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
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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)
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