NCERT Solutions for Class 11 Physics Chapter 13 Oscillations

NCERT
SOLUTIONS
CLASS - 11th

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Class : 11th
Subject : Physics
Chapter : 14
Chapter Name : Oscillations

Q14.1 Which of the following examples represent periodic motion?
(a) A swimmer completing one (return) trip from one bank of a river to the other and back.
(b) A freely suspended bar magnet displaced from ts N-S direction and released.
(c) A hydrogen molecule rotating about its center of mass.
(d) An arrow released from a bow.

Answer. (b) and (c)
(a) The swimmer's motion is not periodic. The motion of the swimmer between the banks of a
river is back and forth. However, it does not have a de nite period. his is
because the time taken by the swimmer during his back and forth journey may not be the same.
(b) The motion Of a freely-suspended magnet, if displaced from its N-S direction and released,
is periodic. This is because the magnet oscillates about its position with a
de nite period of time.
(c) When a hydrogen molecule rotates about its centre of mass, it comes to the same position
again and again after an equal interval of time. Such motion is periodic.
(d) An arrow released from a bow moves only in the forward direction. It does not come
backward. Hence, this motion is not a periodic.

Page : 362 , Block Name : Exercise

Q14.2 Which of the following examples represent (nearly) simple harmonic motion and which
represent periodic but not simple harmonic motion?
(a) the rotation of earth about its axis.
(b) motion of an oscillating mercury column in a U-tube.
(c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly
above the lowermost point.
(d) general vibrations of a polyatomic molecule about its equilibrium position.

Answer. (b) and (c) are SHMs
(a) and (d) are periodic, but not SHMs
(a) During its rotation about its axis, earth comes to the same position again and again in equal
intervals of time. Hence, it is a periodic motion. However, this motion is not simple harmonic.
This is because earth does not have a to and fro motion about its axis.
(b) An oscillating mercury column in a CJ-tube is simple harmonic. This is because the mercury
moves to and fro on the same path, about the xed position, with a certain
period of time.

(c) The ball moves to and fro about the lowermost point of the bowl when released. Also, the
ball comes back to ts initial position in the same period of time, again and
again. Hence, its motion is periodic as well as simple harmonic.
(d) A polyatomic molecule has many natural frequencies of oscillation. Its vibration is the
superposition of individual simple harmonic motions of a number of different
molecules. Hence, it is not simple harmonic, but periodic.

Page : 362 , Block Name : Exercise

Q14.3 depicts four x t plots for linear motion of a particle. Which of the plots represent periodic
motion? What is the period of motion (in case of periodic motion)?

Answer. (b) and (d) are periodic
(a) It is not a periodic motion. his represents a unidirectional, linear uniform motion.
There is no repetition of motion in this case
(b) In this case, the motion of the particle repeats itself after 2 s. Hence, It is a periodic motion,
having a period of 2 s.
(c) It is not a periodic motion. This is because the particle repeats the motion in one position
only. For a periodic motion, the entire motion of the particle must be repeated in equal intervals
of time.
(d) In this case, the motion of the particle repeats itself after 2 s. Hence, it is a periodic motion,

having a period of 2 s .

Page : 362 , Block Name : Exercise

Q14.4 Which of the following functions of time represent
(a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give
period for each case of periodic
motion (ω is any positive constant ) :

(a) sin ωt − cos ωt

3
(b) sin ωt

(c) 3 cos(π/4 − 2ωt)

(d) cos ωt + cos 3ωt + cos 5ωt

2 2
(e) exp(−ω t )

Answer.
(a) SHM

The given function is:

sin ωt − cos ωt

1 1
= √2 [ sin ωt − cos ωt]
√2 √2

π
= √2 sin(ωt − )
4

π
= √2 sin(ωt − )
4

This function represents SHM as it can be written in the form: a sin(ωt + ϕ)
Its period is :
2π

ω

(b) Periodic, but not SHM

The given function is:

3
sin ωt

1
= [3 sin ωt − sin 3ωt]
2

The terms sin ωt and sin ωt individually represent simple harmonic motion (SHM). However,
the superposition of two SHM IS periodic and not simple harmonic.
(c) SHM

The given function is:

π
3 cos[ − 2ωt]
4

π
= 3 cos[2ωt − ]
4

This function represents simple harmonic motion because it can be written in the form:

a cos(ωt + ϕ)

Its period is 2π π
=
2ω ω

(d) Periodic, but not SHM

The given function is cos ωt + cos 3ωt + cos 5ωt . Each individual cosine function

represents SHM. However, the superposition of three simple harmonic motions is

periodic, but not simple harmonic.

(e) Non-periodic motion

2 2
The given function exp (−ω t )
is an exponential function. Exponential functions do not

repeat themselves. Therefore, it is a non-periodic motion.

2 2
(f) The given function 1 + ωt + ω t is non-periodic.

Page : 363 , Block Name : Exercise

Q14.5 A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart.
Take the direction from A to B as the positive direct on and give the signs of velocity,
acceleration and force on the particle when it is
(a) at the end A,
(b) at the end B,
(c) at the mid-point of A3 going towards A,
(d) at 2 cm away from 3 going towards A,
(e) at 3 cm away from A going towards B, and
(f) at 4 cm away from B going towards A.

Answer. (a) Zero, Positive, Positive
(b) Zero, Negative, Negative
(c) Negative, Zero, Zero
(d) Negative, Negative, Negative
(e) Zero, Positive, Positive
(f) Negative, Negative, Negative
The given situation is shown in the following gure. Points A and B are the two end points, with
AB = 10 cm. O is the midpoint of the path.

A particle is in linear simple harmonic motion between the endpoints

(a) At the extreme point A, the particle is at rest momentarily. Hence, its velocity is zero at this
point. Its acceleration is positive as it is directed along AO.
Force is also positive in this case as the particle is directed rightward.

(b) At the extreme point B, the particle is at rest momentarily. Hence, its velocity is zero at this
point. Its acceleration is negative as it is directed along 3. Force is also negative in this case as
the particle is directed leftward.

(c)

The particle is executing a simple harmonic motion. O is the mean position of the particle. Its
velocity at the mean position O is the maximum. The value for velocity is
negative as the particle is directed leftward. The acceleration and force of a particle executing
SHM is zero at the mean position.

(d)

The particle is moving toward point O from the end B. This direction of motion is opposite to
the conventional positive direction, which is from A to B, Hence, the particle's velocity and
acceleration, and the force on it are all negative.

(e)

The particle is moving toward point O from the end A. This direction of motion is from A to B,
which is the conventional positive direction. Hence, the values for velocity,
acceleration, and force are all positive.

(f)

This case is similar to the one given in (d).

Page : 363 , Block Name : Exercise

Q14.6 Which of the following relationships between the acceleration a and the displacement x
of a particle involve simple harmonic motion?
(a) a = 0.7x

2
(b) a = −200x

(c) a = −10x

3
(d) a = 100x

Answer. (c) A motion represents simple harmonic motion if it is governed by the force law

F = −kx

ma = −k

k
∴ a = − x
m

Where,

F is the force

m is the displacement

a is the displacement

a is the acceleration

k is a constant

Among the given equations, only equation a =—10 x is written in the above form with
Hence, this relation represent SHM.

Page : 363 , Block Name : Exercise

Q14.7 The motion of a particle executing simple harmonic motion is described by the
displacement function. x(t) = A cos (wt + Φ).lf the initial (t = 0) position of the particle is 1 cm
and its initial velocity is w cm/s, what are its amplitude and initial phase angle? The angular
frequency of the particle is Tt sl. If instead of the cosine function, we choose the sine function
to describe the SHM: B sin(wt + α), what are the amplitude and initial phase of the particle
with the above initial conditions?

Answer.
Initially, at t = 0 .

Displacement, x = 1cm

Initial velocity, v = ωcm/sec .

−1
Angular frequency, ω = πrad/s

It is given that:

x(t) = A cos(ωt + ϕ)

1 = A cos(ω × 0 + ϕ) = A cos ϕ

A cos ϕ = 1
dx
Velocity, v =
dt

ω = −Aω sin(ωt + ϕ)

1 = −A sin(ω × 0 + ϕ) = −A sin ϕ

A sin ϕ = −1
2 2 2
A (sin ϕ + cos ϕ) = 1 + 1

2
A = 2

∴ A = √2cm

tan ϕ = −1

3π 7π
∴ ϕ = , ,……
4 4

SHM is given as:

x = B sin(ωt + α)

Putting the given values in this equation, we get:

I = B sin[ω × 0 + α]

B sin α = 1

Velocity, v = ωB cos(ωt + α)

Substituting the given values, we get:

π = πB sin α

B sin α = 1
2 2 2
B [sin α + cos α] = 1 + 1

2
B = 2

∴ B = √2cm
B sin α 1
=
B cos α 1

π
tan α = 1 = tan
4

π 5π
∴ α = , ,……
4 4

Page : 364 , Block Name : Exercise

Q14.8 A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A
body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s.
What is the weight of the body?

Answer. Maximum mass that the scale can read, M = 50 kg
Maximum displacement of the spring = Length of the scale, l = 20 cm = 0.2 m
Time period, T = 0.6 s
Maximum force exerted on the spring, F = Mg
where,
2
g = acceleration due to gravity = 9.8m/s

F = 50 × 9.8 = 490

¿Spring constant, k = F

l
=
490

0.2
= 2450Nm
−1

Mass m, is suspended from the balance.

m
T = 2π√
k

Time period,

2 2
T 0.6
∴ m = ( ) × k = ( ) × 2450 = 22.36kg
2π 2×3.14

∴ Weight of the body = mg = 22.36 × 9.8 = 219.167N

Hence, the weight of the body is about 219N .

Page : 364 , Block Name : Exercise

Q14.9 A spring having with a spring constant 1200 N m is mounted on a horizontal table as
−l

shown in Figure A mass of 3 kg is attached to the free end of the spring. The mass is then pulled
sideways to a distance of 2.0 cm and released. Determine
(i) the frequency of oscillations,
(ii) maximum acceleration of the mass, and
(iii) the maximum speed of the mass.

Answer.
−1
Spring constant, k = 1200Nm

Mass, m = 3kg

Displacement, A = 2.0cm = 0.02cm

(i) Frequency of oscillation v, is given by the relation:

1 1 k
v = = √
T 2π m

Where, T is the time period

1 1200
∴ v = √ = 3.18m/s
2×3.14 3

Hence, the frequency of oscillations is 3.18m/s .

(ii) Maximum acceleration (a) is given by the relation:

2
a = ω A

Where,

k
ω = Angular frequency = √
m

A = Maximum displacement

k 1200×0.02 −2
∴ a = A = = 8ms
m 3

2
Hence, the maximum acceleration of the mass is 8.0m/s .

(iii) Maximum velocity, vmax = Aω

k 1200
= A√ = 0.02 × √ = 0.4m/s
m 3

Hence, the maximum velocity of the mass is 0.4m/s .

Page : 364 , Block Name : Exercise

Q14.10 In Exercise 14.9, let us take the position of mass when the spring is unstreched as x =0,
and the direction from eft to right as the positive direction of x axis. Give x as a function of time
t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is

(a) at the mean position,
(b) at the maximum stretched position, and
(c) at the maximum compressed position.
In what way do these functions for SHM differ from each other, amplitude or the initial phase?
in frequency, in

Answer.
(a) x = 2 sin 20t

(b) x = 2 cos 20t

(c) x = −2 cos 20t

The functions have the same frequency and amplitude, but different initial phases.

The functions have the same frequency and amplitude, but different initial phases.
Distance traveled by the mass sideways, A = 2.0 cm
Force constant of the spring, k = 1200Nm
−1

Angular frequency of oscillation:
k
ω = √
m
−1
= 20rads
1200
= √ = √400
3

(a) When the mass is at the mean position, initial phase is 0.
Displacement, x = A sin ωt

= 2 sin 20t

= 2 sin 20t

(b) At the maximum stretched position, the mass is toward the extreme right. Hence,
the initial phase is π

2
π
= 2 sin(20t + )
2

= 2 cos 20t

(c) At the maximum compressed position, the mass is toward the extreme left. Hence,
In initial phase
3π

2

3π
= 2 sin (20t + )
2
=−2 cos 20t

Page : 364 , Block Name : Exercise

Q14.11 Following gures correspond to two circular motions. The radius of the circle, the period
of revolution, the initial position, and the sense of revolution (i.e., clockwise or anticlockwise)
are indicated on each gure.

Obtain the correspond ng simple harmonic motions of the x-projection of the radius vector Of
the revolving particle p, in each case.

Answer. (a) Time period, T = 2 s
Amplitude, A = 3 cm
π
At time, t = 0, the radius vector OP makes an angle with the positive x -axis, i.e.,
2
π
ϕ=+
phase angle 2

Therefore, the equation of simple harmonic motion for the x-projection of OP, at time t, is given
by the displacement equation:
2πt
x = A cos[ + ϕ]
T

2πt π 2πt
= 3 cos( + ) = −3 sin( )
2 2 2

∴ x = −3 sin π/cm

(b) Time period, T = 4s

Amplitude, a = 2m

At time t = 0, OP makes an angle n with the x -axis, in the anticlockwise direction.

Hence, phase angle, O = +n

Therefore, the equation of simple harmonic motion for the x-projection of OP, at time t,

is given as:

2πt 2πt
x = a cos( + ϕ) = 2 cos( + π)
T 4

π
∴ x = −2 cos( t)m
2

Page : 364 , Block Name : Exercise

Q14.12 Plot the corresponding reference circle for each of the following simple harmonic
motions. Indicate the initial (t o) position of the particle, the radius of the circle, and the
angular speed of the rotating particle. For simplicity, the sense of rotation may be xed to be
anticlockwise in every case: (x is in cm and t is in s).
(a) x =-2 sin (3t + n/3)
(b) x = cos (n/6 - t)
(c) x = 3 sin (2nt + n/4)
(d) x = 2 cos nt

Answer.
π
x=−2 sin(3t+ ) π π
3
(a) = +2 cos(3t + + )
3 2

5π
= 2 cos(3t + )
6

2π
x=A cos( t+ϕ)
If this equation is compared with the standard SHM equation T
, , then we get :
Amplitude, A = 2cm

5π ∘
Phase angle, ϕ = = 150
6

2π
Angular velocity, ω = = 3 rad/sec.
T

The motion of the particle can be plotted as shown in the following gure.

(b) x = cos( π

6
− t) = cos(t −
π

6
)

If this equation is compared with the standard SHM equation x = A cos( 2π

T
t + ϕ) , then we
get :
Amplitude, A = 1
π ∘
Phase angle, ϕ = − = −30
6

2π
Angular velocity, ω = = 1rad/s
T

The motion of the particle can be plotted as shown in the following gure.

(c) x = 3 sin(2πt + π

4
)

π π 3π
= −3 cos[(2πt + ) + ] = −3 cos(2πt + )
4 2 4

If this equation is compared with the standard SHM equation
We get:
Amplitude, A = 3 crn
2π
ω = = 2πrad/s
T

The motion of the particle can be plotted as shown in the following gure.

(d) x = 2 cos nt
If this equation is compared with the standard SHM equation x = A cos( , then we
2π
t + ϕ)
T

get
Amplitude, A = 2cm

Phase angle, Φ = 0

Angular velocity, ω = πrad/s

The motion of the particle can be plotted as shown in the following gure.

Page : 364 , Block Name : Exercise

Q14.13 Figure (a) shows a spring of force constant k clamped rigidly at one end and a mass m
attached to its free end. A force F applied at the free end stretches the spring. Figure (b) shows
the same spring with both ends free and attached to a mass m at either end. Each end of the
spring in Figure (b) is stretched by the same force F.

(a) What is the maximum extension of the spring in the two cases?
(b) If the n Figure (a) and the two masses in Fig. (b) are released, what is the period of oscillation
in each case?

Answer. Mien a force F, is applied to the free end of the spring, an extension l, is produced. For
the maximum extension, it can be written as:
F = kl
Where , k is the spring constant
Hence, the maximum extension produced in the spring, l = F

k

For the two svstem
The displacement (x) produced in this case is:
l
x =
2

Net force, F = +2kx

F
∴ I =
k

(b) For the one block system:
For mass (m) of the block, force is written as :
2
d x
F = ma = m
dt2

Where, x is the displacement of the block in time t
2
d x
∴ m 2
= −kx
dt

It is negative because the direction of elastic force is opposite to the direction of
displacement.
2
d x k 2
2
= −( ) x = −ω x
dt m

2 k
Where, ω =
m

k
ω = √
m

2 k
Where, ω =
m

k
ω = √
m

Where,

ω is angular frequency of the oscillation

2π
∴ TIme period of the oscillation, T =
ω

for the two system :
2
d x
F = m 2
dt

2
d x
m 2
= −2kx
dt

It is negative because the direction of elastic force is opposite to the direction of
displacement.

2
d x 2k 2
= −[ ] x = −ω x
dt2 m

where
2k
ω = √
m

2π m
T = = 2π√
ω 2k

Page : 365 , Block Name : Exercise

Q14.14 The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0
m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min,
what is its maximum speed?

Answer.
Angular frequency of the piston, ω = 200rad/min .

Stroke = 1.0m
1.0
A Amplitude, A = = 0.5m
2

vmax = Aω

= 200 × 0.5 = 100m/min

Page : 365 , Block Name : Exercise

Q14.15 The acceleration due to gravity on the surface of moon is 1.7 ms . What is the time
−2

period of a simple pendulum on the surface of moon if its time period on the surface of earth is
3.5 s? (g on the surface of earth is 9.8 ms ) −2

Answer.
′ −2
Acceleration due to gravity on the surface of moon, g = 1.7ms

−2
Acceleration due to gravity on the surface of earth, g = 9.8ms

Time period of a simple pendulum on earth, T = 3.5s

l
T = 2π√
g

Where,

I is the length of the pendulum
2
T
∴ l = 2
× g
(2π)

2
(3.5)
∴ l = 2
× 9.8m
4×(3.14)

The length of the pendulum remains constant.
On moon's surface, time period, T ′ l
= 2π√
g′

9.8
= 3.5√ = 8.4s
1.7

Page : 365 , Block Name : Exercise

Q14.16 Answer the following questions:
(a) Time period of a particle in SHM depends on the force constant k and mass m of the particle
: T = 2π√ A simple pendulum executes SHM approximately. Why then is the time period of
m

k

a pendulum independent of the mass of the pendulum?
(b) The motion Of a simple pendulum s approximately simple harmonic for small angle
oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater
than √2π

g
l
Think of a qualitative argument to apprec ate this result.
(c) A man with a wristwatch on his hand falls from the top Of a tower. Does the watch give
correct time during the free fall?
(d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely
falling under gravity?

Answer.
(a) The time period of a simple pendulum, T = 2π√
m

k

For a simple pendulum, k is expressed in terms of mass m, as:
k xm
m .
= Constant
k

Hence, the time period T, of a simple pendulum is independent of the mass of the bob.

(b) In the case of a simple pendulum, the restoring force acting on the bob of the pendulum is
given as:
F = −mg sin θ

Where,

F = Restoring force

m = Mass of the bob

g = Acceleration due to gravity

θ = Angle of displacement

For small θ, sin θ = θ

For large θ, sin θ is greater than θ

This decreases the effective value of g .

Hence, the time period increases as:

l
T = 2π√ ′
g

Where, I is the length of the simple pendulum

(c) The time shown by the wristwatch of a man falling from the top of a tower is not affected by
the fall. Since a wristwatch does not work on the principle of a simple
pendulum, it is not affected by the acceleration due to gravity during free fall. Its working
depends on spring action.

(d) When a simple pendu um mounted in a cabin falls freely under gravity, its acceleration is

zero. Hence the frequency of oscillation of this simple pendulum is zero.

Page : 365 , Block Name : Exercise

Q14.17 A simple pendulum of length and having a bob of mass M is suspended in a car. The car
is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small
oscillations in a radial direction about its equilibrium position, what will be its time period?

Answer. The bob Of the simple pendulum will experience the acceleration due to gravity and the
centripetal acceleration provided by the circular motion of the car.
Acceleration due to gravity = g
Centripetal acceleration
2
v
=
R

Where,
v is the uniform speed of the car

R is the radius of the track

Effective acceleration (aef f ) is given as:

2
2
v
aef f = √g
2
+ ( )
R

l
T = 2π√
aef f

Where, / is the length of the pendulum

l
= 2π
√ 2 v
4
g +
2
R

Page : 365 , Block Name : Exercise

Q14.18 Cylindrical piece of cork of density of base area A and height h oats in a liquid of
density . The cork is depressed slightly and then released. Show that the cork oscillates up and
hρ
down simple harmonically with a period T = 2π√ ρ,g
where p is the density of cork. (Ignore
damping due to viscosity of the liquid).

Answer.
Base area of the cork = A

Height of the cork = h

Density of the liquid = ρ1

Density of the cork = ρ

In equilibrium :
Weight of the cork = Weight of the liquid displaced by the oating cork
Let the cork be depressed slightly by x. As a result, some extra water of a certain volume is
displaced. Hence, an extra upthrust acts upward and provides the restoring
force to the cork.

F = −( Volume × Density × g)

Volume = Area × Distance through which the cork is depressed

Volume = Ax

∴ F = −A × Pl g … (i)

According to the force law:

F = kx
F
k =
x

Where, k is a constant

F
k = = −Aρ1 g
x

The time period of the oscillations of the cork:

m
T = 2π√
k

Where,

m = Mass of the cork

= volume of the cork × Density

= Ahp

Hence, the expression for the time period becomes:

Ahρ hρ
T = 2π√ = 2π√
Aρi g ρj g

Page : 366 , Block Name : Exercise

Q14.19 One end of a U-tube containing mercury is connected to a suction pump and the other
end to atmosphere. A small pressure difference is maintained between the two columns. Show
that, when the suction pump is removed, the column of mercury in the IJ tube executes simple
harmonic motion.

Answer.
Area of cross-section of the U-tube = A

Density of the mercury column = ρ

Acceleration due to gravity = g

Restoring force, F = Weight of the mercury column of a certain height

F = −( Volume × Density × g)

F = −(A × 2h × ρ × g) = −2Aρgh = −k × Displacement in one of the arms (h)

2h is the height of the mercury column in the two arms

F
k is a constant, given by k = − = 2Aρg
h

where, m is the mass of the mercury column
Let 1 be the length of the total mercury in the U-tube.
Mass Of mercury, m = Volume Of mercury x Density of mercury
Alρ l
T = 2π√ = 2π√
2Aρg 2g

Hence, the mercury column executes simple harmonic motion with time period 2π√ 1

2g

Page : 366 , Block Name : Exercise

Q14.20 An air chamber of volume V has a neck area of cross section a into which a ball of mass
m just ts and can move up and down without any friction (Fig. 14.33). Show that when the ball
is pressed down a little and released, it executes SHM. Obtain an expression for the time period
of oscillations assuming pressure-volume variations of air to be isothermal [see Figure].

Answer.
Volume of the air chamber = V

Area of cross-section of the neck = a

Mass of the ball = m

The pressure inside the chamber is equal to the atmospheric pressure.

Let the ball be depressed by x units. As a result of this depression, there would be a

decrease in the volume and an increase in the pressure inside the chamber.

Decrease in the volume of the air chamber, ΔV = ax
Change in volume
=
Original volume

ΔV ax
⇒ =
V V

In this case, stress is the increase in pressure. The negative sign indicates that pressure
increases with a decrease in volume.
−Bax
p =
V

The restoring force acting on the ball,

F = p × a

−Bax
= ⋅ a
V

2
−Ba x
=
V

In simple harmonic motion, the equation for restoring force is:

F = −kx … (ii)

Where, k is the spring constant

Comparing equations (i) and (iii), we get:

In simple harmonic motion, the equation for restoring force is:

F = −kx … (ii)

Where, k is the spring constant

Comparing equations (i) and (ii), we get:

2
Ba
k =
V

Time period,

Vm
T = 2π√
2
Ba

Page : 366 , Block Name : Additional Exercise

Q14.21 You are riding in an automobile of mass 3000 kg. Assuming that you are examining the
oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire
automobile is placed on it. Also, the amplitude Of oscillation decreases by 50% during one
complete oscillation. Estimate the values of
(a) the spring constant k and (b) the damping constant b for the spring and shock absorber
system Of one wheel, assuming that each wheel supports 750 kg.

Answer.
(a) Mass of the automobile, m = 3000kg

Displacement in the suspension system, x = 15cm = 0.15m

There are 4 springs in parallel to the support of the mass of the automobile.

The equation for the restoring force for the system:

F = −4kx = mg

Where, k is the spring constant of the suspension system

m
T = 2π√
4k

mg 3000×10 4
k = = = 5000 = 5 × 10 N/m
4x 4×0.15

4
Spring constant, k = 5 × 10 N/m

(b) Each wheel support a mass M = 3000
= 750kg
4

For damping factor b, the equation for displacement is written as:
− log /2M
x = x0 e

The amplitude of oscillation decreases by 50% .
x0
∴ x =
2
bt
log 2 =
e 2M

2M log 2
e
∴ b =
t

m 3000
Time period, t = 2π√ = 2π√ 4
= 0.7691s
4k 4×5×10

2×750×0.693
∴ b = = 1351.58kg/s
0.7691

Therefore, the damping constant of the spring is 1351.58kg/s .

Page : 366 , Block Name : Additional Exercise

Q14.22 Show that for a particle in linear SHM the average kinetic energy over a period of
oscillation equals the average potential energy over the same period.

Answer.
x = A sin ωt

Where,

A = Amplitude of oscillation

k
ω = Angular frequency = √
M

The velocity of the particle is:

dx
v = = Aω cos ωt
dt

The kinetic energy of the particle is:

1 2 1 2 2 2
Ek = MN = MA ω cos ωt
2 2
1 2 1 2 2 2
Ek = Mv = MA ω cos at
2 2

The potential energy of the particle is:

1 2 1 2 2 2
Ep = kx = Mω A sin ωt
2 2

For time period T , the average kinetic energy over a single cycle is given as:

1 T
(Ek ) = ∫ Ek dt
hug T 0

1 T 1 2 2 2
= ∫ MA ω cos ωtdt
T 0 2

1 T (1+cos 2ωt)
2 2
= MA ω ∫ dt
2T 0 2

T
1 2 2 sin 2ωt
= M A ω [t + ]
4T 2ω
0

1 2 2
= M A ω (T )
4T

1 2 2
= MA ω
4

It can be inferred from equations (i) and (ii) that the average kinetic energy for a given time
period is due to the avera e potential energy for the same time period.

Page : 366 , Block Name : Additional Exercise

Q14.23 A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is
twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5
s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire.
(Torsional spring constant a is de ned by the relation J = -a θ, where J is the restoring couple
and e the angle of twist).

Answer. Mass of the circular disc, m = 10 kg
Radius of the disc, r 15 cm 0.15 m
The torsional oscillations of the disc has a time period, T = 1.5 s
The moment of inertia of the disc is:
1 2
I = mr
2

1 2
= × (10) × (0.15)
2

2
= 0.1125kgm

I
T = 2π√
α

Time period,

a is the torsional constant.

2
4π I
α = 2
T

2
4×(π) ×0.1125
= 2
(1.5)

= 1.972Nm/rad

Hence, the torsional spring constant of the wire is 1.972Nmrad −1

Page : 366 , Block Name : Additional Exercise

Q14.24 A body describes simple harmonic motion with amplitude of 5 cm and a period of 0.2 s.
Find the acceleration and velocity of the body when the displacement is
(a) 5 cm, (b) 3 cm (c) 0 cm

Answer.
Amplitude, A = 5cm = 0.05m

Time period, T = 0.2s

(a) For disolacement x = 5cm = 0.05m

Acceleration is given by:

2
a = −ω x

2
2π
= −( ) x
T

2
2π
= −( ) x
T

2
2π
= −( ) ×
T

2
2π
= −( ) × 0.05
T

2 2
= −5π m/s

Velocity is given by:

2 2
v = ω √A − x

2π
= √(0.05)2 − (0.05)2
T

= 0

When the displacement of the body is 5cm, its acceleration is

2 2
−5π m/s and velocity is 0.

(b) For displacement, x = 3cm = 0.03m

Acceleration is given by:

2
a = −ω x
2
2π
= −( ) x
T

2
2π
= −( ) 0.03
0.2

2 2
= −3π m/s

Velocity is given by:

2 2
v = ω √A − x

2π
= √A 2 − x2
T

2π
= √(0.05)2 − (0.03)2
T
2π
= √(0.05)2 − (0.03)2
T

2π
= × 0.04
0.2

= 0.4nm/s

(c) For displacement, x = 0

Acceleration is given by:

2
a = −ω x = 0

Velocity is given by:

2 2
v = ω √A − x
2π
= √(0.05)2 − 0
0.2

= 0.5πm/s

When the displacement of the body is 0, its acceleration is 0 and velocity is 0.5nm/s .

Page : 366 , Block Name : Additional Exercise

Q14.25 A mass attached to a spring is free to oscillate, with angular velocity in a horizontal
plane without friction or damping. It is pulled to a distance xo and pushed towards the centre
with a velocity vo at time t = O Determine the amplitude of the resulting oscillations in terms Of
the parameters ω, x and v . [Hint: Start with the equation x = a cos(ωt + θ) and note that the
0 0

initial velocity is negative.]

Answer.
x = A cos(ωt + θ)

Where,

A is the amplitude

x is the displacement

θ is the phase constant

dx
velocity, v = = −Aω sin(at + θ)
dt
dx
v = = −Aω sin(at + θ)
dt

At t = 0, x = x0

x0 = A cos θ = x0 … (i)

Squaring and adding equations (i) and (ii), we get:

2
v
2 2 2 2 0
A (cos θ + sin θ) = x + ( )
0 ω2

v0 2
2
∴ A = √x + ( )
0 ω

v0 2
2
∴ A = √x + ( )
0 ct

2
Hence, the amplitude of the resulting oscillation is ∴ A = √x 2 v0
+ ( )
0 ct

Page : 366 , Block Name : Additional Exercise