e−b′t

))t=0

+ e−bt(b′A1e

b′t − b′A2e−b′t

))t=0

3

v0 = −b(A1 + A2) + (b′A1 − b′A2)

v0 = −bx0 + b′(A1 − A2)

A1 − A2 =x0b+ v0

b′(9)

Solving (8) and (9), we get

A1 =x0

2

(1 +

b+ v0x0

b′

)=x0

2

(1 +

b+ v0x0√

b2 − ω2

)

A2 =x0

2

(1−

b+ v0x0

b′

)=x0

2

(1−

b+ v0x0√

b2 − ω2

)So, we get the final solution

x =x0

2e−bt

((1 +

b+ v0x0√

b2 − ω2

)e√b2−ω2t +

(1−

b+ v0x0√

b2 − ω2

)e−√b2−ω2t

)(10)

x = x01

2e−bt

((1 +

b+ v0x0√

b2 − ω2

)e√b2−ω2t +

(1−

b+ v0x0√

b2 − ω2

)e−√b2−ω2t (11)

The variation of x with time t is shown in fig.

The motion of the particle is clearly aperiodic. The displacement falls off and

asymptotically approaches zero with increasing time. This is the over damped motion

and is exhibited by a pendulum moving in a highly viscous liquid or by an aperiodic

moving coil galvanometer.

Case II: For b < ω

then√b2 − ω2 =

√−(ω2 − b2) = j

√ω2 − b2 = jω′. From (7) we get

x = x0

(1

2e−bt

((1 +

b+ v0x0

jω′

)ejω

′t +(

1−b+ v0

x0

jω′

)e−jω

′t

)

4

x = x0

(e−bt

2

((ejω

′t + e−jω′t)

+b+ v0

x0

jω′

(ejω

′t − e−jω′t)))

x = x0

(e−bt

((ejω

′t + e−jω′t

2

)+b+ v0

x0

ω′

(ejω

′t − e−jω′t

2j

)))

x = x0

(e−bt

(cosω′t+

b+ v0x0

ω′sinω′t

))(12)

Let AcosΘ = 1 and AsinΘ =b+

v0x0

ω′ , So

A2 = 1 +b2

ω′2= 1 +

(b+ v0

x0

)2

ω2 − b2

and

Θ = tan−1b+ v0

x0√ω2 − b2

So we get from (12)

x = x0e−btA

(cosω′tcosΘ + sinΘsinω′t

)

x = x0Ae−btcos

(√ω2 − b2t−Θ

)x = Re−btcos

(√ω2 − b2t−Θ

)(13)

where R = x0A Eq.(13) shows that the motion is oscillatory with an angular frequency

ω =√ω2

0 − b2 and an amplitude Re−bt. Since the amplitude decays exponentially with

time, the motion is damped oscillatory. The variation of x with time t is shown in

fig.

Here, the oscillations soon die down as the amplitude is proportional to e−bt.

5

If the particle is initially displaced to x0 and then released, v0 = 0.,Then

A =ω0√ω2

0 − b2

So we get

x =x0ω0√ω2

0 − b2e−btcos

(√ω2 − b2t−Θ′

)

Θ′ = tan−1 b√ω2 − b2

Case II: Critical damping (b→ ω0)

Let b→ ω0, so we take√ω2 − b2 = δ. So we get from (11)

x =x0

2e−bt

((1 +

b+ v0x0√

b2 − ω2

)e√b2−ω2t +

(1−

b+ v0x0√

b2 − ω2

)e−√b2−ω2t

x =x0

2e−bt

((1 +

b+ v0x0√

b2 − ω2

)(1 + δt) +

(1−

b+ v0x0√

b2 − ω2

)(1− δt)

x = x0e−bt(

1 +(b+

v0

x0

)t

)This shows that the motion is aperiodic, but x approaches zero quicker than the over

damped case. The motion is now said to be critically damped, the variation of x with

t being displayed by fig.

Q. Derive the differential equation of damped motion from consideration of the

energy of the system.

We have kinetic energy of the system

EK.E. =1

2m(dx

dt

)2

and the potential energy

EP.E. =1

2sx2

6

Let the particle be described an element of displacement δx, then the loss of ki-

netic and potential energy of the particle will be equal to the work done against the

frictional force. Hence

−δ(EK.E. + EP.E.

)= dw = k

dx

dtδx

−δ(

1

2m(dx

dt

)2

+1

2sx2

)= k

dx

dtδx

dx

dt

(d2x

dt2+ 2b

dx

dt+ ω2

0x

)= 0

But dxdt6= 0 for all values of t. Hence,

d2x

dt2+ 2b

dx

dt+ ω2

0x = 0

Forced Vibration:

A vibrating system gradually loses its amplitude since energy is dissipated due to

frictional forces. To maintain the system in vibration, energy must be supplied from

outside. If an external periodic force is applied to the vibrating system, the system

tends to vibrate with its own natural frequency. But the applied driving force tries

to impress its own frequency on the vibrating system. Initially, the system vibrates

with both the frequencies. The natural vibrations die out in course of time due to

the prevailing resisting forces and the system finally in the steady state, vibrates with

the frequency of the driving force with a constant amplitude.

Such vibration where the system oscillates with a frequency the same as that of an

externally impressed periodic force is known as the forced vibration of the system.

Analytical treatment of forced vibration:

Let a particle of mass m, capable of executing a damped simple harmonic mo-

tion, be subjected to an external simple harmonic forced of constant amplitude and

frequency. Let x is the displacement of the particle from its mean position at time t.

7

The forces acting on the particle are the following:

(i) The restoring force Fs = −sx tending to bring the particle back to its mean

position, s being the stiffness factor.

(ii) The resisting force Fr = −k dxdt

, k being the resisting force per unit velocity.

(iii) The driving periodic force Fd = Fcosωt, where F is the amplitude and ω is

the angular frequency of the force.

The net force in the +ve x-direction is

F = Fd + Fr + Fs

F = Fcosωt− kdxdt− sx

From Newton’s laws of motion, we get

md2x

dt2= Fcosωt− kdx

dt− sx

d2x

dt2=F

mcosωt− k

m

dx

dt− s

mx

d2x

dt2+ 2b

dx

dt+ ω2

0x = fcosωt (1)

where 2b = k/m and ω20 = s/m and f = F

m. The quantity ω0 is the undamped natural

angular frequency of the particle and b is the decay constant.

For the natural motion of the particle, the differential equation is

d2x

dt2+ 2b

dx

dt+ ω2

0x = 0

So the complementary function is

x1 = Re−btcos(√

ω2 − b2t−Θ)

(2)

8

where R and Θ are arbitrary constants to be determined from the initial conditions.

Let xr is the displacement for the driving force Fcosωt. So the equation of motion

for this cased2x

dt2+ 2b

dx

dt+ ω2

0x = fcosωt (3)

If xi is the displacement for the driving force Fsinωt. So the equation of motion for

this cased2x

dt2+ 2b

dx

dt+ ω2

0x = fcosωt (4)

Multiplying (4) by i(√−1) and adding (3), we get Let xr is the displacement for the

driving force Fcosωt. So the equation of motion for this case

d2(xr + ixi)

dt2+ 2b

d(xr + ixi)

dt+ ω2

0(xr + ixi) = fcosωt+ ifsinωt

d2X

dt2+ 2b

dX

dt+ ω2

0X = feωt (5)

where X = xr + ixi. Let the particular integral of (5)

X = X0eiωt

where X0 is a complex quantity. Substituting in (5), we get

X0

(− ω2 + 2ibω + ω2

0

)eiωt = feiωt

X0 =f

(ω20 − ω2) + i(2bω)

X0 =f(

(ω20 − ω2)− i(2bω)

)(ω2

0 − ω2)2 + (2bω)2(6)

Let

acosφ = ω20 − ω2

and

asinφ = 2bω

9

Hence,

a =√

(ω20 − ω2)2 + 4b2ω2

and

tanφ =2bω

(ω20 − ω2)

From (6) we get]

X0 =f(acosφ− iasinφ)

)(ω2

0 − ω2)2 + 4b2ω2

X0 =fa(cosφ− isinφ)

)(ω2

0 − ω2)2 + 4b2ω2

X0 =fe−iφ√

(ω20 − ω2)2 + 4b2ω2

So, the solution

X = X0eiωt

X =fe−iφ√

(ω20 − ω2)2 + 4b2ω2

eiωt

X =f√

(ω20 − ω2)2 + 4b2ω2

ei(ωt−φ)

xr + ixi =f√

(ω20 − ω2)2 + 4b2ω2

(cos(ωt− φ) + isin(ωt− φ)

)

So for the driving force for Fcosωt, the particular integral is

xr =f√

(ω20 − ω2)2 + 4b2ω2

cos(ωt− φ)

xr =F

m√

(ω20 − ω2)2 + 4b2ω2

cos(ωt− φ)

xr =F√

(mω20 −mω2)2 + 4b2m2ω2

cos(ωt− φ)

xr =F√

k2ω2 + (s−mω2)2cos(ωt− φ)

10

Similarly, for the driving force Fsinωt, we get

xi =F√

k2ω2 + (s−mω2)2sin(ωt− φ)

So, the general solution

x = x1 + xr

x = Re−btcos(√

ω2 − b2t−Θ)

+F√

k2ω2 + (s−mω2)2cos(ωt− φ)

The first term on the right hand side gives the natural damped simple harmonic

motion of a particle. The second term gives the forced vibration. Initially, both the

terms are operative and their superposition gives an irregular motion. After a few

time constants, the natural vibration dies out since its amplitude falls off exponentially

with time with the decay constant b. Thus, finally only the forced vibration given by

the second term persists. This is the steady state motion. SO, in steady state

x =F√

k2ω2 + (s−mω2)2cos(ωt− φ)

Amplitude or displacement resonance:

When the displacement amplitude is a maximum for some frequency of the driver,

we have the phenomenon of amplitude or displacement resonance.

We have at the steady-state displacement amplitude of forced vibration as

A =F√

k2ω2 + (mω2 − s)2(1)

The amplitude A is maximum for that value of ω for which the denominator on the

right-hand side of (1) is a minimum. So, we get

k2ω2 + (s−mω2)2 = minimum

d

dω

(k2ω2 + (s−mω2)2

)= 0

11

2k2ω + 4(mω2 − s)ωm = 0

k2 + 2(mω2 − s)m = 0

ω2 =s

m− k2

2m2

ω2 = ω20 − 2b2

Substituting the value of ω2 in (1), we get the maximum displacement amplitude

Am =F/m

2b√

(ω20 − b2)

The variation of the displacement amplitude with the angular frequency of the driver

different values of the decay constant is shown in fig.

Velocity resonance:

When the velocity amplitude attains a maximum for a certain frequency of the

driving force, we have velocity resonance.

If the forcing function is Fcosωt, the steady-state displacement

x =F√

k2ω2 + (s−mω2)2sin(ωt− φ)

So, the velocity

v =dx

dt=

Fω√k2 + (mω2 − s

ω)2cos(ωt− φ)

12

when ωm− sω

= 0, then v attains its maximum value

vm =F

k

Hence velocity resonance occurs when

ωm− s

ω= 0

ωm =s

ω

ω = ω0

i.e. when the forcing frequency equals the undamped natural frequency of the forced

system. Plot of the velocity amplitude of the forcing system versus angular frequency

of the driver for differen values of the decay constant is shown in fig.

Power relations in Forced vibration:

We have work done

W = (Fcosωt)dx

In the steady state, the power of the driver in forced vibration is

P = (Fcosωt)dx

dt

F =F 2ω√

k2ω2 + (s−mω2)2cosωtcos(ωt− φ)

F =F 2ω√

k2ω2 + (s−mω2)2

(cos2ωtcosφ+ sinωtcosωtsinφ

)

P =F 2ω√

k2ω2 + (s−mω2)2

(cos2ωtcosφ+

1

2sin2ωtsinφ

)

The average power over a complete cycle is

Pav =1

T

∫ T

0Pdt

13

Pav =1

T

∫ T

0

F 2ω√k2ω2 + (s−mω2)2

(cos2ωtcosφ+

1

2sin2ωtsinφ

)dt

Pav =1

T

F 2ω√k2ω2 + (s−mω2)2

(T

2cosφ+ 0

)

Pav =1

2

F 2ω√k2ω2 + (s−mω2)2

cosφ

Again

cosφ =k√

k2ω2 + (s−mω2)2

So, average power

Pav =F 2ωk

2(k2ω2 + (s−mω2)2

)

At resonance we have√k2 + (s−mω2)2 = k. So the average power at resonance

(Pav)res =F 2

2k

Sharpness of resonance: We have average power

Pav =F 2k

2

(k2 +

(mω − s

ω

)2) (1)

At resonance i.e. at ω = ω0 =√

sm

(Pav)max =F 2

2k

If Pav is plotted against ω, clearly, Pav will attain its maximum value. and will fall

off on either side of this frequency as shown in fig.

14

Hence, resonance is sharp when damping (k) is small.

For large damping, the resonance is broad or flat which is shown in fig.

The sharpness of resonance gives the rapidity with which the average power Pav

drops off as ω differs from its value at resonance. Hence, higher sharpness of resonance

is for low damping.

Average power at resonance

(Pav)res =F 2

2k(2)

Dividing (1) and (2), we get

Pav

Pav

)res

=k2(

k2 +(mω − s

ω

)2)

Pav

Pav

)res

=k2(

k2 + (mω − sω

)2

)Pav

Pav

)res

=k2(

k2 +m2

(ω − ω2

0

ω

)2)

Pav

Pav

)res

=4b2(

4b2 + ω20

(ωω0− ω2

0

ω

)2)

15

Pav

Pav

)res

=4b2(

4b2 + ω20∆2

)where ∆ = ω

ω0− ω0

ω.

At ω = ω0, i.e. at ∆ = 0,Pav

Pav

)res

= 1

Bellow or above resonance

∆ 6= 0

andPav

Pav

)res

< 1

The sharpness of resonance is quantitatively defined as the reciprocal of |∆| at which

Pav

Pav

)res

=1

2

The above fig. shows that for a given value of k or b, there are two values of ω,

namely, ω1 and ω2 for which the average power of Pav is half its value at resonance.

These frequencies are called the half-power frequencies. At the half-power frequencies,

16

we getPav

Pav

)res

=4b2(

4b2 + ω20∆2

)

1

2=

4b2(4b2 + ω2

0∆2

)ω2

0∆2 = 4b2

ω0∆ = ±2b

This gives the sharpness of resonance as

Sr =1

|∆|=ω0

2b

ω − ω20

ω= ±2b

ω2 ∓ 2bω − ω20 = 0

So, two half-power frequencies are

ω1 =√b2 + ω2

0 − b

and

ω2 =√b2 + ω2

0 + b

So, we get

ω2 − ω1 = 2b

Hence, the sharpness of resonance

Sr =1

|∆|=ω0

2b=

ω0

ω2 − ω1

Again

ω1ω2 =(√

b2 + ω20 − b

)(√b2 + ω2

0 − b)

ω1ω2 =(b2 + ω2

0

)− b2

ω1ω2 = ω20

17

Doppler Effect:

The pitch of a note i.e. its frequency as perceived by a listener, appears to change

when there is a relative motion between the source of sound and the listener. The

apparent inequality between the emitted and the perceived frequencies is referred to

as Doppler effect.

(i) Source moving, observer at rest:

Let O be the observer at rest. S be the source moving with a velocity Vs along

SO. The waves of frequency n propagate towards the observer with a velocity V , if

the source is stationary.

So, the original wavelength of sound

λ =V

n

Now, in t = 1 sec the wave travels the distance

SO = V t = V

Let the source S moves to S’ in t = 1 sec.,

SS ′ = Vst = Vs

So, the waves of n frequency will occupy the distance in t = 1 sec

S ′O = V − Vs

18

So, the changed wavelength

λ′ =S ′O

n=V − Vsn

As the velocity of sound remain constant, the apparent frequency of the note is

n′ =V

λ′=

V

V − Vsn (1)

Hence, the apparent change in frequency

n′ − n =(

V

V − Vs− 1

)n

n′ − n =Vs

V − Vsn

So, the fractional change in frequency

n′ − nn

=Vs

V − Vs

Case I: If the source moves away from the observer, Vs is -ve, so apparent frequency

of the note

n′ =V

V + Vsn

Case II: When the wind blows with velocity Vw in the direction of sound, the sound

velocity is increased to V + Vw. So, we get the apparent frequency from (1)

n′ =V + Vw

V + Vw − Vsn

Case II: When the wind blows with velocity Vw in the opposite direction of sound,

the sound velocity is decreased to V − Vw. So, we get the apparent frequency from

(1)

n′ =V − Vw

V − Vw − Vsn

II. Observer moving, source at rest:

Let the observer O moves with velocity VO away from the source.

19

If the observer is at rest, then the waves crossing O in t = 1 sec would occupy a

length OA = V . So, the original wavelength of sound

λ =OA

n=V

n

Now, in t = 1 sec the observer moves to O’. Hence

OO′ = V0

So, in t = 1 sec. the observer receives the waves occupying the length

O′A = V − VO

So, the apparent frequency of the note is

n′ =V − VO

λ=V − VOV

n (2)

Case I: If the observer approaches the source, VO is -ve, so apparent frequency of

the note

n′ =V + VOV

n

Case II: When the wind blows with velocity Vw in the direction of sound, the sound

velocity is increased to V + Vw. So, we get the apparent frequency from (1)

n′ =V + Vw − VOV + Vw

n

Case II: When the wind blows with velocity Vw in the opposite direction of sound,

the sound velocity is decreased to V − Vw. So, we get the apparent frequency from

(1)

n′ =V − Vw − VOV − Vw

n

(III) Both source and observer moving:

When the observer is at rest and the source moves, the apparent frequency is

n1 =V

V − Vsn (1)

20

When the source of emitting frequency n1 is stationary, but the observer is moving

towards the source, the apparent frequency is

n′ =V + VOV

n1 (2)

n′ =V + VOV − Vs

n (3)

Case I: When the wind blows with velocity Vw in the direction of sound, the sound

velocity is increased to V + Vw. So, we get the apparent frequency from (3)

n′ =V + Vw + VOV + Vw − Vs

n (4)

21