question 14.1: which of the following examples represent...

Question 14.1:

Which of the following examples represent periodic motion?

A swimmer completing one (return) trip from one bank of a river to the other and back.

A freely suspended bar magnet displaced from its N

A hydrogen molecule rotating about its center of mass.

An arrow released from a bow.

Answer

Answer: (b) and (c)

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 atime taken by the swimmer during his back and forth journey may not be the same.

The motion of a freely-suspended magnet, if displaced from its Nreleased, is periodic. This is because the magnet oscilldefinite period of time.

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.

An arrow released from a bow moves onbackward. Hence, this motion is not a periodic.

Question 14.2:

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

the rotation of earth about its axis.

motion of an oscillating mercury column in a U

motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point.

Which of the following examples represent periodic motion?


A freely suspended bar magnet displaced from its N-S direction and released.

A hydrogen molecule rotating about its center of mass.

An arrow released from a bow.

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 definite period. This is because the time taken by the swimmer during his back and forth journey may not be the same.

suspended magnet, if displaced from its N-S direction and released, is periodic. This is because the magnet oscillates about its position with a

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.

An arrow released from a bow moves only in the forward direction. It does not come backward. Hence, this motion is not a periodic.

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

rotation of earth about its axis.

motion of an oscillating mercury column in a U-tube.

motion of a ball bearing inside a smooth curved bowl, when released from a point slightly


S direction and released.

The swimmer’s motion is not periodic. The motion of the swimmer between the banks of definite period. This is because the

time taken by the swimmer during his back and forth journey may not be the same.

S direction and ates about its position with a

When a hydrogen molecule rotates about its centre of mass, it comes to the same position

ly in the forward direction. It does not come

Which of the following examples represent (nearly) simple harmonic motion and which

motion of a ball bearing inside a smooth curved bowl, when released from a point slightly

GOD

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GOD

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UNIT-10 (CHAPTERS - 14 & 15)

general vibrations of a polyatomic molecule about its equilibrium position.

Answer

Answer: (b) and (c) are SHMs

and (d) are periodic, but not SHMs

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.

An oscillating mercury column in a Umercury moves to and fro on the same path, about the fixed position, with a certain period of time.

The ball moves to and fro about the lowermost point of the bowl when released. Also, the ball comes back to its initial position in the same period of time, again and again. Hence,its motion is periodic as well as simple harmonic.

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.

Question 14.3:

Figure 14.27 depicts four x-t represent periodic motion? What is the period of motion (in case of periodic motion)?

(a)

(b)

vibrations of a polyatomic molecule about its equilibrium position.

are SHMs

are periodic, but not SHMs

During its rotation about its axis, earth comes to the same position again and again in 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.

An oscillating mercury column in a U-tube is simple harmonic. This is because the on the same path, about the fixed position, with a certain period

The ball moves to and fro about the lowermost point of the bowl when released. Also, the ball comes back to its initial position in the same period of time, again and again. Hence,its motion is periodic as well as simple harmonic.

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.

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)?

During its rotation about its axis, earth comes to the same position again and again in 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.

tube is simple harmonic. This is because the on the same path, about the fixed position, with a certain period

The ball moves to and fro about the lowermost point of the bowl when released. Also, the ball comes back to its initial position in the same period of time, again and again. Hence,

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.

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)?

(c)

(d)

Answer

Answer: (b) and (d) are periodic

It is not a periodic motion. This represents a unidirectional, linear uniform motion. There is no repetition of motion in this case.

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.

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.

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.

Question 14.4:

Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) nonmotion (ω is any positive constant):

sin ωt – cos ωt

sin3 ωt

3 cos (π/4 – 2ωt)

cos ωt + cos 3ωt + cos 5ωt

exp (–ω2t2)

1 + ωt + ω2t2

Answer

SHM

The given function is:

This function represents SHM as

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):

This function represents SHM as it can be written in the form:

Which of the following functions of time represent (a) simple harmonic, (b) periodic but motion? Give period for each case of periodic

Its period is:

Periodic, but not SHM


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.

SHM


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

Its period is:

Periodic, but not SHM

The given function is represents SHM. However, the periodic, but not simple harmonic.

Non-periodic motion

The given function repeat themselves. Therefore, it is a non

The given function 1 + ωt + ω

individually represent simple harmonic motion (SHM). However, the superposition of two SHM is periodic and not simple harmonic.


. Each individual cosine function represents SHM. However, the superposition of three simple harmonic motions is periodic, but not simple harmonic.

is an exponential function. Exponential functions do not repeat themselves. Therefore, it is a non-periodic motion.

ω2t2 is non-periodic.

individually represent simple harmonic motion (SHM). However, the superposition of two SHM is periodic and not simple harmonic.


. Each individual cosine function superposition of three simple harmonic motions is

is an exponential function. Exponential functions do not

Question 14.5:

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 direction and give the signs of velocity, acceleration and force on the particle when it is

at the end A,

at the end B,

at the mid-point of AB going towards A,

at 2 cm away from B going towards A,

at 3 cm away from A going towards B, and

at 4 cm away from B going towards A.

Answer

Zero, Positive, Positive

Zero, Negative, Negative

Negative, Zero, Zero

Negative, Negative, Negative

Zero, Positive, Positive

Negative, Negative, Negative

Explanation:

The given situation is shown in the following figure. 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 end points

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 dire

Force is also positive in this case as the particle is directed rightward.

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 B.

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 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 acceleration, and force are all positive.

(f)

This case is similar to the one given in

Its acceleration is positive as it is directed along AO.

Force is also positive in this case as the particle is directed rightward.

At the extreme point B, the particle is at rest momentarily. Hence, its velocity is zero at

Its acceleration is negative as it is directed along B.

is also negative in this case as the particle is directed leftward.

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

the particle is directed leftward. The acceleration and force of a particle executing SHM is zero at the mean position.

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.

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.

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

At the extreme point B, the particle is at rest momentarily. Hence, its velocity is zero at

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

the particle is directed leftward. The acceleration and force of a particle

The particle is moving toward point O from the end B. This direction of motion is direction, which is from A to B. Hence, the

The particle is moving toward point O from the end A. This direction of motion is from A direction. Hence, the values for velocity,

Question 14.6:

Which of the following relationships between the acceleration a particle involve simple harmonic motion?

a = 0.7x

a = –200x2

a = –10x

a = 100x3

Answer

A motion represents simple harmonic motion if it is governed by the force law:

F = –kx

ma = –k

Where,

F is the force

m is the mass (a constant for a body)

x is the displacement

a is the acceleration

k is a constant

Among the given equations, only equation

Hence, this relation represents SHM.

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


is the mass (a constant for a body)

Among the given equations, only equation a = –10 x is written in the above form with

Hence, this relation represents SHM.

and the displacement x of


is written in the above form with

Question 14.7:

The motion of a particle executing simple harmonic motion is described by the displacement function,

x (t) = A cos (ωt + φ).

If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is π s–1. If instead of the cosine function, we choose the sine function to describe the SHM: x = B sin (ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.

Answer

Initially, at t = 0:

Displacement, x = 1 cm

Initial velocity, v = ω cm/sec.

Angular frequency, ω = π rad/s–1

It is given that:

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

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

SHM is given as:

Putting the given values in this equation, we get:

Velocity,

Substituting the given values, we get:

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

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

Question 14.8:

A spring balance has a scale that reads from 0 to 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,

Maximum displacement of the spring = Length of the scale,

Time period, T = 0.6 s

Maximum force exerted on the spring,

Where,

g = acceleration due to gravity = 9.8 m/s

F = 50 × 9.8 = 490

∴Spring constant,

Mass m, is suspended from the balance.

Time period,

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?

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

Maximum force exerted on the spring, F = Mg

= acceleration due to gravity = 9.8 m/s2

om the balance.

50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a

= 20 cm = 0.2 m

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

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

Question 14.9:

A spring having with a spring constant 1200 N mshown in Fig. 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 (iii) the maximum speed of the mass.

Answer

Spring constant, k = 1200 N m

Mass, m = 3 kg

Displacement, A = 2.0 cm = 0.02 cm

Frequency of oscillation v,is given by the relation:

Where, T is the time period

= 22.36 × 9.8 = 219.167 N

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

A spring having with a spring constant 1200 N m–1 is mounted on a horizontal table as shown in Fig. 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.

= 1200 N m–1

= 2.0 cm = 0.02 cm

,is given by the relation:

is mounted on a horizontal table as shown in Fig. A mass of 3 kg is attached to the free end of the spring. The mass is then

the mass, and

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

Maximum acceleration (a) is given by the relation:

a = ω2 A

Where,

ω = Angular frequency =

A = Maximum displacement

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

Maximum velocity, vmax = Aω

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

Question 14.10:

In Exercise 14.9, let us take the position of mass when the spring is unstreched as and the direction from left to right as the positive direction of of time t for the oscillating mass if at the moment we start the stopwatch (is

at the mean position,

at the maximum stretched position, and

at the maximum compressed position.

In what way do these functions for SHM differ from each other, in frequenamplitude or the initial phase?

oscillations is 3.18 m/s.

) is given by the relation:

Hence, the maximum acceleration of the mass is 8.0 m/s2.

ω

velocity of the mass is 0.4 m/s.

In Exercise 14.9, let us take the position of mass when the spring is unstreched as and the direction from left to right as the positive direction of x-axis. Give x as a function

for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass

at the maximum stretched position, and

at the maximum compressed position.

In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

In Exercise 14.9, let us take the position of mass when the spring is unstreched as x = 0, as a function

= 0), the mass

cy, in

Answer

x = 2sin 20t

x = 2cos 20t

x = –2cos 20t

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

Distance travelled by the mass sideways, A = 2.0 cm

Force constant of the spring, k = 1200 N m–1

Mass, m = 3 kg

Angular frequency of oscillation:

= 20 rad s–1

When the mass is at the mean position, initial phase is 0.

Displacement, x = Asin ωt

= 2sin 20t

At the maximum stretched position, the mass is toward the extreme right. Hence, the

initial phase is .

Displacement,

= 2cos 20t

At the maximum compressed position, the mass is toward the extreme left. Hence, the

initial phase is .

Displacement,

= –2cos 20

The functions have the same frequency

initial phases .

Question 14.11:

Figures 14.29 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

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

Answer

Time period, T = 2 s

Amplitude, A = 3 cm

2cos 20t

The functions have the same frequency and amplitude (2 cm), but different

Figures 14.29 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 figure.

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

and amplitude (2 cm), but different

Figures 14.29 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-

projection of the radius

At time, t = 0, the radius vector OP makes an angle

angle

Therefore, the equation of simple harmonic motion for thegiven by the displacement equation:

Time period, T = 4 s

Amplitude, a = 2 m

At time t = 0, OP makes an angle π with the phase angle, Φ = + π

Therefore, the equation of simple harmonic motion for the given as:

Question 14.12:

Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (

x = –2 sin (3t + π/3)

x = cos (π/6 – t)

= 0, the radius vector OP makes an angle with the positive x-axis, i.e., phase

Therefore, the equation of simple harmonic motion for the x-projection of OP, at timegiven by the displacement equation:

= 0, OP makes an angle π with the x-axis, in the anticlockwise direction. Hence,

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

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

nticlockwise in every case: (x is in cm and t is in s).

axis, i.e., phase

projection of OP, at time t, is

axis, in the anticlockwise direction. Hence,

projection of OP, at time t, is

reference circle for each of the following simple harmonic = 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

x = 3 sin (2πt + π/4)

x = 2 cos πt

Answer

If this equation is compared with the standard SHM equation , then we get:

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




Amplitude, A = 3 cm

Phase angle, = 135°

Angular velocity,


x = 2 cos πt

If this equation is compared with the standard SHM equationwe get:

Amplitude, A = 2 cm

Phase angle, Φ = 0

Angular velocity, ω = π rad/s


Question 14.13:

Figure 14.30 (a) shows a spring of force constant m attached to its free end. A force 14.30 (b) shows the same spring with both ends free and attached to a mass


If this equation is compared with the standard SHM equation

Angular velocity, ω = π rad/s


Figure 14.30 (a) shows a spring of force constant k clamped rigidly at one end and a mass attached to its free end. A force F applied at the free end stretches the spring. Figure

14.30 (b) shows the same spring with both ends free and attached to a mass m

, then

clamped rigidly at one end and a mass applied at the free end stretches the spring. Figure

m at either

end. Each end of the spring in Fig. 14.30(b) is stretched by the same force F.

What is the maximum extension of the spring in the two cases?

If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case?

Answer

For the one block system:

When 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,

For the two block system:

The displacement (x) produced in this case is:

Net force, F = +2 kx

For the one block system:

For mass (m) of the block, force is written as:

Where, x is the displacement of the block in time t

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

Where,

ω is angular frequency of the oscillation

∴Time period of the oscillation,

For the two block system:

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

Where,

Angular frequency,

∴Time period,

Question 14.14:

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

Stroke = 1.0 m

Amplitude,

The maximum speed (vmax) of the piston is give by the relation:

Question 14.15:

The acceleration due to gravity on the surface of moon is 1.7 msperiod of a simple pendulum on the surface of moon if its time period on the surface of

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?

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

) of the piston is give by the relation:

The acceleration due to gravity on the surface of moon is 1.7 ms–2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of

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

. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of

earth is 3.5 s? (g on the surface of earth is 9.8 ms

Answer

Acceleration due to gravity on the surface of moon,

Acceleration due to gravity on the surface of earth,

Time period of a simple pendulum on earth,

Where,

l is the length of the pendulum

The length of the pendulum remains constant.

On moon’s surface, time period,

Hence, the time period of the simple pendulum on the surface of moon is 8.4 s.

Question 14.16:

on the surface of earth is 9.8 ms–2)

Acceleration due to gravity on the surface of moon, = 1.7 m s–2

to gravity on the surface of earth, g = 9.8 m s–2

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

is the length of the pendulum

The length of the pendulum remains constant.

On moon’s surface, time period,

the simple pendulum on the surface of moon is 8.4 s.the simple pendulum on the surface of moon is 8.4 s.

Answer the following questions:

Time period of a particle in SHM depends on the force constant k and mass m of the particle:

. A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?

The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is

greater than . Think of a qualitative argument to appreciate this result.

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?

What is the frequency of oscillation of a simple pendulum mounted in a cabinthat is freely falling under gravity?

Answer

The time period of a simple pendulum,

For a simple pendulum, k is expressed in terms of mass m, as:

k m

= Constant

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

In the case of a simple pendulum, the restoring force acting on the bob of the pendulum is given as:

F = –mg sinθ

Where,

F = Restoring force

m = Mass of the bob

g = Acceleration due to gravity

θ = Angle of displacement

For small θ, sinθ

For large θ, sinθ is greater than

This decreases the effective value of

Hence, the time period increases as:

Where, l is the length of the simple pendulum

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.

When a simple pendulum mounted in a cabin falls freely under zero. Hence the frequency of oscillation of this simple pendulum is zero.

Question 14.17:

A simple pendulum of length is moving on a circular track of radius 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

is greater than θ.

This decreases the effective value of g.

Hence, the time period increases as:

is the length of the simple pendulum

The time shown by the wristwatch of a man falling from the top of a tower is not affected 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

When a simple pendulum mounted in a cabin falls freely under gravity, its acceleration is zero. Hence the frequency of oscillation of this simple pendulum is zero.

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

The bob of the simple pendulum will experience the acceleration due to gravity and the etal acceleration provided by the circular motion of the car.

The time shown by the wristwatch of a man falling from the top of a tower is not affected 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

gravity, its acceleration is

is suspended in a car. The car . If the pendulum makes

small oscillations in a radial direction about its equilibrium position, what will be its time

The bob of the simple pendulum will experience the acceleration due to gravity and the

Acceleration due to gravity =

Centripetal acceleration

Where,

v is the uniform speed of the car

R is the radius of the track

Effective acceleration (aeff) is given as:

Time period,

Where, l is the length of the pendulum

∴Time period, T

Question 14.18:

Cylindrical piece of cork of density of base area

density . The cork is depressed slightly and then released. Show that the cork up and down simple harmonically with a period

where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).

Answer

Acceleration due to gravity = g

is the uniform speed of the car

) is given as:

is the length of the pendulum

Cylindrical piece of cork of density of base area A and height h floats in a liquid of

. The cork is depressed slightly and then released. Show that the cork up and down simple harmonically with a period

is the density of cork. (Ignore damping due to viscosity of the liquid).

floats in a liquid of

. The cork is depressed slightly and then released. Show that the cork oscillates

is the density of cork. (Ignore damping due to viscosity of the liquid).

Base area of the cork = A

Height of the cork = h

Density of the liquid =

Density of the cork = ρ

In equilibrium:

Weight of the cork = Weight of the liquid displaced by the floating cork

Let the cork be depressed slightly by x. As a result, some extra water of a certain volume is displaced. Hence, an extra up-thrust acts upward and provides the restoring force to the cork.

Up-thrust = Restoring force, F = Weight of the extra water displaced

F = –(Volume × Density × g)

Volume = Area × Distance through which the cork is depressed

Volume = Ax

∴ F = – A x g … (i)

According to the force law:

F = kx

Where, k is a constant

The time period of the oscillations of the cork:

Where,

m = Mass of the cork

= Volume of the cork × Density

= Base area of the cork × Height of the cork × Density of the cork

= Ahρ

Hence, the expression for the time period becomes:

Question 14.19:

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 simple harmonic motion.

Answer

Area of cross-section of the U

Density of the mercury column =

Acceleration due to gravity =

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

F = –(Volume × Density × g)

F = –(A × 2h × ρ ×g) = –2Aρ

Where,

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

k is a constant, given by

= Volume of the cork × Density

= Base area of the cork × Height of the cork × Density of the cork

Hence, the expression for the time period becomes:

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 U-tube executes

section of the U-tube = A

Density of the mercury column = ρ

Acceleration due to gravity = g

= Weight of the mercury column of a certain height

)

Aρgh = –k × Displacement in one of the arms (h)

is the height of the mercury column in the two arms

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

tube executes

Time period,

Where,

m is the mass of the mercury column

Let l be the length of the total mercury in the U

Mass of mercury, m = Volume of mercury × Density of mercury

= Alρ

∴

Hence, the mercury column executes simple harmonic motion with time period

Question 14.20:

An air chamber of volume V m just fits 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 pressureisothermal [see Fig. 14.33].

Answer

is the mass of the mercury column

be the length of the total mercury in the U-tube.

= Volume of mercury × Density of mercury

Hence, the mercury column executes simple harmonic motion with time period

V has a neck area of cross section a into which a ball of mass just fits 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 oscillations assuming pressure-volume variations of air to be

Hence, the mercury column executes simple harmonic motion with time period .

into which a ball of mass just fits 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 volume variations of air to be

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

Volumetric strain

Bulk Modulus of air,

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

The restoring force acting on the ball,

F = p × a

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:

Time period,

Question 14.21:

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 byduring one complete oscillation. Estimate the values of (a) the spring constant the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.

Answer

Mass of the automobile, m = 3000 kg

Displacement in the suspension system,

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 sus

Time period,

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 byduring one complete oscillation. Estimate the values of (a) the spring constant

for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.

= 3000 kg

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

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

The equation for the restoring force for the system:

is the spring constant of the suspension system

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)

for the spring and shock absorber system of one wheel, assuming

And = 5000 = 5 × 10

Spring constant, k = 5 × 104 N/m

Each wheel supports a mass,

For damping factor b, the equation for displacement is written as:

The amplitude of oscillation decreases by 50%.

∴

Where,

Time period,

= 1351.58 kg/s

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

Question 14.22:

Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average

= 5000 = 5 × 104 N/m

N/m

Each wheel supports a mass, M = = 750 kg

, the equation for displacement is written as:

The amplitude of oscillation decreases by 50%.

= 0.7691 s

= 1351.58 kg/s

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

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.Show that for a particle in linear SHM the average kinetic energy over a period of

Answer

The equation of displacement of a particle executing SHM at an instant t is given as:

Where,

A = Amplitude of oscillation

ω = Angular frequency

The velocity of the particle is:

The kinetic energy of the particle is:

The potential energy of the particle is:

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

And, average potential energy over one cycle is given as:

It can be inferred from equations (time period is equal to the average potential energy for the same time period.

Question 14.23:

A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is

And, average potential energy over one cycle is given as:

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


hat the average kinetic energy for a given time period is equal to the average potential energy for the same time period.


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 α is defined by the relation restoring couple and θ 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,

The moment of inertia of the disc is:

I

= × (10) × (0.15)2

= 0.1125 kg m2

Time period,

α is the torsional constant.

= 1.972 Nm/rad

Hence, the torsional spring constant of the wire is 1.972 Nm rad

Question 14.24:

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 α is defined by the relation J = –α θ, where J

the angle of twist).

= 10 kg

= 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:

Hence, the torsional spring constant of the wire is 1.972 Nm rad–1.

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

J is the

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 = 5 cm = 0.05 m

Time period, T = 0.2 s

For displacement, x = 5 cm = 0.05 m

Acceleration is given by:

a = –

Velocity is given by:

v

= 0

When the displacement of the body is 5 cm, its acceleration is –5π2 m/s2 and velocity is 0.

For displacement, x = 3 cm = 0.03 m


a =


v

= 0.4 π m/s

When the displacement of the body is 3 cm, its acceleration is –3π m/s2 and velocity is 0.4π m/s.

For displacement, x = 0


a = 0


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

Question 14.25:

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 a distance centre with a velocity v0 at time oscillations in terms of the parameters ω, cos (ωt+θ) and note that the initial velocity is negative.]

Answer

The displacement equation for an oscillating mass is given by:

x =

Where,

A is the amplitude

x is the displacement

θ is the phase constant

Velocity,

At t = 0, x = x0

x0 = Acosθ = x0 … (i)

And,


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 a distance x0 and pushed towards the

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

) and note that the initial velocity is negative.]

The displacement equation for an oscillating mass is given by:


A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal and pushed towards the

itude of the resulting . [Hint: Start with the equation x = a

… (ii)

Squaring and adding equations (

Hence, the amplitude of the resulting oscillation is

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

Hence, the amplitude of the resulting oscillation is .

Question 15.1:

A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?

Answer

Mass of the string, M = 2.50 kg

Tension in the string, T = 200 N

Length of the string, l = 20.0 m

Mass per unit length,

The velocity (v) of the transverse wave in the string is given by the relation:

∴Time taken by the disturbance to reach the other end

Question 15.2:

A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 m s–1? (g= 9.8 m s

Answer

A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?

= 2.50 kg

= 200 N

= 20.0 m

) of the transverse wave in the string is given by the relation:

Time taken by the disturbance to reach the other end, t =

A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed

? (g= 9.8 m s–2)

A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the

A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed

Height of the tower, s = 300 m

Initial velocity of the stone, u

Acceleration, a = g = 9.8 m/s

Speed of sound in air = 340 m/s

The time ( ) taken by the stone to strike the water in the pond can be calculated using the second equation of motion, as:

Time taken by the sound to reach the top of the tower,

Therefore, the time after which the splash is heard,

= 7.82 + 0.88 = 8.7 s

Question 15.3:

A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20 °C = 343 m s–1.

Answer

Length of the steel wire, l = 12 m

Mass of the steel wire, m = 2.10 kg

= 300 m

u = 0

= g = 9.8 m/s2

Speed of sound in air = 340 m/s

) taken by the stone to strike the water in the pond can be calculated using the ion, as:

Time taken by the sound to reach the top of the tower,

Therefore, the time after which the splash is heard,

A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the the wire so that speed of a transverse wave on the wire equals the speed of sound in dry

= 12 m

= 2.10 kg

) taken by the stone to strike the water in the pond can be calculated using the

A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry

Velocity of the transverse wave,

Mass per unit length,

For tension T, velocity of the transverse wave can be obtained using the relation:

∴T = v2 µ

= (343)2 × 0.175 = 20588.575

Question 15.4:

Use the formula to explain why the speed of sound in air

is independent of pressure,

increases with temperature,

increases with humidity.

Answer

Take the relation:

Velocity of the transverse wave, v = 343 m/s

, velocity of the transverse wave can be obtained using the relation:

× 0.175 = 20588.575 ≈ 2.06 × 104 N

to explain why the speed of sound in air

, velocity of the transverse wave can be obtained using the relation:

Now from the ideal gas equation for n = 1:

PV = RT

For constant T, PV = Constant

Since both M and γ are constants, v = Constant

Hence, at a constant temperature, the speed of sound in a gaseous medium is independent of the change in the pressure of the gas.

Take the relation:

For one mole of an ideal gas, the gas equation can be written as:

PV = RT

P = … (ii)

Substituting equation (ii) in equation (i), we get:

Where,

Mass, M = ρV is a constant

γ and R are also constants

We conclude from equation (

Hence, the speed of sound in a gas is directly proportional to the square root of the temperature of the gaseous medium, i.e., thein the temperature of the gaseous medium and vice versa.

Let be the speeds of sound in moist air and dry air respectively.

Let be the densities of moist air and dry air respectively.

Take the relation:

Hence, the speed of sound in moist air is:

And the speed of sound in dry air is:

On dividing equations (i) and (

However, the presence of water vapour reduces the density of air, i.e.,

Hence, the speed of sound in moist air is greater than medium, the speed of sound increases with humidity.

We conclude from equation (iv) that .

Hence, the speed of sound in a gas is directly proportional to the square root of the temperature of the gaseous medium, i.e., the speed of the sound increases with an increase in the temperature of the gaseous medium and vice versa.

be the speeds of sound in moist air and dry air respectively.

be the densities of moist air and dry air respectively.

, the speed of sound in moist air is:

And the speed of sound in dry air is:

) and (ii), we get:

However, the presence of water vapour reduces the density of air, i.e.,

Hence, the speed of sound in moist air is greater than it is in dry air. Thus, in a gaseous medium, the speed of sound increases with humidity.

Hence, the speed of sound in a gas is directly proportional to the square root of the speed of the sound increases with an increase

it is in dry air. Thus, in a gaseous

Question 15.5:

You have learnt that a travelling wave in one dimension is represented by a function y = f (x, t)where x and t must appear in the combination x – v t or x + v t, i.e. y = f (x ± v t). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave:

(x – vt)2

Answer

Answer: No;

Does not represent a wave

Represents a wave

Does not represent a wave

The converse of the given statement is not true. The essential requirement for a function to represent a travelling wave is that it should remain finite for all values of x and t.

Explanation:

For x = 0 and t = 0, the function (x – vt)2 becomes 0.

Hence, for x = 0 and t = 0, the function represents a point and not a wave.

For x = 0 and t = 0, the function

Since the function does not converge to a finite value for x = 0 and t = 0, it represents a travelling wave.

For x = 0 and t = 0, the function

Since the function does not converge to a finite value for represent a travelling wave.

Question 15.6:

A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air is 340 m s

Answer

Frequency of the ultrasonic sound, ν = 1000 kHz = 10

Speed of sound in air, va = 340 m/s

The wavelength (λr) of the reflected sound is given by the relation:

Frequency of the ultrasonic sound, ν = 1000 kHz = 10

Speed of sound in water, vw = 1486 m/s

The wavelength of the transmitted sound is given as:

= 1.49 × 10–3 m

= 0, the function

Since the function does not converge to a finite value for x = 0 and t = 0, it does not

A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water wavelength of (a) the reflected sound, (b) the transmitted sound?

Speed of sound in air is 340 m s–1 and in water 1486 m s–1.

Frequency of the ultrasonic sound, ν = 1000 kHz = 106 Hz

= 340 m/s

reflected sound is given by the relation:

Frequency of the ultrasonic sound, ν = 1000 kHz = 106 Hz

= 1486 m/s

The wavelength of the transmitted sound is given as:

= 0, it does not

A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water wavelength of (a) the reflected sound, (b) the transmitted sound?

Question 15.7:

A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km soperating frequency of the scanner is 4.2 MHz.

Answer

Speed of sound in the tissue,

Operating frequency of the scanner, ν = 4.2 MHz = 4.2

The wavelength of sound in the tissue is given as:

Question 15.8:

A transverse harmonic wave on a string is described by

Where x and y are in cm and

Is this a travelling wave or a stationary wave?

If it is travelling, what are the speed and direction of its propagation?

What are its amplitude and frequency?

What is the initial phase at the

What is the least distance between two successive crests in the wave?

Answer

ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s–1? The operating frequency of the scanner is 4.2 MHz.

Speed of sound in the tissue, v = 1.7 km/s = 1.7 × 103 m/s

Operating frequency of the scanner, ν = 4.2 MHz = 4.2 × 106 Hz

The wavelength of sound in the tissue is given as:

A transverse harmonic wave on a string is described by

are in cm and t in s. The positive direction of x is from left to right.

Is this a travelling wave or a stationary wave?

If it is travelling, what are the speed and direction of its propagation?

What are its amplitude and frequency?

What is the initial phase at the origin?

What is the least distance between two successive crests in the wave?

ultrasonic scanner to locate tumours in a tissue. What is the ? The

is from left to right.

Answer:

Yes; Speed = 20 m/s, Direction = Right to left

3 cm; 5.73 Hz

3.49 m

Explanation:

The equation of a progressive wave travelling from right to left is given by the displacement function:

y (x, t) = a sin (ωt + kx + Φ) … (i)

The given equation is:

On comparing both the equations, we find that equation (ii) represents a travelling wave, propagating from right to left.

Now, using equations (i) and (ii), we can write:

ω = 36 rad/s and k = 0.018 m–1

We know that:

and

Also,

v = νλ

Hence, the speed of the given travelling wave is 20 m/s.

Amplitude of the given wave,

Frequency of the given wave:

On comparing equations (i) and (

The distance between two successive crests or troughs is equal to the wavelength of the wave.

Wavelength is given by the relation:

Question 15.9:

For the wave described in Exercise 15.8, plot the displacement (= 0, 2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?

Answer

Hence, the speed of the given travelling wave is 20 m/s.

Amplitude of the given wave, a = 3 cm

Frequency of the given wave:

) and (ii), we find that the initial phase angle,

The distance between two successive crests or troughs is equal to the wavelength of the

Wavelength is given by the relation:

For the wave described in Exercise 15.8, plot the displacement (y) versus (t) graphs for 0, 2 and 4 cm. What are the shapes of these graphs? In which aspects does the

oscillatory motion in travelling wave differ from one point to another: amplitude,

The distance between two successive crests or troughs is equal to the wavelength of the

) graphs for x 0, 2 and 4 cm. What are the shapes of these graphs? In which aspects does the

oscillatory motion in travelling wave differ from one point to another: amplitude,

All the waves have different phases.

The given transverse harmonic wave is:

For x = 0, the equation reduces to:

Now, plotting y vs. t graphs using the different values of

t(s)

0

y(cm)

3

For x = 0, x = 2, and x = 4, the phases of the three waves will get changed. This is because amplitude and frequency are invariant for any change in waves are shown in the given figure.

All the waves have different phases.

transverse harmonic wave is:

= 0, the equation reduces to:

graphs using the different values of t, as listed in the given table.

0 –3

= 4, the phases of the three waves will get changed. This is because amplitude and frequency are invariant for any change in x. The y-t plots of the three waves are shown in the given figure.

, as listed in the given table.

0

= 4, the phases of the three waves will get changed. This is because plots of the three

Question 15.10:

For the travelling harmonic wave

y (x, t) = 2.0 cos 2π (10t – 0.0080x + 0.35)

Where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of

4 m,

0.5 m,

,

Answer

Equation for a travelling harmonic wave is given as:

y (x, t) = 2.0 cos 2π (10t – 0.0080x + 0.35)

= 2.0 cos (20πt – 0.016πx + 0.70 π)

Where,

Propagation constant, k = 0.0160 π

Amplitude, a = 2 cm

Angular frequency, ω= 20 π rad/s

Phase difference is given by the relation:

For x = 4 m = 400 cm

Φ = 0.016 π × 400 = 6.4 π rad

For 0.5 m = 50 cm

Φ = 0.016 π × 50 = 0.8 π rad

For

For

Question 15.11:

The transverse displacement of a string (clamped at its both ends) is given by

Where x and y are in m and t ×10–2 kg.

Answer the following:

Does the function represent a travelling wave or a stationary wave?

Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of ea

Determine the tension in the string.

Answer

The general equation representing a stationary wave is given by the displacement function:


t in s. The length of the string is 1.5 m and its mass is 3.0

Does the function represent a travelling wave or a stationary wave?

Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave?

Determine the tension in the string.

The general equation representing a stationary wave is given by the displacement


and its mass is 3.0

Interpret the wave as a superposition of two waves travelling in opposite directions. What

The general equation representing a stationary wave is given by the displacement

y (x, t) = 2a sin kx cos ωt

This equation is similar to the given equation:

Hence, the given function represents a stationary wave.

A wave travelling along the positive x-direction is given as:

The wave travelling along the negative x-direction is given as:

The superposition of these two waves yields:

The transverse displacement of the string is given as:

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

∴Wavelength, λ = 3 m

It is given that:

120π = 2πν

Frequency, ν = 60 Hz

Wave speed, v = νλ

= 60 × 3 = 180 m/s

The velocity of a transverse wave travelling in a string is given by the relation:

Where,

Velocity of the transverse wave,

Mass of the string, m = 3.0 × 10

Length of the string, l = 1.5 m

Mass per unit length of the string,

Tension in the string = T

From equation (i), tension can be obtained as:

T = v2μ

= (180)2 × 2 × 10–2

= 648 N

Question 15.12:

For the wave on a string described in Exercise 15.11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (ii) What is the amplitude of a point 0.375 m a

Answer


Velocity of the transverse wave, v = 180 m/s

= 3.0 × 10–2 kg

= 1.5 m

Mass per unit length of the string,

), tension can be obtained as:

For the wave on a string described in Exercise 15.11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (ii) What is the amplitude of a point 0.375 m away from one end?


For the wave on a string described in Exercise 15.11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (ii)

Answer:

(i)

Yes, except at the nodes


No

0.042 m

Explanation:

(i)

All the points on the string oscillate with the same frequency, except at the nodes which have zero frequency.

All the points in any vibrating loop have the same phase, except at the nodes.

All the points in any vibrating loop have different amplitudes of vibration.

The given equation is:

For x = 0.375 m and t = 0

Question 15.13:

Given below are some functions of longitudinal) of an elastic wave. State which of these represent (i) a traveling wave, (ii) a



All the points on the string oscillate with the same frequency, except at the nodes which have zero frequency.

vibrating loop have the same phase, except at the nodes.


Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a traveling wave, (ii) a

All the points on the string oscillate with the same frequency, except at the nodes

vibrating loop have the same phase, except at the nodes.


represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a traveling wave, (ii) a

stationary wave or (iii) none at all:

y = 2 cos (3x) sin (10t

y = 3 sin (5x – 0.5t) + 4 cos (5

y = cos x sin t + cos 2

Answer

The given equation represents a stationary wave because the harmonic terms appear separately in the equation.

The given equation does not contain any harmonic term. Therefore, it does not represent either a travelling wave or a stationary wave.

The given equation represents a travelling wave as the harmonic terms the combination of kx – ωt.

The given equation represents a stationary wave because the harmonic terms appear separately in the equation. This equation actually represents the superposition of two stationary waves.

Question 15.14:

A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of th4.0 × 10–2 kg m–1. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?

Answer

Mass of the wire, m = 3.5 × 10

stationary wave or (iii) none at all:

t)

) + 4 cos (5x – 0.5t)

+ cos 2x sin 2t

The given equation represents a stationary wave because the harmonic terms appear separately in the equation.

The given equation does not contain any harmonic term. Therefore, it does not represent travelling wave or a stationary wave.

The given equation represents a travelling wave as the harmonic terms kx and ω

The given equation represents a stationary wave because the harmonic terms appear separately in the equation. This equation actually represents the superposition of

A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10–2 kg and its linear mass density is

. What is (a) the speed of a transverse wave on the string, and (b) the

= 3.5 × 10–2 kg

The given equation represents a stationary wave because the harmonic terms kx and ωt

The given equation does not contain any harmonic term. Therefore, it does not represent

and ωt are in

The given equation represents a stationary wave because the harmonic terms kx and ωtappear separately in the equation. This equation actually represents the superposition of

A wire stretched between two rigid supports vibrates in its fundamental mode with a kg and its linear mass density is

. What is (a) the speed of a transverse wave on the string, and (b) the

Linear mass density,

Frequency of vibration, ν = 45 Hz

∴Length of the wire,

The wavelength of the stationary wave (relation:

For fundamental node, n = 1:

λ = 2l

λ = 2 × 0.875 = 1.75 m

The speed of the transverse wave in the string is given

v = νλ= 45 × 1.75 = 78.75 m/s

The tension produced in the string is given by the relation:

T = v2µ

= (78.75)2 × 4.0 × 10–2 = 248.06 N

Question 15.15:

A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.

Answer

vibration, ν = 45 Hz

The wavelength of the stationary wave (λ) is related to the length of the wire by the

= 1:

The speed of the transverse wave in the string is given as:

= 45 × 1.75 = 78.75 m/s

The tension produced in the string is given by the relation:

= 248.06 N

long tube open at one end, with a movable piston at the other end, shows with a fixed frequency source (a tuning fork of frequency 340 Hz) when the

tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.

) is related to the length of the wire by the

long tube open at one end, with a movable piston at the other end, shows with a fixed frequency source (a tuning fork of frequency 340 Hz) when the

tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of

Frequency of the turning fork, ν = 340 Hz

Since the given pipe is attached with a piston at one end, it will behave as a pipe with one end closed and the other end open, as shown in the given figure.

Such a system produces odd harmonics. The fundamental note in a closed pipe is gby the relation:

Where,

Length of the pipe,

The speed of sound is given by the relation:

= 340 × 1.02 = 346.8 m/s

Question 15.16:

A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod is given to be 2.53 kHz. What is the speed of sound in steel?

Answer

Length of the steel rod, l = 100 cm = 1 m

rk, ν = 340 Hz

Since the given pipe is attached with a piston at one end, it will behave as a pipe with one end closed and the other end open, as shown in the given figure.

Such a system produces odd harmonics. The fundamental note in a closed pipe is g

The speed of sound is given by the relation:

= 340 × 1.02 = 346.8 m/s

A steel rod 100 cm long is clamped at its middle. The fundamental frequency of the rod is given to be 2.53 kHz. What is the speed of sound in

= 100 cm = 1 m

Since the given pipe is attached with a piston at one end, it will behave as a pipe with one

Such a system produces odd harmonics. The fundamental note in a closed pipe is given

A steel rod 100 cm long is clamped at its middle. The fundamental frequency of the rod is given to be 2.53 kHz. What is the speed of sound in

Fundamental frequency of vibration, ν = 2.53 kHz = 2.53

When the rod is plucked at its middle, an antinode(N) are formed at its two ends, as shown in the given figure.

The distance between two successive nodes is

The speed of sound in steel is given by the relation:

v = νλ

= 2.53 × 103 × 2

= 5.06 × 103 m/s

= 5.06 km/s

Question 15.17:

A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (Speed of sound in air is 340 m s

Answer

Answer: First (Fundamental); No

Length of the pipe, l = 20 cm = 0.2 m

Fundamental frequency of vibration, ν = 2.53 kHz = 2.53 × 103 Hz

When the rod is plucked at its middle, an antinode (A) is formed at its centre, and nodes (N) are formed at its two ends, as shown in the given figure.

The distance between two successive nodes is .

The speed of sound in steel is given by the relation:

A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (Speed of sound in air is 340 m s–1).

First (Fundamental); No

= 20 cm = 0.2 m

(A) is formed at its centre, and nodes

A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both

Source frequency = nth normal mode of frequency, ν

Speed of sound, v = 340 m/s

In a closed pipe, the nth normal mode of frequency is given by the relation:

Hence, the first mode of vibration frequency is resonantly excited by the given source.

In a pipe open at both ends, the

Since the number of the mode of vibration (not produce a resonant vibration in an open pipe.

Question 15.18:

Two sitar strings A and B playing the note ‘of frequency 6 Hz. The tension in the string A is slightly reduced and the beat is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?

normal mode of frequency, νn = 430 Hz

normal mode of frequency is given by the relation:

the first mode of vibration frequency is resonantly excited by the given source.

In a pipe open at both ends, the nth mode of vibration frequency is given by the relation:

Since the number of the mode of vibration (n) has to be an integer, the given sournot produce a resonant vibration in an open pipe.

Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency

the first mode of vibration frequency is resonantly excited by the given source.

mode of vibration frequency is given by the relation:

) has to be an integer, the given source does

’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency

Answer

Frequency of string A, fA = 324 Hz

Frequency of string B = fB

Beat’s frequency, n = 6 Hz

Frequency decreases with a decrease in the is directly proportional to the square root of tension. It is given as:

Question 15.19:

Explain why (or how):

In a sound wave, a displacement node is a pressure antinode and vice versa,

Bats can ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”,

A violin note and sitar note may have the same frequency, yet we can distinguish between the two notes,

Solids can support both longitudinal and transverse waves, but onlycan propagate in gases, and

= 324 Hz

Frequency decreases with a decrease in the tension in a string. This is because frequency is directly proportional to the square root of tension. It is given as:

In a sound wave, a displacement node is a pressure antinode and vice versa,

ertain distances, directions, nature, and sizes of the obstacles without any

A violin note and sitar note may have the same frequency, yet we can distinguish between

Solids can support both longitudinal and transverse waves, but only longitudinal waves

tension in a string. This is because frequency

ertain distances, directions, nature, and sizes of the obstacles without any

A violin note and sitar note may have the same frequency, yet we can distinguish between

longitudinal waves

The shape of a pulse gets distorted during propagation in a dispersive medium.

Answer

A node is a point where the amplitude of vibration is the minimum and pressure is the maximum. On the other hand, an athe maximum and pressure is the minimum.

Therefore, a displacement node is nothing but a pressure antinode, and vice versa.

Bats emit very high-frequency ultrasonic sound waves. These waves get reflecttoward them by obstacles. A bat receives a reflected wave (frequency) and estimates the distance, direction, nature, and size of an obstacle with the help of its brain senses.

The overtones produced by a sitar and a violin, and the strengths of thedifferent. Hence, one can distinguish between the notes produced by a sitar and a violin even if they have the same frequency of vibration.

Solids have shear modulus. They can sustain shearing stress. Since fluids do not have any definite shape, they yield to shearing stress. The propagation of a transverse wave is such that it produces shearing stress in a medium. The propagation of such a wave is possible only in solids, and not in gases.

Both solids and fluids have their respective bulkstress. Hence, longitudinal waves can propagate through solids and fluids.

A pulse is actually is a combination of waves having different wavelengths. These waves travel in a dispersive medium with different velocitiesmedium. This results in the distortion of the shape of a wave pulse.

Question 15.20:

A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air. (i) What is the train (a) approaches the platform with a speed of 10 m swith a speed of 10 m s–1? (ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 m s

Answer


A node is a point where the amplitude of vibration is the minimum and pressure is the maximum. On the other hand, an antinode is a point where the amplitude of vibration is the maximum and pressure is the minimum.


frequency ultrasonic sound waves. These waves get reflecttoward them by obstacles. A bat receives a reflected wave (frequency) and estimates the distance, direction, nature, and size of an obstacle with the help of its brain senses.

The overtones produced by a sitar and a violin, and the strengths of these overtones, are different. Hence, one can distinguish between the notes produced by a sitar and a violin even if they have the same frequency of vibration.

Solids have shear modulus. They can sustain shearing stress. Since fluids do not have any shape, they yield to shearing stress. The propagation of a transverse wave is such

that it produces shearing stress in a medium. The propagation of such a wave is possible only in solids, and not in gases.

Both solids and fluids have their respective bulk moduli. They can sustain compressive stress. Hence, longitudinal waves can propagate through solids and fluids.

A pulse is actually is a combination of waves having different wavelengths. These waves travel in a dispersive medium with different velocities, depending on the nature of the medium. This results in the distortion of the shape of a wave pulse.

A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air. (i) What is the frequency of the whistle for a platform observer when the train (a) approaches the platform with a speed of 10 m s–1, (b) recedes from the platform

? (ii) What is the speed of sound in each case? The speed of be taken as 340 m s–1.


A node is a point where the amplitude of vibration is the minimum and pressure is the ntinode is a point where the amplitude of vibration is


frequency ultrasonic sound waves. These waves get reflected back toward them by obstacles. A bat receives a reflected wave (frequency) and estimates the distance, direction, nature, and size of an obstacle with the help of its brain senses.

se overtones, are different. Hence, one can distinguish between the notes produced by a sitar and a violin

Solids have shear modulus. They can sustain shearing stress. Since fluids do not have any shape, they yield to shearing stress. The propagation of a transverse wave is such

that it produces shearing stress in a medium. The propagation of such a wave is possible

moduli. They can sustain compressive

A pulse is actually is a combination of waves having different wavelengths. These waves , depending on the nature of the

A train, standing at the outer signal of a railway station blows a whistle of frequency 400 frequency of the whistle for a platform observer when the

, (b) recedes from the platform ? (ii) What is the speed of sound in each case? The speed of

(a)Frequency of the whistle, ν = 400 Hz

Speed of the train, vT= 10 m/s

Speed of sound, v = 340 m/s

The apparent frequency by the relation:

The apparent frequency by the relation:

The apparent change in the frequency of sound is caused by the relative motions of the source and the observer. These relative motions produce no effect on the speed of Therefore, the speed of sound in air in both the cases remains the same, i.e., 340 m/s.

Question 15.21:

A train, standing in a station-wind starts blowing in the direction from the1. What are the frequency, wavelength, and speed of sound for an observer standing on the station’s platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 m scan be taken as 340 m s–1.

Answer

(a)Frequency of the whistle, ν = 400 Hz

= 10 m/s

of the whistle as the train approaches the platform is given

of the whistle as the train recedes from the platform is given

The apparent change in the frequency of sound is caused by the relative motions of the source and the observer. These relative motions produce no effect on the speed of Therefore, the speed of sound in air in both the cases remains the same, i.e., 340 m/s.

-yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with at a speed of 10 m s. What are the frequency, wavelength, and speed of sound for an observer standing on

the station’s platform? Is the situation exactly identical to the case when the air is still and the yard at a speed of 10 m s–1? The speed of sound in still air

of the whistle as the train approaches the platform is given

of the whistle as the train recedes from the platform is given

The apparent change in the frequency of sound is caused by the relative motions of the source and the observer. These relative motions produce no effect on the speed of sound. Therefore, the speed of sound in air in both the cases remains the same, i.e., 340 m/s.

yard, blows a whistle of frequency 400 Hz in still air. The yard to the station with at a speed of 10 m s–

. What are the frequency, wavelength, and speed of sound for an observer standing on the station’s platform? Is the situation exactly identical to the case when the air is still and

? The speed of sound in still air

For the stationary observer: 400 Hz; 0.875 m; 350 m/s

For the running observer: Not exactly identical

For the stationary observer:

Frequency of the sound produced by the whistle, ν = 400 Hz

Speed of sound = 340 m/s

Velocity of the wind, v = 10 m/s

As there is no relative motion between the source and the observer, the frequency of the sound heard by the observer will be the same as that produced by the source, i.e., 400 Hz.

The wind is blowing toward the observer. Hence, the effective speed of the sound increases by 10 units, i.e.,

Effective speed of the sound, ve = 340 + 10 = 350 m/s

The wavelength (λ) of the sound heard by the observer is given by the relation:

For the running observer:

Velocity of the observer, vo = 10 m/s

The observer is moving toward the source. As a result of the relative motions of the source and the observer, there is a change in frequency ( ).

This is given by the relation:

Since the air is still, the effective speed of sound = 340 + 0 = 340 m/s

The source is at rest. Hence, the wavelength of the sound will not change, i.e., λ remains

0.875 m.

Hence, the given two situations are not exactly identical.

Question 15.22:

A travelling harmonic wave on a string is described by

What are the displacement and velocity of oscillation of a point at = 1 s? Is this velocity equal to the velocity of wave propagation?

Locate the points of the string which have the same transverse displacements and velocity as the x = 1 cm

Answer

The given harmonic wave is:

For x = 1 cm and t = 1s,

Hence, the given two situations are not exactly identical.

A travelling harmonic wave on a string is described by

What are the displacement and velocity of oscillation of a point at x = 1 cm, and 1 s? Is this velocity equal to the velocity of wave propagation?

Locate the points of the string which have the same transverse displacements and 1 cm point at t = 2 s, 5 s and 11 s.

The given harmonic wave is:

= 1 cm, and t

Locate the points of the string which have the same transverse displacements and

The velocity of the oscillation at a given point and time is given as:

Now, the equation of a propagating wave is given by:

∴Hence, the velocity of the wave oscillation at x = 1 cm and t = 1 s is not equal to the velocity of the wave propagation.

Propagation constant is related to wavelength as:

Therefore, all the points at distances nλ , i.e. ± 12.56 m, ± 25.12 m, … and so on for x = 1 cm, will have the same displacement as the x = 1 cm points at t

= 2 s, 5 s, and 11 s.

Question 15.23:

A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium. (a) Does the pulse have a definite (i(b) If the pulse rate is 1 after every 20 s, (that is the whistle is blown for a split of second

after every 20 s), is the frequency of the note produced by the whistle equal to Hz?

Answer

(i)No

(ii)No

(iii)Yes

No

Explanation:

The narrow sound pulse does not have a fixed wavelength or frequency. However, the speed of the sound pulse remains the same, which is equal to the speed of sound in that medium.

The short pip produced after every

or 0.05 Hz. It means that 0.05 Hz is the frequency of the repetition of the pip of the whistle.

Question 15.24:

One end of a long string of linear mass density 8.0 × 10electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible (fork end) of the string x = 0 has zero transverse displacement (

A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium. (a) Does the pulse have a definite (i) frequency, (ii) wavelength, (iii) speed of propagation? (b) If the pulse rate is 1 after every 20 s, (that is the whistle is blown for a split of second

after every 20 s), is the frequency of the note produced by the whistle equal to

The narrow sound pulse does not have a fixed wavelength or frequency. However, the speed of the sound pulse remains the same, which is equal to the speed of sound in that

The short pip produced after every 20 s does not mean that the frequency of the whistle is

or 0.05 Hz. It means that 0.05 Hz is the frequency of the repetition of the pip of the

One end of a long string of linear mass density 8.0 × 10–3 kg m–1 is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At t = 0, the left end

= 0 has zero transverse displacement (y = 0) and is moving along

A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium. (a) ) frequency, (ii) wavelength, (iii) speed of propagation?

(b) If the pulse rate is 1 after every 20 s, (that is the whistle is blown for a split of second

after every 20 s), is the frequency of the note produced by the whistle equal to or 0.05

The narrow sound pulse does not have a fixed wavelength or frequency. However, the speed of the sound pulse remains the same, which is equal to the speed of sound in that

20 s does not mean that the frequency of the whistle is

or 0.05 Hz. It means that 0.05 Hz is the frequency of the repetition of the pip of the

is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming

= 0, the left end = 0) and is moving along

positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as function of x and t that describes the wave on the string.

Answer

The equation of a travelling wave propagating along the positive y-direction is given by the displacement equation:

y (x, t) = a sin (wt – kx) … (i)

Linear mass density,

Frequency of the tuning fork, ν = 256 Hz

Amplitude of the wave, a = 5.0 cm = 0.05 m … (ii)

Mass of the pan, m = 90 kg

Tension in the string, T = mg = 90 × 9.8 = 882 N

The velocity of the transverse wave v, is given by the relation:

Substituting the values from equations (ii), (iii), and (iv) in equation (i), we get the displacement equation:

y (x, t) = 0.05 sin (1.6 × 103t

Question 15.25:

A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km hof sound reflected by the submarine? Take the speed of sound in water to be 1450 m s

Answer

Operating frequency of the SONAR system, ν = 40 kHz

Speed of the enemy submarine,

Speed of sound in water, v = 1450 m/s

The source is at rest and the observer (enemy submarine) is moving toward it. Hence, the apparent frequency ( ) received and reflected by the submarine is given by the relation:

The frequency ( ) received by the enem

Question 15.26:

Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience

– 4.84 x) m

A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h–1. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 m s

Operating frequency of the SONAR system, ν = 40 kHz

Speed of the enemy submarine, ve = 360 km/h = 100 m/s

= 1450 m/s

The source is at rest and the observer (enemy submarine) is moving toward it. Hence, the ) received and reflected by the submarine is given by the relation:

) received by the enemy submarine is given by the relation:


A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. An enemy . What is the frequency

of sound reflected by the submarine? Take the speed of sound in water to be 1450 m s–1.

The source is at rest and the observer (enemy submarine) is moving toward it. Hence, the ) received and reflected by the submarine is given by the relation:

y submarine is given by the relation:


both transverse (S) and longitudinal (P) sound waves. Typically the speed of S wave is about 4.0 km s–1, and that of P wave is 8.0 km s–1. A seismograph records P and S waves from an earthquake. The first P wave arrives 4 min before the first S wave. Assuming the waves travel in straight line, at what distance does the earthquake occur?

Answer

Let vSand vP be the velocities of S and P waves respectively.

Let L be the distance between the epicentre and the seismograph.

We have:

L = vStS (i)

L = vPtP (ii)

Where,

tS and tP are the respective times taken by the S and P waves to reach the seismograph from the epicentre

It is given that:

vP = 8 km/s

vS = 4 km/s

From equations (i) and (ii), we have:

vS tS = vP tP

4tS = 8 tP

tS = 2 tP (iii)

It is also given that:

tS – tP = 4 min = 240 s

2tP – tP = 240

tP = 240

And tS = 2 × 240 = 480 s

From equation (ii), we get:

L = 8 × 240

= 1920 km

Hence, the earthquake occurs at a distance of 1920 km from the seismograph.

Question 15.27:

A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?

Answer

Ultrasonic beep frequency emitted by the bat, ν = 40 kHz

Velocity of the bat, vb = 0.03

Where, v = velocity of sound in air

The apparent frequency of the sound striking the wall is given as:

This frequency is reflected by the stationary wall (

The frequency ( ) of the received sound is given by the relation:


A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?

cy emitted by the bat, ν = 40 kHz

= 0.03 v

= velocity of sound in air

The apparent frequency of the sound striking the wall is given as:

This frequency is reflected by the stationary wall ( ) toward the bat.

) of the received sound is given by the relation:


A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the

171 [XI – Physics]

UNIT X

OSCILLATIONS AND WAVES

Periodic Motion : It is that motion which is identically repeated after afixed interval of time. The fixed interval of time after which the motion isrepeated is called period of motion.

For example, the revolution of earth around the sun

Oscillatory Motion or Vibratory Motion : It is that motion in which abody moves to and for or back and forth repeatedly about a fixed point(called mean position), in a definite interval of time.

Simple Harmonic Motion : It is a special type of periodic motion, inwhich a particle moves to and fro repeatedly about a mean (i.e., equilibrium)position and the magnitude of force acting on the particle at any instantis directly proportional to the displacement of the particle from the mean(i.e., equilibrium) position at that instant i.e. F = –k y.

where k is known as force constant. Here, –ve sign shows that therestoring force (F) is always directed towards the mean position.

Displacement in S.H.M : The displacement of a particle executing S.H.Mat an instant is defined as the distance of the particle from the meanposition at that instant. It can be given by the relation

y = a sin t or y = a cos t

The first relation is valid when the time is measured from the meanposition and the second relation is valid when the time is measured fromthe extreme position of the particle executing S.H.M.

Velocity in S.H.M : It is defined as the time rate of change of thedisplacement of the particle at the given instant. Velocity in S.H.M. isgiven by

2 2 2dy dV a sin t a cos t a 1 sin t a 1 y a

dt dt

[XI – Physics] 172

2 2a y

Acceleration in S.H.M : It is defined as the time rate of change of thevelocity of the particle at the given instant, i.e.,

dv

adt

2 2da cos t a sin t y

dt

Time period in S.H.M is given by

displacement inertia factorT 2 , or T 2

acceleration spring factor

Frequency of vibration in S.H.M.,

1 1 acceleration 1 spring factor, or v

T 2 displacement 2 inertia factor

Total energy in S.H.M

= 2 2 2 2 2 2 21 1 1P.E. K.E. m y m a y m a a constant.

2 2 2

Expression for time period

(i) in case of simple pendulum T 2 g

(iv) Oscillations of a loaded spring T 2 m k

where, m is the mass of body attached at the free end of spring and Kis the force constant of spring.

Spring constant (K) of a spring : It is defined as the force per unitextension or compression of the spring.

(i) The spring constant of the combination of two springs in series is

1 2

1 2

k kK

k k


(ii) The spring constant of the combination of two springs in parallel's

K = k1 + k2

Undamped oscillations : When a simple harmonic system oscillateswith a constant amplitude (which does not change) with time, its oscillationsare called undamped oscillations.

Damped oscillations : When a simple harmonic system oscillates witha decreasing amplitude with time, its oscillations are called dampedoscillations.

Free, forced and resonant oscillations

(a) Free oscillations : When a system oscillates with its own naturalfrequency without the help of an external periodic force, itsoscillations are called free oscillations.

(b) Forced oscillations : When a system oscillates with the help ofan external periodic force of frequency, other than its own naturalfrequency, its oscillations are called forced oscillations.

(c) Resonant oscillations : When a body oscillates with its ownnatural frequency, with the help of an external periodic force whosefrequency is the same as that of the natural frequency of theoscillating body, then the oscillations of the body are calledresonant oscillations.

A Wave Motion is a form of disturbance which travels through a mediumon account of repeated periodic vibrations of the particles of the mediumabout their mean position, the motion being handed on from one particleto the adjoining particle.

A material medium is a must for propagation of waves. It should possessthe properties of inertia, and elasticity. The two types of wave motionare :

(i) Transverse wave motion that travels in the form of crests and troughs.

(ii) Longitudinal wave motion that travels in the form of compressionsand rarefactions.

Speed of longitudinal waves in a long solid rod is y

where, Y is Young's modulus of the material of solid rod and P is density

of the material. The speed of longitudinal waves in a liquid is given by


k , where K is bulk modulus of elasticity of the liquid.

The expression for speed of longitudinal waves in a gas, as suggestedby Newton and modified late by Laplace is

pP

where C C

P is pressure exerted by the gas.

The speed of transverse waves over a string is given by T m

where, T is tension in the string and m is mass of unit length of the string.

Equation of plane progressive waves travelling with a velocity alongpositive direction of X-axis is

2

y a sin t x

where, is wavelength of the wave, a is amplitude of particle, and x isthe distance from the origin.

Superposition principle enables us to find the resultant of any number

of waves meeting at a point. If 1 2 3 ny , y , y , .......y

are displacements at a

point due to n waves, the resultant displacement y

at that point is given

by 1 2 ny y y .......y

On a string, transverse stationary waves are formed due to super-imposition of direct and the reflected transverse waves.

The wavelength of nth mode of vibration of a stretched string is

n2 Ln

and its frequency, vn = n v1

This note is called nth harmonic or (n – 1)th overtone.

Nodes are the points, where amplitude of vibration is zero, In the nth

mode of vibration, there are (n + 1) nodes located at distances (from oneend)

L 2Lx 0, , .....L

n n


Antinodes are the points, where amplitude of vibration is maximum. In thenth mode of vibration, there are n antinodes, located at distances (from one end)

2n 1L 3L 5 L, , , ........ L

2 2 n 2n 2n

x =n

In an organ pipe closed at one end, longitudinal stationary waves areformed.

frequency of nth mode of vibration, vn = (2n – 1)v1

In an organ pipe open at both ends, antinodes are formed at the twoends, separated by a node in the middle in the first normal mode ofvibration and so on. The fundamental frequency in this case is twice thefundamental frequency in a closed organ pipe of same length.

In an open organ pipe, all harmonics are present, whereas in a closedorgan pipe, even harmonics are missing.

Beats : Beats is the phenomenon of regular variation in the intensity ofsound with time when two sources of nearly equal frequencies are soundedtogether.

If v1 and v2 are the frequencies of two sources producing beats, then timeBeat frequency = v1 – v2

Doppler's Effect : Whenever there is a relative motion between a sourceof sound and listener, the apparent frequency of sound heard is differentfrom the actual frequency of sound emitted by the source.

If is actual frequency of sound emitted and ', the apparent frequency,

Lo

s

v v'

v v

v = Speed of sound through medium

vL = velocity of listener, Vs = Velocity of source velocities in the directionsource of listener (SL) should be taken positive and those opposite shouldbe taken negative.

ONE MARK QUESTIONS

1. How is the time period effected, if the amptitude of a simple pendulum isincreased?


2. Define force constant of a spring.

3. At what distance from the mean position, is the kinetic energy in simpleharmonic oscillator equal to potential energy?

4. How is the frequency of oscillation related with the frequency of changein the of K.E and P.E of the body in S.H.M.?

5. What is the frequency of total energy of a particle in S.H.M.?

6. How is the length of seconds pendulum related with acceleration duegravity of any planet?

7. If the bob of a simple pendulum is made to oscillate in some fluid ofdensity greater than the density of air (density of the bob density of thefluid), then time period of the pendulum increased or decrease.

8. How is the time period of the pendulum effected when pendulum is takento hills or in mines?

9. A transverse wave travels along x-axis. The particles of the medium mustmove in which direction?

10. Define angular frequency. Give its S.I. unit.

11. Sound waves from a point source are propagating in all directions. Whatwill be the ratio of amplitudes at distances of x meter and y meter fromthe source?

12. Does the direction of acceleration at various points during the oscillationof a simple pendulum remain towards mean position?

13. What is the time period for the function f(t) = sin t +cos t may representthe simple harmonic motion?

14. When is the swinging of simple pendulum considered approximately SHM?

15. Can the motion of an artificial satellite around the earth be taken asSHM?

16. What is the phase relationship between displacement, velocity andaccelection in SHM?

17. What forces keep the simple pendulum in motion?

18. How will the time period of a simple pendulum change when its length isdoubled?


19. What is a harmonic wave function?

20. If the motion of revolving particle is periodic in nature, give the nature ofmotion or projection of the revolving particle along the diameter.

21. In a forced oscillation of a particle, the amplitude is maximum for afrequency w1 of the force, while the energy is maximum for a frequencyw2 of the force. What is the relation between w1 and w2?

22. Which property of the medium are responsible for propagation of wavesthrough it?

23. What is the nature of the thermal change in air, when a sound wavepropagates through it?

24. Why does sound travel faster in iron than in water or air?

25. When will the motion of a simple pendulum be simple harmonic?

26. A simple harmonic motion of acceleration ‘a’ and displacement ‘x’ isrepresented by a + 42x = 0. What is the time period of S.H.M?

27. What is the main difference between forced oscillations and resonance?

28. Define amplitude of S.H.M.

29. What is the condition to be satisfied by a mathematical relation betweentime and displacement to describe a periodic motion?

30. Why the pitch of an organ pipe on a hot summer day is higher?

31. Under what conditions does a sudden phase reversal of waves onreflection takes place?

32. The speed of sound does not depend upon its frequency. Give an examplein support of this statement.

33. If an explosion takes place at the bottom of lake or sea, will the shockwaves in water be longitudinal or transverse?

34. Frequency is the most fundamental property of wave, why?

35. How do wave velocity and particle velocity differ from each other?

36. If any liquid of density higher than the density of water is used in aresonance tube, how will the frequency change?


37. Under what condition, the Doppler effect will not be observed, if thesource of sound moves towards the listener?

38. What physical change occurs when a source of sound moves and thelistener is stationary?

39. What physical change occurs when a source of sound is stationary andthe listener moves?

40. If two sound waves of frequencies 480 Hz and 536 Hz superpose, willthey produce beats? Would you hear the beats?

41. Define non dispessive medium.

2 MARKS QUESTIONS

1. Which of the following condition is not sufficient for simple harmonicmotion and why?

(i) acceleration and displacement

(ii) restoring force and displacement

2. The formula for time period T for a loaded spring, displacementT 2

acceleration

Does-the time period depend on length of the spring?

3. Water in a U-tube executes S.H.M. Will the time period for mercury filledup to the same height in the tube be lesser of greater than that in caseof water?

4. There are two springs, one delicate and another hard or stout one. Forwhich spring, the frequency of the oscillator will be more?

5. Time period of a particle in S.H.M depends on the force constant K and

mass m of the particle 1 mT

2 k

. A simple pendulum for small angular

displacement executes S.H.M approximately. Why then is the time periodof a pendulum independent of the mass of the pendulum?

6. What is the frequency of oscillation of a simple pendulum mounted in acabin that is falling freely?

7. Why can the transverse waves not be produced in air?


8. The velocity of sound in a tube containing air at 27°C and pressure of 76cm of Hg is 330 ms–1. What will be its velocity, when pressure is increasedto 152 cm of mercury and temperature is kept constant?

9. Even after the breakup of one prong of tunning fork it produces a roundof same frequency, then what is the use of having a tunning fork with twoprongs?

10. Why is the sonometer box hollow and provided with holes?

11. The displacement of a particle in S.H.M may be given by

y = a sin(t + )

show that if the time t is increased by 2/, the value of y remains thesame.

12. What do you mean by the independent behaviour of waves?

13. Define wave number and angular wave number and give their S.I. units.

14. Why does the sound travel faster in humid air?

15. Use the formula p

v

to explain, why the speed of sound in air

(a) is independent of pressure

(b) increase with temperature

16. Differentiate between closed pipe and open pipe at both ends of samelength for frequency of fundamental note and harmonics.

17. Bats can ascertain distances, directions, nature and size of the obstaclewithout any eyes, explain how?

18. In a sound wave, a displacement node is a pressure antinode and vice-versa. Explain, why.

19. How does the frequency of a tuning fork change, when the temperatureis increased?

20. Explain, why can we not hear an echo in a small room?

21. What do you mean by reverberation? What is reverberation time?


3 MARKS QUESTIONS

1. Show that for a particle in linear simple harmonic motion, the accelerationis directly proportional to its displacement of the given instant.

2. Show that for a particle in linear simple harmonic motion, the averagekinetic energy over a period of oscillation, equals the average potentialenergy over the same period.

3. Deduce an expression for the velocity of a particle executing S.H.M whenis the particle velocity (i) Maximum (ii) minimum?

4. Draw (a) displacement time graph of a particle executing SHM with phaseangle equal to zero (b) velocity time graph and (c) acceleration timegraph of the particle.

5. Show that a linear combination of sine and cosine function likex(t) = a sint + b cost represents a simple harmonic Also, determine itsamplitude and phase constant.

6. Show that in a S.H.M the phase difference between displacement andvelocity is /2, and between displacement and acceleration is .

7. Derive an expression for the time period of the horizontal oscillations ofa massless loaded spring.

8. Show that for small oscillations the motion of a simple pendulum is simpleharmonic. Derive an expression for its time period.

9. Distinguish with an illustration among free, forced and resonant oscillations.

10. In reference to a wave motion, define the terms

(i) amplitude (ii) time period

(iii) frequency (iv) angular frequency

(v) wave length and wave number.

11. What do you understand by phase of a wave? How does the phasechange with time and position.

12. At what time from mean position of a body executive S.H.M. kinetic energyand potential energy will be equal?


LONG ANSWER QUESTIONS

1. Derive expressions for the kinetic and potential energies of a simpleharmonic oscillator. Hence show that the total energy is conserved inS.H.M. in which positions of the oscillator, is the energy wholly kinetic orwholly potential?

2. One end of a U-tube containing mercury is connected to a section pumpand the other end is connected to the atmosphere. A small pressuredifference is maintained between the two columns. Show that when thesuction pump is removed, the liquid in the U-tube executes S.H.M.

3. Discuss the Newton's formula for velocity of sound in air. What correctionwas applied to it by Laplace and why?

4. What are standing waves? Desire and expression for the standing waves.Also define the terms node and antinode and obtain their positions.

5. Discuss the formation of harmonics in a stretched string. Show that incase of a stretched string the first four harmonics are in the ratio 1:2:3:4,

6. Give the differences between progressive and stationary waves.

7. If the pitch of the sound of a source appears to drop by 10% to a movingperson, then determine the velocity of motion of the person. Velocity ofsound = 30 ms–1.

8. Give a qualitative discussion of the different modes of vibration of anopen organ pipe.

9. Describe the various modes of vibrations of a closed organ pipe.

10. What are beats? How are they produced? Briefly discuss one applicationfor this phenomenon.

11. Show that the speed of sound in air increases by 61 cms–1 for every 1°Crise of temperature.

NUMERICALS

1. The time period of a body executing S.H.M is 1s. After how much time

will its displacement be 12 of its amplitude.


2. A particle is moving with SHM in a straight line. When the distance of theparticle from the equilibrium position has values x1 and x2, thecorresponding value of velocities are u1 and u2. Show that the time periodof oscillation is given by

1 22 22 12 21 2

–2

–

x xt

u u

3. Find the period of vibrating particle (SHM), which has accelesation of 45cm s–2, when displacement from mean position is 5 cm

4. A 40 gm mass produces on extension of 4 cm in a vertical spring. A massof 200 gm is suspended at its bottom and left pulling down. Calculate thefrequency of its vibration.

5. The acceleration due to gravity on the surface of the moon is 1.7 ms–2.What is the time period of a simple pendulum on the moon, if its timeperiod on the earth is 3.5 s? [g = 9.8 ms–2]

6. A particle executes simple harmonic motion of amplitude A.

(i) At what distance from the mean position is its kinetic energyequal to its potential energy?

(ii) At what points is its speed half the maximum speed?

7. A particle executes S.H.M of amplitude 25 cm and time period 3s. Whatis the minimum time required for the particle to move between two points12.5 cm on either side of the mean position?

8. The vertical motion of a huge piston in a machine is approximately S.H.Mwith a frequency of 0.5 s–1. A block of 10kg is placed on the piston. Whatis the maximum amplitude of the piston's S.H.M. for the block and pistonto remain together?

9. At what temperature will the speed of sound be double its value at 273°K?

10. A spring balance has a scale that reads from 0 to 50 kg. The length ofthe scale is 20 cm. A body suspended from this spring, when displacedand released, oscillates with a period of 0.60 s. What is the weight of thebody?

11. If the pitch of the sound of a source appears to drop by 10% to a movingperson, then determine the velocity of motion of the person. Velocity ofsound = 330 ms–1.


12. You are riding in an automobile of mass 3000 kg. Assuming that you areexamining the oscillation characteristics of its suspension system. Thesuspension sags 15 cm when the entire automobile is placed on it. Also,the amplitude of oscillation decreases by 50% during one completeoscillation, Estimate the values of (a) the spring constant and (b) thedamping constant ‘b’ for the spring and shock absorber system of onewheel assuming that each wheel supports 750 kg.

13. A string of mass 2.5 kg is under a tension of 200N. The length of thestretched string is 20m. If a transverse jerk is struck at one end of thestring, how long does the disturbance take to reach the other end?

14. Which of the following function of time represent (a) periodic and (n) non-periodic motion? Give the period for each case of periodic motion. [w isany positive constant].

(i) sin t + cos t

(ii) sin t + cos 2t + sin 4t

(iii) e–t

(iv) log (t)

15. The equation of a plane progressive wave is given by the equationy = 10 sin 2 (t – 0.005x) where y and x are in cm and t in seconds.Calculate the amplitude, frequency, wave length and velocity of the wave.

16. A tuning fork arrangement (pair) produces 4 beats s–1 with one fork offrequency 288 cps. A little was ix placed on the unknown fork and it thenproduces 2 beats s–1. What is the frequency of the unknown fork?

Unknown frequency = Known frequency ± Beat frequency

= 288 ± 4 = 292 or 284 Hz.

On loading with wax, the frequency decreases, the beat frequency alsodecreases to 2.

Unknown frequency = 292 cps (higher one).

17. A pipe 20 cm long is closed at one end, which harmonic mode of the pipeis resonantly excited by a 430 Hz source? Will this same source can bein resonance with the pipe, if both ends are open? Speed of sound = 340ms–1.


18. The length of a wire between the two ends of a sonometer is 105 cm.Where should the two bridges be placed so that the fundamentalfrequencies of the three segments are in the ratio of 1 : 3 : 15 ?

19. The transverse displacement of a string (clamped at its two ends) isgiven by

y(x,t) = 0.06 sin23

x cos(120t)

where x,y are in m and t is in s. The length of the string is 1.5 m and itsmass is 3.0 × 10–2 kg. Answer the following.

(a) Does the function represent a travelling or a stationary wave?

(b) Interpret the wave as a superposition of two waves travelling inopposite directions. What are the wavelength frequency and speedof propagation of each wave?

(c) Determine the tension in the string.

20. A wire stretched between two rigid supports vibrates in its fundamentalmode with a frequency 45 Hz. The mass of the wire is 3.5 × 10–2 kg andits linear density is 4.0 × 10–2 kg m–. What is (a) the speed of transversewave on the string and (b) the tension in the string?

21. A steel sod 100 cm long is clamped at its middle. The fundamentalfrequency of longitudinal vibrations of the rod as given to be 2.53 kHz.What is the speed of sound in steel?

ANSWERS OF ONE MARK QUESTIONS

1. No effect on time period when amptitude of pendulum is increased ordecreased.

2. The spring constant of a spring is the change in the force it exerts,divided by the change in deflection of the spring.

3. Not at the mid point, between mean and extreme position, it will be at

x a 2 .

4. P.E. or K.E. completes two vibrations in a time during which S.H.M


completes one vibration or the frequency of P.E. or K.E. is double than thatof S.H.M

5. The frequency of total energy of particle is S.H.M is zero because itremains constant.

6. Length of the seconds pendulum proportional to (acceleration due togravity)

7. Increased

8. As T 1

,g T will increase.

9. In the y–z plane or in plane perpendicular to x-axis.

10. It is the angle covered per unit time or it is the quantity obtained bymultiplying frequency by a factor of 2.

= 2v, S.I. unit is rad s–1

11. Intensity = amplitude2

21

distance

required ratio = y/x

12. No, the resultant of Tension in the string and weight of bob is not always

towards the mean position.

13. T = 2/

14. Swinging through small angles.

15. No, it is a circular and periodic motion but not SHM.

16. In SHM, –The velocity leads the displacement by a phase /2 radians

and acceleration leads the velocity by a phase /2 radians.

17. The component of weight (mg sin)

18. 2 times, as T

19. A harmonic wave function is a periodic function whose functional form issine or cosine.


20. S.H.M

21. Both amplitude and energy of the particle can be maximum only in thecase of resonance, for resonance to occur 1 = 2.

22. Properties of elasticity and inertia.

23. When the sound wave travel through air adiabatic changes take place inthe medium.

24. Sound travel faster in iron or solids because iron or solid is highly elasticas compared to water (liquids) or air (gases).

25. When the displacement of bob from the mean position is so small thatsin

26. a = –42x = –2x = 2

2 2T 1s

2

27. The frequency of external periodic force is different from the naturalfrequency of the oscillator in case of forced oscillation but in resonancetwo frequencies are equal.

28. The maximum displacement of oscillating particle on either side of itsmean position is called its amplitude.

29. A periodic motion repeats after a definite time interval T.

So, y(t) = y(t + T) = y(t + 2T) etc.

30. On a hot day, the velocity of sound will be more since (frequencyproportional to velocity) the frequency of sound increases and hence itspitch increases.

31. On reflection from a denser medium, a wave suffers a sudden phasereversal.

32. If sounds are produced by different musical instruments simultaneously,then all these sounds are heard at the same time.

33. Explosion at the bottom of lake or sea create enormous increase inpressure of medium (water). A shock wave is thus a longitudinal wavetravelling at a speed which is greater than that of ordinary wave.


34. When a wave passes through different media, velocity and wavelengthchange but frequency does not change.

35. Wave velocity is constant for a given medium and is given by v = vButparticle velocity changes harmonically with time and it is maximum atmean position and zero at extreme position.

36. The frequency of vibration depends on the length of the air column andnot on reflecting media, hence frequency does not change.

37. Doppler effect will not be observed, if the source of sound moves towardsthe listener with a velocity greater than the velocity of sound. Same isalso true if listener moves with velocity greater than the velocity of soundtowards the source of sound.

38. Wave length of sound changes.

39. The number of sound waves received by the listener changes.

40. Yes, the sound waves will produce 56 beats every second. But due topersistence of hearing, we would not be able to hear these beats.

41. A medium in which speed of wave motion is independent of frequency ofwave is called non-dispersive medium. For sound, air is non dispersivemedium.

ANSWERS OF TWO MARKS QUESTIONS

1. Condition (i) is not sufficient, because direction of acceleration is notmentioned. In SHM, the acceleration is always in a direction opposite tothat of the diplacement.

2. Although length of the spring does not appear in the expression for thetime period, yet the time period depends on the length of the spring. Itis because, force constant of the spring depends on the length of thespring.

3. The time period of the liquid in a U-tube executing S.H.M. does notdepend upon density of the liquid, therefore time period will be same,when the mercury is filled up to the same height in place of water in theU-tube.

4. We have, 1 k 1 g

v2 m 2


So, when a hard spring is loaded with a mass m. The extension l will belesser w.r.t. delicate one. So frequency of the oscillation of the hardspring will be more and if time period is asked it will be lesser.

5. Restoring force in case of simple pendulum is given by

mgF y K mg /

l

So force constant itself proportional to m as the value of k is substitutedin the formula, m is cancelled out.

6. The pendulum is in a state of weight less ness i.e. g = 0. The frequencyof pendulum

1 gv 0

2 l

7. For air, modulus of rigidity is zero or it does not possess property ofcohession. Therefore transverse waves can not be produced.

8. At a given temperature, the velocity of sound is independent of pressure,so velocity of sound in tube will remain 330ms–1.

9. Two prongs of a tunning fork set each other in resonant vitorations andhelp to maintain the vibrations for a longer time.

10. When the stem of the a tunning fork gently pressed against the top ofsonometer box, the air enclosed in box also vibrates and increases theintensity of sound. The holes bring the inside air incontact with the outsideair and check the effect of elastic fatigue.

11. The displacement at any time t is

y = a sin(t + )

displacement at any time (t + 2/) will be

y = a sin [(t + 2/) + ] = [sin{t + ) + 2}]

y = a sin(t + ) [ sin(2 + ) = sin]

Hence, the displacement at time t and (t + 2/) are same.

12. When a number of waves travel through the same region at the sametime, each wave travels independently as if all other waves were absent.


This characteristic of wave is known as independent behaviour of waves.For example we can distinguish different sounds in a full orchestra.

13. Wave number is the number of waves present in a unit distance of

medium. 1 . S.I. unit is m–1.

Angular wave number or propagation constant is 2/It represents phase

change per unit path difference and denoted by k = 2/ S.I. unit of kis rad m–1.

14. Because the density of water vapour is less than that of the dry air hencedensity of air decreases with the increase of water vapours or humidityand velocity of sound inversely proportional to square root of density.

15. Given,

Pv

(a) Let V be the volume of 1 mole of air, then

M Mor V

V

for 1 mole of air PV = RT

PM P RTRT or

M

RT

vM

(i)

So at constant temperature v

is constant as , R and M are constant

(b) From equation (i) we know that v T, so with the increase intemperature velocity of sound increases.

16. (i) In a pipe open at both ends, the frequency of fundamental noteproduced is twice as that produced by a closed pipe of samelength.

(ii) An open pipe produces all the harmonics, while in a closed pipe,the even harnonics are absent.


17. Bats emit ultrasonic waves of very small wavelength (high frequencies)and so high speed. The reflected waves from an obstacle in their pathgive them idea about the distance, direction, nature and size of theobstacle.

18. At the point, where a compression and a rarefaction meet, the displacementis minimum and it is called displacement node. At this point, pressuredifference is maximum i.e. at the same point it is a pressure antinode. Onthe other hand, at the mid point of compression or a rarefaction, thedisplacement variation is maximum i.e. such a point is pressure node, aspressure variation is minimum at such point.

19. As the temperature increases, the length of the prong of the tunning forkincreases. This increases the wavelength of the stationary waves set up

in the tuning fork. As frequency, 1

, so frequency of the tunning fork

decreases.

20. For an echo of a simple sound to be heard, the minimum distance betweenthe speaker and the walls should be 17m, so in any room having lengthless than 17 m, our ears can not distinguish between sound receiveddirectly and sound received after reflection.

21. The phenomenon of persistence or prolongation of sound after the sourcehas stopped emitting sound is called reverberation. The time for whichthe sound persists until it becomes inaudible is called the reverberationtime.

SOLUTION / HINTS OF NUMERICALS

1. Soln : y = r sinwt 2

r sin tT

Here y = 1

r and T 1s3

1 2r r sin t 2 t 4

T2

1

t s8

2. When x = x1, v = u1


When x = x2, v = u2

As 2 2–v A x

2 2 2 2 2 21 1 1 1– oru A x u A x ...(i)

and 2 2 2 2 2 22 2 2 2– oru A x u A x ...(ii)

Substracting (ii) from (i), we get

2 2 2 2 2 2 2 2 2 11 2 1 2 2 1– – – – –u u A x A x x x

or

1 22 21 22 22 1–

u u

x x

1 22 22 12 21 2

22

–

x xT

u u

3. S oln : H ere y = 5 cm and acceleration a = 45 cm s–2

We know a = 2y

45 = 2 × 5 or = 3 rad s–1

And2 2

T 2.095 s3

4. Soln :

Here mg' = 40 g = 40 × 980 dyne ; l = 4 cm.

say k is the force constant of spring, then

mg = kl or k = mg/

140 980k 9800 dyne cm

4

when the spring is loaded with mass m = 200 g

1 k 1 98002 m 2 200


= 1.113 s–1

5. Soln : Here on earth, T = 3.5 s ; g = 9.8 ms–2

for simple pendulum

T 2g

3.5 29.8

(i)

on moon, g´ = 1.7 ms–2 and if T´ is time period

then T´ 21.7

(ii)

Dividing eq(ii) by eq.(i), we get

T ' 9.8 9.8or T ' 3.5 8.4 s

3.5 1.7 1.7

7. Soln :

Given, r = 25 cm; T = 3s ; y = 12.5 cm

The displacement y = r 2

sin tT

2 212.5 25 sin t or t or t 0.25 s.

3 3 6

The minimum time taken by the particle 2t = 0.5 s

8. Soln : Given, 10.5 s g = 9.8 ms–1

a = 2y = (2)2y = 422y

amax at the extreme position i.e. r = y

amax = 422r and amax = g to remain in contact.

or 2 2 2 2

g 9.8r 0.993 m

4 4 0.5

9. Say 1 in the velocity of sound at T1 = 273°k and 2 = 21 at temperatureT2


Now 2 2 1 2

2 1 1

v T 2 Tv T 273

or T2 = 4 × 273 = 1092°K

10. Here m = 50 kg, l = 0.2 m

we know mg = kl or 1mg 50 9.8k 2450 Nm

0.2

T = 0.60 s and M is the mass of the body, then using

2

2

2450 0.60MT 2 M 22.34 kg

k 4

Weight of body Mg = 22.34 × 9.8 = 218.93 N.

11. Soln : Speed of wave (transverse) in stretched string

Tv

m T is Tension, m is mass per unit length

13

400200 ms

8 100.8

12. Apparent frequency heard by the person is given by

0 0– ´´ or

v v v vv v

But 1´ 90 9, 330 ms

100 10v

03309

10 330v

or 09

330 330 29710

v

v0 = 330 – 297 = 33 ms–1

As the pitch of the sound appears to decrease, so the person is movingaway from the observer.


13. Soln : Given T = 200N, length of sting l = 20m

total mass of the string = 2.5 kg

mass per unit length of the string

12.5m 0.125 kg m

20

1T 200Now 40 ms

m 0.125

Hence time taken by the transverse wave to reach other end

20t 0.5 s

v 40

14. Soln : Given speed of sound in air, = 343 ms–1

length of wire, l = 12.0 m, total mass of wire M = 2.10 kg

1M 2.10m 0.175 kgm

l 12.0

Now

2 42TT m 343 0.175 20, 588.6N 2.06 10 N

m

15. Solu, here y = 10 sin 2 (t – 0.005x)

2

y 10 sin 200 t x200 (i)

The equation of a travelling wave is given by

y = a

2sin vt x (ii)

Comparing the equation (i) and (ii), we have

= 10 cm, = 200 cm and = 200 ms–1

NOW

z

2001 H

200


16. Solu, L = 100 cm T = 20 kg = 20 × 1000 × 980 dyne

14.9m 0.049 g cm

100

Now the frequency of fundamental note produced,

1 T2L m

z1 20 1000 980

100H2 100 0.049

17. Solu : The frequency of nth mode of vibration of a pipe closed at one endis given by

n

2n 14L

river = 340 ms–1, L = 20 cm = 0.2 m ; n = 430 Hz

2n 1 340

430 n 14 0.2

Therefore, first mode of vibration of the pipe is excited, for open pipesince n must be an integer, the same source can not be in resonancewith the pipe with both ends open

18. Total length of the wire, L = 105 cm

1 : 2 : 3 = 1 : 3 : 15

Let L1, L2 and L3 be the length of the three parts. As 1L

1 2 31 1 1

: : : : 15 : 5 : 11 3 15

L L L

Sum of the ratios = 15 + 5 + 1 = 21

1 215 5

105 75 cm; 105 25 cm;21 21

L L

31

105 5 cm21

L


Hence the bridges should be placed at 75 cm and (75 + 25 =) 100 cmfrom one end.

19.

2

y x, t 0.06 sin x cos 120 t3 (i)

(a) The displacement which involves harmonic functions of x and tseparately represents a stationary wave and the displacement,which is harmonic function of the form (vt ± x), represents atravelling wave. Hence, the equation given above represents astationary wave.

(b) When a wave pulse

12

y a sin vt x travelling along x-

axis is superimposed by the reflected pulse.

y2 = –a 2sin vt x

from the other end, a stationery wave is

formed and is given by

y = y1 + y2 =

2 22a sin cos vt (ii)

comparing the eqs (i) and (ii) we have

2 2or 3 m

3

and

2v = 120 or v = 60 = 60 × 3 = 180 ms–1

Now frequency zv 180

60 H3

(c) Velocity of transverse wave in a string is given by

Tv

m

Here–2

–2 –13 10m 2 10 kgm

1.5

Also v = 180 ms–1


T = v2 m = (180)2 × 2 × 10–2 = 648N

20. Solution : frequency of fundamental mode, = 45Hz

Mass of wire M = 3.5 × 10–2 kg; mass per unit length, m = 4.0 × 10–2

kgm–1

2

2M 3.5 10

Length of wire L 0.875 mm 4.0 10

(a) for fundamental mode

L or 2L 0.875 2 1.75 m2

velocity v = = 45 × 1.75 = 78.75 ms–1

(b) The velocity of transverse wave

2 22Tv T v m 78.75 4.0 10 248.6N

m

21. Solution :

Given : = 2.53 kHz = 2.53 × 103Hz

(L) Length of steel rod = 100 cm = 1m.

when the steel rod clamped at its middle executes longitudinal vibrationsof its fundamental frequency, then

L or 2L 2 1 2m

2

The speed of sound in steel

= = 2.53 × 103 × 2 = 5.06 × 103 ms–1

question 14.1: which of the following examples represent...

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