NCERT Solutions for Class 11 Physics Chapter 13 Oscillations

Q. 13.1 Which of the following examples represents 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 its N-S direction and released.

(c) A hydrogen molecule rotating about its centre of mass.

(d) An arrow released from a bow

Answer:

(a) The motion is not periodic, though it is to and fro.

(b) The motion is periodic.

(c) The motion is periodic.

(d) The motion is not periodic.

Q. 13.2 Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?

(a) The rotation of the Earth about its axis.

(b) Motion of an oscillating mercury column in a U-tube.

(c) The 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:

(a) Periodic but not S.H.M.

(b) S.H.M.

(c) S.H.M.

(d) Periodic but not S.H.M. [A polyatomic molecule has a number of natural frequencies, so its vibration is a superposition of SHM’s of a number of different frequencies. This is periodic but not SHM.

Q. 13.3 Fig. 13.18 depicts four x-t plots for the linear motion of a particle. Which of the plots represents periodic motion? What is the period of motion (in case of periodic motion)?

Fig 13.18

Answer:

The x-t plots for linear motion of a particle in Fig. 13.18 (b) and (d) represent periodic motion, with both having a period of motion of two seconds.

Q. 13.4 (a) Which of the following functions of time represent

(a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give a period for each case of periodic motion ( ω is any positive constant):

sinωt−cosωt

Answer:

sin⁡ωt−cos⁡ωt=2(12sin⁡ωt−12cos⁡ωt)=2(cos⁡π4sin⁡ωt−sin⁡π4cos⁡ωt)=2sin⁡(ωt−π4)

Since the above function is of form Asin(ωt+ϕ) it represents SHM with a time period of 2πω

Q. 13.4 (b) 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):

sin3ωt

Answer:

sin3ωt=3sinωt−4sin3ωt
sin3ωt=14(3sinωt−sin3ωt)

The two functions individually represent SHM, but their superposition does not give rise to SHM. The motion will definitely be periodic with a period of 2πω

Q. 13.4 (c) Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give a period for each case of periodic motion ( ω is any positive constant):

3cos(π/4−2ωt)

Answer:

The function represents SHM with a period of πω

Q.13.4 (d) 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):

cosωt+cos3ωt+cos5ωt

Answer:

Here, each individual function is SHM. But superposition is not SHM. The function represents periodic motion but not SHM.

period=LCM(2πω,2π3ω,2π5ω)=2πω

Q. 13.4 (e) 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):

exp(−ω2t2)

Answer:

The given function is exponential and, therefore, does not represent periodic motion.

Q. 13.4 (f) Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give a period for each case of periodic motion (ω is any positive constant):

1+ωt+ω2t2

Answer:

The given function does not represent periodic motion.

Q. 13.5 (a) A particle is in linear simple harmonic motion between two points, A and B, 10cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is at the end A,

Answer:

Velocity is zero. Force and acceleration are in the positive direction.

Q. 13.5 (b) A particle is in linear simple harmonic motion between two points, A and B , 10cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is at the end B,

Answer:

Velocity is zero. Acceleration and force are negative.

Q. 13.5 (c) A particle is in linear simple harmonic motion between two points, A and B, 10cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is at the mid-point of AB going towards A

Answer:

Velocity is negative, that is, towards A, and its magnitude is maximum. Acceleration and force are zero.

Q. 13.5 (d) A particle is in linear simple harmonic motion between two points, A and B , 10cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is at 2cm away from B going towards A

Answer:

The velocity is negative. Acceleration and force are also negative.

Q.13.5 (e) A particle is in linear simple harmonic motion between two points, A and B, 10cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it isat 3cm away from A going towards B

Answer:

Velocity is positive. Acceleration and force are also positive.

Q. 13.5 (f) A particle is in linear simple harmonic motion between two points, A and B, 10cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is at 4cm away from B going towards A.

Answer:

Velocity, acceleration and force are all negative

Q. 13.6 Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?

(a) a=0.7x

(b) a=−200x2

(c) a=−10x

(d) a=100x3

Answer:

Only the relation given in (c) represents simple harmonic motion as the acceleration is proportional in magnitude to the displacement from the midpoint, and its direction is opposite to that of the displacement from the mean position.

Q. 13.7 The motion of a particle executing a simple harmonic motion is described by the displacement function, x(t)=Acos(ωt+ϕ). If the initial (t=0) position of the particle is 1cm and its initial velocity is ωcm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is πs−1. If instead of the cosine function, we choose the sine function to describe the SHM: x=Bsin(ωt+α), what are the amplitude and initial phase of the particle with the above initial conditions?

Answer:

ω=π rad s−1

x(t)=Acos(πt+ϕ)

at t = 0

x(0)=Acos(π×0+ϕ)
1=Acosϕ....(i)

v=dx(t)dt
v(t)=−Aπsin(πt+ϕ)

at t = 0

v(0)=−Aπsin(π×0+ϕ)
ω=−Aπsinϕ
1=−Asinϕ....(ii)

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

12+12=(Acosϕ)2+(−Asinϕ)2
2=A2cos2ϕ+A2sin2ϕ
2=A2
A=2

Dividing equation (ii) by (i), we get

tan⁡ϕ=−1
ϕ=3π4,7π4,11π4......

x(t)=Bsin(πt+α)

at t = 0

x(0)=Bsin(π×0+α)
1=Bsinα....(iii)

v=dx(t)dt
v(t)=Bπcos(πt+α)

at t = 0

v(0)=Bπcos(π×0+α)
ω=Bπcosα
1=Bcosα....(iv)

Squaring and adding equations (iii) and (iv), we get

12+12=(Bsinα)2+(Bcosα)2
2=B2sin2α+B2cos2α
2=B2
B=2

Dividing equation (iii) by (iv), we get

tan⁡α=1
α=π4,5π4,9π4......

Q. 13.8 A spring balance has a scale that reads from 0 to 50kg. The length of the scale is 20cm, A body suspended from this balance, when displaced and released, oscillates with a period of 0.6s. What is the weight of the body?

Answer:

The spring constant of the spring is given by

k= Weight of Maximum mass the scale can read Maximum displacement of the scale k=50×9.820×10−2k=2450Nm−1

The time period of a spring attached to a body of mass m is given by

T=2πmkm=T2k4π2m=(0.6)2×24504π2m=22.34 kgw=mgw=22.34×9.8w=218.95 N

Q. 13.9 (i) A spring having with a spring constant 1200Nm−1 is mounted on a horizontal table as shown in Figure 13.19. A mass of 3kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0cm and released.

Fig 13.19

Determine

(i) the frequency of oscillations,

Answer:

The frequency of oscillation of an object of mass m attached to a spring of spring constant k is given by

ν=12πkmν=12π×12003ν=3.183 Hz

Q. 13.9 (ii) A spring having with a spring constant 1200Nm−1 is mounted on a horizontal table as shown in Figure 13.19. A mass of 3kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0cm and released.

Fig 13.19

Determine

(ii) maximum acceleration of the mass, and

Answer:

A body executing S.H.M experiences maximum acceleration at the extreme points

amax=FAmamax=kAmamax=1200×0.23amax=8 ms−2

Q. 13.9 (iii) A spring having with a spring constant 1200Nm−1 is mounted on a horizontal table as shown in Figure 13.9. A mass of 3kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0cm and released.

Fig 13.19

Determine

(iii) the maximum speed of the mass.

Answer:

Maximum speed occurs at the mean position and is given by

vmax=Aωvmax=0.02×2π×3.18vmax=0.4 ms−1

Q. 13.10 (b) In Exercise 13.9, let us take the position of mass when the spring is unstretched as x=0, and the direction from left 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 at the maximum stretched position

Answer:

Amplitude is A = 0.02 m

Time period is ω

ω=kmω=12003ω=20rad/s (b) At t=0 the mass is at the maximum stretched position. x(0)=Aϕ=π2x(t)=0.02sin⁡(20t+π2)x(t)=0.02cos⁡(20t)

Here x is in metres and t is in seconds.

Q. 13.10 (c) In Exercise 13.9, let us take the position of mass when the spring is unstretched as x=0, and the direction from left 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 at the maximum compressed position.

Answer:

Amplitude is A = 0.02 m

Time period is ω

ω=kmω=12003ω=20rad/s (c) At t=0 the mass is at the maximum compressed position. x(0)=−Aϕ=3π2x(t)=0.02sin⁡(20t+3π2)x(t)=−0.02cos⁡(20t)

Here x is in metres and t is in seconds.

The above functions differ only in the initial phase and not in amplitude or frequency.

Q. 13.11 Figures 13.20 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 anti-clockwise) are indicated on each figure.

Fig 13.20

Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.

Answer:

(a) Let the required function be x(t)=asin⁡(±ωt+ϕ)

Amplitude =3 cm=0.03 m
T=2 s

ω=2πTω=πrad s

Since initial position x(t)=0,ϕ=0
As the sense of revolution is clockwise

x(t)=0.03sin⁡(−ωt)x(t)=−0.03sin⁡(πt)

Here x is in metres and t is in seconds.
(b) Let the required function be x(t)=asin⁡(±ωt+ϕ)

Amplitude =2 m

T=4 s

ω=2πTω=π2rad s

Since initial position x(t)=−A,ϕ=3π2
As the sense of revolution is anti-clockwise

x(t)=2sin⁡(ωt+3π2)x(t)=−2cos⁡(π2t)

Here x is in metres and t is in seconds.

Q. 13.12 (a) Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t=0) 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 fixed to be anticlockwise in every case: (x is in cm and t is in s).

x=−2sin(3t+π/3)

Answer:

x=−2sin⁡(3t+π/3)x=2cos⁡(3t+π3+π2)x=2cos⁡(3t+5π6)

The initial position of the particle is x(0)

x(0)=2cos⁡(0+5π6)x(0)=2cos⁡(5π6)x(0)=−3 cm

The radius of the circle i.e. the amplitude is 2 cm
The angular speed of the rotating particle is ω=3rads−1
Initial phase is

ϕ=5π6ϕ=150∘

The reference circle for the given simple Harmonic motion is

Q. 13.12 (b) Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t=0) 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 fixed to be anticlockwise in every case: (x is in cm and t is in s).

x=cos(π/6−t)

Answer:

x(t)=cos(π6−t)
x(t)=cos(t−π6)

The initial position of the particle is x(0)

x(0)=cos(0−π6)
x(0)=cos(π6)
x(0)=32cm

The radius of the circle i.e. the amplitude is 1 cm

The angular speed of the rotating particle is ω=1rad s−1

Initial phase is

ϕ=−π6ϕ=−30o

The reference circle for the given simple Harmonic motion is

Q. 13.12 (c) Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t=0) 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 fixed to be anticlockwise in every case: (x is in cm and t is in s).

x=3sin(2πt+π/4)

Answer:

x=3sin⁡(2πt+π/4)x=−3cos⁡(2πt+π4+π2)=3cos⁡(2πt+π+π4+π2)=3cos⁡(2πt+3π2+π4)=3cos⁡(2πt+7π4)

At t= 0

phase=7π4

The reference circle is as follows

Q. 13.12 (d) Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t=0) 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 fixed to be anticlockwise in every case: (x is in cm and t is in s).

x=2cosπt

Answer:

x(t)=2cos(πt)

The initial position of the particle is x(0)

x(0)=2cos(0)
x(0)=2cm

The radius of the circle i.e. the amplitude, is 2 cm

The angular speed of the rotating particle is ω=πrad s−1

Initial phase is

ϕ=0o

The reference circle for the given simple Harmonic motion is

Q. 13.14 The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0m If the piston moves with simple harmonic motion with an angular frequency of 200rad/min, what is its maximum speed?

Answer:

Amplitude of SHM = 0.5 m

angular frequency is

ω=200rad/minω=3.33rad/s

If the equation of SHM is given by

x(t)=Asin⁡(ωt+ϕ)

The velocity would be given by

v(t)=dx(t)dtv(t)=d(Asin⁡(ωt+ϕ))dtv(t)=Aωcos⁡(ωt+ϕ)

The maximum speed is therefore

vmax=Aωvmax=0.5×3.33vmax=1.67 ms−1

Q. 13.15 The acceleration due to gravity on the surface of moon is 1.7ms−2 What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5s ? (g on the surface of earth is 9.8 m s–2)

Answer:

The time period of a simple pendulum of length l executing S.H.M is given by

T=2πlg ge=9.8 m s−2 gm=1.7 m s−2

The time period of the pendulum on the surface of Earth is Te=3.5 s
The time period of the pendulum on the surface of the moon is Tm

TmTe=gegmTm=Te×gegmTm=3.5×9.81.7Tm=8.4 s

Q .13.16 A simple pendulum of length l 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:

Acceleration due to gravity = g (in downward direction)

Centripetal acceleration due to the circular movement of the car = a c

ac=v2R (in the horizontal direction)

Effective acceleration is

g′=g2+ac2

g′=g2+v4R2

The time period is T'

T′=2πlg′
T′=2πlg2+v4R2

Q. 13.17 A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρı . The cork is depressed slightly and then released

Show that the cork oscillates up and down simply harmonically with a period T=2πhρρ1g where ρ is the density of the cork. (Ignore damping due to the viscosity of the liquid).

Answer:

Let the cork be displaced by a small distance x in the downward direction from its equilibrium position, where it is floating.

The extra volume of fluid displaced by the cork is Ax

Taking the downwards direction as positive, we have

ma=−ρ1gAxρAha=−ρ1gAx d2x dt2=−ρ1gρhx Comparing with a=−kx we have k=ρ1gρhT=2πkT=2πρhρ1g

Q. 13.18 One end of a U-tube containing mercury is connected to a suction pump and the other end to the 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 U-tube executes a simple harmonic motion.

Answer:

Let the height of each mercury column be h.

The total length of mercury in both columns = 2h.

Let the cross-sectional area of the mercury column be A.

Let the density of mercury be ρ

When either of the mercury columns dips by a distance x, the total difference between the two columns becomes 2x.

The weight of this difference is 2Axρg

This weight drives the rest of the column to the original mean position.

Let the acceleration of the column be a. Since the force is restoring

2hAρ(−a)=2xAρga=−ghx

d2xdt2=−ghx which is the equation of a body executing S.H.M

The time period of the oscillation would be

T=2πhg

Q. 2 (a) You are riding in an automobile of mass 3000kg. . Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50o/o during one complete oscillation. Estimate the values of the spring constant K

Answer:

Mass of automobile (m) = 3000 kg

There are a total of four springs.

Compression in each spring, x = 15 cm = 0.15 m

Let the spring constant of each spring be k

4kx=mg
k=3000×9.84×0.15
k=4.9×104 N

Q. 2 (b) You are riding in an automobile of mass 3000kg . Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50o/o during one complete oscillation. Estimate the values of the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750kg. .

Answer:

The amplitude of oscillation decreases by 50 % in one oscillation i.e. in one time period.

T=2πmkT=2π×30004×4.9×104T=0.77 s

For the damping factor b, we have

x=x0e(−bt2m)x=x0/2t=0.77 s m=750 kge−0.75%2×750=0.5ln⁡(e−0.7752×780)=ln⁡0.50.77b1500=ln⁡2b=0.693×15000.77b=1350.2287 kg s−1

Q .4 A circular disc of mass 10kg 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 1.5s. The radius of the disc is 15cm . Determine the torsional spring constant of the wire. (Torsional spring constant a is defined by the relation J=−aθ , where J is the restoring couple and θ the angle of twist).

Answer:

J=−aθ

Moment of Inertia of the disc about the axis passing through its centre and perpendicular to it is

I=MR22J=I d2θ dt2−aθ=MR22 d2θ dt2 d2θ dt2=−2aMR2θ

The period of Torsional oscillations would be

T=2πMR22aa=2π2MR2T2a=2π2×10×(0.15)2(1.5)2a=1.97 N mrad−1

Q. 5 (a) A body describes simple harmonic motion with an amplitude of 5cm and a period of 0.2s. Find the acceleration and velocity of the body when the displacement is 5cm

Answer:

A=5 cm=0.05 m T=0.2 sω=2πTω=2π0.2ω=10πrad s−1

At displacement x acceleration is a=−ω2x

At displacement x velocity is v=ωA2−x2

(a)At displacement 5 cm

v=10π(0.05)2−(0.05)2v=0a=−(10π)2×0.05a=−49.35ms−2

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

Answer:

A=5 cm=0.05 m T=0.2 sω=2πTω=2π0.2ω=10πrad s−1

At displacement x, acceleration is a=−ω2x

At displacement x, velocity is v=ωA2−x2

(b)At displacement 3 cm

v=10π(0.05)2−(0.03)2v=10π0.0016v=10π×0.04v=1.257 ms−1a=−(10π)2×0.03a=−29.61 ms−2

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

Answer:

A=5 cm=0.05 m T=0.2 sω=2πTω=2π0.2ω=10πrad s−1

At displacement x, acceleration is a=−ω2x

At displacement x, velocity is v=ωA2−x2

(c)At displacement 0 cm

v=10π(0.05)2−(0)2v=10π×0.05v=1.57 ms−1a=−(10π)2×0a=0

Q. 6 A mass attached to a spring is free to oscillate, with angular velocity ω , in a horizontal plane without friction or damping. It is pulled to distance x0 and pushed towards the centre with a velocity v0 at time t=0 Determine the amplitude of the resulting oscillations in terms of the parameters ω , x0 and v0 . [Hint : Start with the equation x=acos(ωt+θ) and note that the initial velocity is negative.]

Answer:

At the maximum extension of the spring, the entire energy of the system would be stored as the potential energy of the spring.

Let the amplitude be A

12kA2=12mv02+12kx02A=x02+mkv02

The angular frequency of a spring-mass system is always equal to km
Therefore

A=x02+v02ω2