ncert11.1

32 g is the molecular mass of oxygen. Hence, the unknown gas is oxygen.

Q. A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly
true only if there are no external influences. A gas column under gravity, for example, does not
have uniform density (and pressure). As you might expect, its density decreases with height.
The precise dependence is given by the so-called law of atmospheres

n2 = n1 exp [-mg (h2 – h1)/ kBT]
Where n2, n1 refer to number density at heights h2 and h1
respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension

in a liquid column: n2 = n1 exp [-mg NA(ρ - P′) (h2 –h1)/ (ρRT)]
Where ρ is the density of the suspended particle, and ρ’ that of surrounding medium. [NA is Avogadro’s

number, and R the
universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the
suspended particle.]

Answer
According to the law of atmospheres, we have:

n2 = n1 exp [-mg (h2 – h1) / kBT] … (i) where,

n1 is the number density at height h1, and n2 is the number density at height h2
mg is the weight of the particle suspended in the gas column Density of the medium = ρ'

Density of the suspended particle = ρ Mass of one suspended particle = m' Mass of the medium
displaced = m Volume of a suspended particle = V
According to Archimedes’ principle for a particle suspended in a liquid column, the effective
weight of the suspended particle is given as:

Weight of the medium displaced – Weight of the suspended particle
= mg – m'g
= mg - V ρ' g = mg - (m/ρ)ρ'g
= mg(1 - (ρ'/ρ) ) ....(ii)

Gas constant, R = kBN

kB = R / N ....(iii)
Substituting equation (ii) in place of mg in equation (i) and then using equation (iii), we get:

n2 = n1 exp [-mg (h2 – h1) / kBT]

= n1 exp [-mg (1 - (ρ'/ρ) )(h2 – h1)(N/RT) ]
= n1 exp [-mg (ρ - ρ')(h2 – h1)(N/RTρ) ]

Q. Given below are densities of some solids and liquids. Give rough estimates of the

size of their atoms:
[Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known
value of Avogadro’s number. You should, however, not take the actual numbers you obtain for

various atomic sizes too literally. Because of the crudeness of the tight packing approximation,

the results only indicate that atomic sizes are in the range of a few Å].

Ans.
If r is the radius of the atom, then volume of each atom = 4/3 π r3
Volume of all atoms in one mole of substance = 4/3 π r3 × N = M/ρ
∴ r = [ 3M / 4πρN]1/3

For Carbon,

-3
M = 12.01 × 10 Kg

ρ = 2.22 × 3 Kg -3
10 m

Similarly,

for gold, r = 1.59 Å

for liquid nitrogen, r = 1.77 Å for lithium, r = 1.73 Å

for liquid fluorine, r = 1.88 Å

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

Ans.
(a) The swimmer's motion is not periodic. Though the motion of a swimmer is to and fro but will not have a

definite period.

(a) The motion of a freely-suspended magnet, if displaced from its N-S direction and released, is periodic because
the magnet oscillates about its position with a definite period of time.

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

14. Oscillations

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

Answer

A.It is priodic but not simple harmonic motion because it is not to and fro about a fixed point.

B.It is a simple harmonic motion because the mercury moves to and fro on the same path, about the
fixed position, with a certain period of time.

C.It is simple harmonic motion because 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.

(d) A polyatomic molecule has many natural frequencies of oscillation. Its vibration is the superposition of
individual simple harmonic motions of a number of different molecules. Hence, it is not simple harmonic, but
periodic.
Q. depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion?
What is the period of motion (in case of periodic motion)
Ans.

(a) It is not a periodic motion. This represents a unidirectional, linear uniform motion. There is no repetition

of motion in this case

(b) In this case, the motion of the particle repeats itself after 2 s. Hence, it is a periodic motion, having a period
of 2 s.
(c) It is not a periodic motion. This is because the particle repeats the motion in one position only. For a

periodic motion, the entire motion of the particle must be repeated in equal intervals of time.

(d) In this case, the motion of the particle repeats itself after 2 s. Hence, it is a periodic motion, having a

period of 2 s.

14.4. Which of the following functions of time represent (a) simple harmonic, (b) periodic but not

simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion (ω is

any positive constant):

A.sin ωt – cos ωt

3 ωt
B.sin

C.3 cos (π/4 – 2ωt)
D.cos ωt + cos 3ωt + cos 5ωt
E.exp (–ω2t2)

Answer

(a) SHM
The given function is: sin ωt – cos ωt

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

(b) Periodic but not SHM The given function is:

3 ωt = 1/4 [3 sin ωt - sin3ωt]
sin

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.

Ites period is: 2π/ω

(c) SHM
The given function is:

This function represents simple harmonic motion because it can be written in the form: a cos (ωt + Φ)Its
period is: 2π/2ω = π/ω
(d) Periodic, but not SHM
The given function is cosωt + cos3ωt + cos5ωt. Each individual cosine function represents SHM.
However, the superposition of three simple harmonic motions is periodic, but not simple harmonic.
(e) Non-periodic motion

The given function exp(-ω2t2) is an exponential function. Exponential functions do not repeat themselves.
Therefore, it is a non-periodic motion.

(f) The given function 1 + ωt + ω2t2 is non-periodic.

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

A.at the end A,
B.at the end B,
C.at the mid-point of AB going towards A,
D.at 2 cm away from B going towards A,
E.at 3 cm away from A going towards B, and
F.at 4 cm away from B going towards A.
Ans.

From above figure, where A and B represent the two

extreme positions of a SHM. FOr velocity, the direction from A to B is taken to b positive. The acceleration and

the force, along AP are taken as positive and alsong Bp are taken as negative.

(a) At the end A, the particle executing SHM is momentarily at rest being its extreme position of motion. Therefore,
its velocity is zero. Acceleration is positive becaue it is directed along AP, Force is also Positive since the force is
directed along AP.

B.At the end B, velocity is zero. Here, accleration and force are negative as they are directed along BP.

C.At the mid point of AB going towards A, the particle is at its mean psoiton P, with a tendancy to move along PA.
Hence, velocity is positive. Both accleration and force are zero.

D.At 2 cm away from B going towards A, the particle is at Q, with a tendancy to move along QP, which is
negative direction. Therefore, velocity, acceleration and force all are positive.

E.At 3 sm away from A going towards B, the particle is at R, with a tendancy to move along RP, which is
positive direction. Here, velocity, acceleration all are positive.

F.At 4 cm away from A going towards A, the particles is at S, with a tendancy to move along SA, which is
negative direction. Therefore, velocity is negative but acclereation is directed towards mean position,
along SP. Hence it is positive and also force is positive similarly.

Q.. Which of the following relationships between the acceleration a and the displacement x of

a particle involve simple harmonic motion?

A.a = 0.7x
B.a = –200x2
C.a = –10x

3
D.a = 100x

Ans.ted automaticalIn SHM, acceleration a is related to displacement by the relation of the form a = -kx, which is for
relation (c)

Q. 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.

Ans.
Intially, at t = 0; Displacement, x = 1 cm Intial velocity, v = ω cm/ sec.
Angular frequency, ω = π rad/s–1 It is given that,

x(t) = A cos(ωt + Φ)

1 = A cos(ω × 0 + Φ) = A cos Φ

A cosΦ = 1 ...(i)

Velocity, v= dx/dt
ω = -A ωsin(ωt + Φ)

1 = -A sin(ω × 0 + Φ) = -A sin Φ A sin Φ = -1 ...(ii)

Squaring and adding equations (i) and (ii), we get: 2 2 Φ + 2 Φ) = 1 + 1
2 A (sin cos

A =2

∴ A = √2 cm

Dividing equation (ii) by equation (i), we get
tanΦ = -1

∴ Φ = 3π/4 , 7π/4,......

SHM is given as: x = Bsin (ωt + α)

Putting the given values in this equation, we get: 1 = Bsin[ω × 0 + α] = 1 + 1

Bsin α = 1 ...(iii)

Velocity, v = ωBcos (ωt + α) Substituting the given values, we get:

π = πBsin α

Bsin α = 1 ...(iv)

Squaring and adding equations (iii) and (iv), we get: 2 2 α + 2 α] = 1 + 1
2 B [sin cos

B =2

∴ B = √2 cm

Dividing equation (iii) by equation (iv), we get: Bsin α / Bcos α = 1/1

tan α = 1 = tan π/4

∴ α = π/4, 5π/4,......

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

Answer

Maximum mass that the scale can read, M = 50 kg
Maximum displacement of the spring = Length of the scale, l = 20 cm = 0.2 m
Time period, T = 0.6 s

Maximum force exerted on the spring, F = Mg where,

2
g = acceleration due to gravity = 9.8 m/s F = 50 × 9.8 = 490
-1
∴ Spring constant, k = F/l = 490/0.2 = 2450 N m. Mass m, is suspended from the balance.

∴Weight of the body = mg = 22.36 × 9.8 = 219.167 N Hence, the weight of the body is about 219 N.

–1
14.9. A spring having with a spring constant 1200 N m 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.

Answer

–1
Spring constant, k = 1200 N m
Mass, m = 3 kg

PDDisplacement, A = 2.0 cm = 0.02 cm
(i) Frequency of oscillation v, is given by the relation:

Hence, the frequency of oscillations is 3.18 cycles per second.

(ii) Maximum acceleration (a) is given by the relation: a = ω2 A
where,

2
Hence, the maximum acceleration of the mass is 8.0 m/s .
(iii) Maximum velocity, vmax = Aω

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

Q.In Exercise 14.9, let us take the position of mass when the spring is unstreched
as x = 0, and the direction from left to right as the positive direction of x-axis. Give
x as a function of time t
for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is
A.at the mean position,
B.at the maximum stretched position, and
C.at the maximum compressed position.
In what way do these functions for SHM differ from each other, in frequency, in
amplitude or the initial phase?

Answer
Distance travelled by the mass sideways, a = 2.0 cm Angular frequency of oscillation:

(a) As time is noted from the mean position, hence using x = a sin ω t, we have x = 2 sin 20 t
(b) At maximum stretched position, the body is at the extreme right position, with an intial phase of π/2 rad.
Then,

(c) At maximum compressed position, the body is at left position, with an intial phase of 3 π/2 rad.
Then,

The functions neither differ in amplitude nor in frequency. They differ in intial phase.
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 anti-clockwise) 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.
Ans.
(a) Time period, t = 2 s Amplitude, A = 3 cm
At time, t = 0, the radius vector OP makes an angle π/2 with the positive x-axis,.e., phase angle Φ = +π/2
Therefore, the equation of simple harmonic motion for the x-projection of OP, at the time t, is given by
the displacement equation:

(b) Time Period, t = 4 s Amplitude, a = 2 m
At time t = 0, OP makes an angle π with the x-axis, in the anticlockwise direction, Hence, phase angle
Φ = +π Therefore, the equation of simple harmonic motion for the x-projection of OP, at the time t, is
given as:

14.12. Plot the corresponding reference circle for each of the following simple harmonic motions.
Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of
the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every
case: (x is in cm and t is in s).
= x = –2 sin (3t + π/3)
= x = cos (π/6 – t)
= x = 3 sin (2πt + π/4)
= x = 2 cos πt
Ans.

(a)

Amplitude, A = 2 cm
Phase angle, Φ = 5π/6 = 150°.
Angular velocity = ω = 2π/T = 3rad/sec.
The motion of the particle can be plotted as shown in fig. 10(a).
(b)

Amplitude, A = 1

Phase angle, Φ = -π/6 = -30°. Angular velocity, ω = 2π/T = 1 rad/s.
The motion of the particle can be plotted as shown in fig. 10(b).

(c)

Amplitude, A = 3 cm
Phase angle, Φ = 3π/4 = 135°
Angular velocity, ω = 2π/T = 2 rad/s.
(d)

The motion of the particle can be plotted as shown in fig. 10(c).
Amplitude, A = 2 cm Phase angle, Φ = 0
Angular velocity, ω = π rad/s.
The motion of the particle can be plotted as shown in fig. 10(d).
14.13. Figure 14.30 (a) shows a spring of force constant k clamped rigidly at one end and a mass m
attached to its free end. A force F applied at the free end stretches the spring. Figure 14.30 (b) shows
the same spring with both ends free and attached to a mass m at either end. Each end of the spring in
Fig. 14.30(b) is stretched by the same force F.

(a) What is the maximum extension of the spring in the two cases?
(b) 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

(a) The maximum extension of the spring in both cases will = Flk, where k is the spring constant of the
springs used.
(b) In Fig.14.30(a) if x is the extension in the spring, when mass
m is returning to its mean position after being released free, then restoring force on the mass iss F = -kx, i.e., F ∝ x

As, this F is directed towards mean position of the mass, hence the mass attached to the spring will
execute SHM.
Spring factor = spring constant = k
inertia factor = mass of the given mass = m As time period,

In Fig.14.30(b), we have a two body system of spring constant k and reduced mass, µ = m × m / m + m =
m/2.
Inertia factor = m/2 Spring factor = k

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?

Ans.
Angular frequency of the piston, ω = 200 rad/ min. Stroke = 1.0 m
Amplitude, A = 1.0/2 = 0.5 m
The maximum speed (vmax) of piston is given by the relation: vmax = Aω = 200 × 0.5 = 100m/min.

14.15. 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 earth is 3.5 s? (g on the surface of earth is 9.8 ms–2)

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

–2
Acceleration due to gravity on the surface of earth, g = 9.8 m s Time period of a simple pendulum on earth,
T = 3.5 s

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

14.16. Answer the following questions:
(a) Time period of a particle in SHM depends on the force constant k and mass m of the
particle:
T = 2π √m/k. A simple pendulum executes SHM approximately. Why then is the time period
of a pendulum independent of the mass of the pendulum?
(b) 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 2π √l/g
Think of a qualitative argument to appreciate this result.
(c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give
correct time during the free fall?
(d) What is the frequency of oscillation of a simple pendulum mounted in a cabinthat is

freely falling under gravity?

Ans.
(a) For a simple pendulum, force constant or spring factor k is proportional to mass m, therefore, m
cancels out in denominator as well as in numerator. That is why the time period of simple pendulum is
independent of the mass of the bob.

(b) In the case of a simple pendulum, the restoring force acting on the bob of the pendulum is given as:
F = –mg sinθ where,

F = Restoring force m = Mass of the bob
g = Acceleration due to gravity θ = Angle of displacement

For small θ, sinθ ≈ θ

For large θ, sinθ is greater than θ. This decreases the effective value of g. Hence, the time
period increases as:
T = 2π √l/g

(c) Yes, because the working of the wrist watch depends on spring action and it has nothing to do with
gravity.
(d) Gravity disappears for a man under free fall, so frequency is zero.

14.17. A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is
moving on a circular track of radius R with a uniform speed v. If the pendulum makes small
oscillations in a radial direction about its equilibrium position, what will be its time period?
Ans.
The bob of the simple pendulum will experience the acceleration due to gravity and the centripetal
acceleration provided by the circular motion of the car.
Acceleration due to gravity = g

2
Centripetal acceleration = v /R where,

v is the uniform speed of the car R is the radius of the track
Effective acceleration (g') is given as:

Q.Cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρ1. The
cork is depressed slightly and then released. Show that the cork oscillates up and down simple
harmonically with a period where ρ is the density of cork. (Ignore damping due to viscosity of the
liquid).
Ans.
Base area of the cork = A
Height of the cork = h
Density of the liquid = ρ1
Density of the cork = ρ
In equilibrium:
Weight of the cork = Weight of the liquid displaced by the 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 ρ1 g .....(i) Accroding to the force law:
F = kx k = F/x

where, k is constant k = F/x = -Aρ1 g ....(ii)

The time period of the oscillations of the cork:
T = 2π √m/k ....(iii) 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:--

Q. One end of a U-tube containing mercury is connected to a suction pump and the other end
to atmosphere. A small pressure difference is maintained between the two columns. Show
that, when the suction pump is removed, the column of mercury in the U-tube executes
simple harmonic motion.
Ans.
Area of cross-section of the U-tube = A Density of the mercury column = ρ
Acceleration due to gravity = g
Restoring force, F = Weight of the mercury column of a certain height
= = –(Volume × Density × g)
== –(A × 2h × ρ × g) = –2Aρgh = –k × Displacement in one of the arms (h)
Where,
2h is the height of the mercury column in the two arms k is a constant, given by k = -F/h = 2Aρg

where,
M is mass of mercury column

Let l be the length of the total mercury in the U-tube.
Mass of mercury, m = Volume of mercury × Density of mercury =Alρ

Hence, the mercury column executes simple harmonic motion with time period 2π √l/2g .

Q.An air chamber of volume V has a neck area of cross section a into which a ball of mass 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 pressure-volume variations of air to be isothermal [see Fig. 14.33].

Ans.
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, V = ax Volumetric strain = Change in volume/original volume
⇒ V/V = ax/V

Bulk modulus of air, B = Stress/Strain = -p/ax/V

In this case, stress is the increase in pressure. The negative sign indicates that pressure increases with decrease in
volume.
p = -Bax/V
The restoring force acting on the ball, F = p × a
= -Bax/V .a

2
= -Bax /V ......(i)
In simple harmonic motion, the equation for restoring force is: F = -kx .....(ii)

2
where, k is the spring constant Comparing equations (i) and (ii), we get: k = Ba /V
Time Period,

Q.You are riding in an automobile of mass 3000 kg.Assuming that you are
examining the oscillation characteristics of its suspension system. The
suspension sags 15 cm when the entire automobile is placed on it. Also, the

amplitude of oscillation decreases by 50% during one complete oscillation.
Estimate the values of (a) the spring constant k and
(b) the damping constant b for the spring and shock absorber system of one
wheel, assuming that each wheel supports 750 kg.

Ans.
(a) Mass of the automobile, m = 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:
F = –4kx = mg
Where, k is the spring constant of the suspension system Time period, T = 2π √m/4k

44

and k = mg/4x = 3000 × 10/ 4 × 0.15 = 5000 = 5 × 10 Nm Spring Constant, k = 5 × 10 Nm

(b) Each wheel supports a mass, M = 3000/4 = 750 kg
For damping factor b, the equation for displacement is written as:

-bt/2M
x = x0e

The amplitude of oscilliation decreases by 50 %. ∴ x = x0/2
-bt/2M

x0/2 = x0e

loge2 = bt/2M

∴ b = 2M loge2 / t
where,

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

Answer

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

x = Asin ωt where,

A = Amplitude of oscillation ω = Angular frequency = √k/M

The velocity of the particle is:v = dx/dt = Aωcosωt The kinetic energy of the particle is:

Ek = 1/2 2 = 1/2 2 ω2 2 ωt
Mv MA cos

The portential energy of the particle is: Ep = 1/2 2 = 1/2 2 ω2 2 2 ωt
kx M A sin

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:

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.

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

by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius
of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant α

is defined by the relation J = –α θ, where J is the restoring couple and θ the angle of twist).

Ans.

Mass of the circular disc, m = 10 kg Radius of the disc, r = 15 cm = 0.15 m

The torsional oscillations of the disc has a time period, T = 1.5 s The moment of inertia of the disc is:
2

I = 1/2mr
2

= 1/2 × (10) × (0.15)

= 0.1125 kg/m2
Time Period, T = 2π √I/α

= is the torisonal constant.

= = 4π2 I / 2 2
(π)2 T (1.5)

= 4 × × 0.1125 /

= 1.972 Nm/rad
–1

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

Q. 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

r = 5 cm = 0.05 m T = 0.2 s
ω = 2π/T = 2π/0.2 = 10 π rad/s

When displacement is y, then acceleration, A = -ω2 y

Case (a) When y = 5cm = 0.05 m

A = -(10π)2 × 0.05 = -5π2 2
m/s

Case (b) When y = 3 cm = 0.03 m

A = -(10π)2 × 0.03 = -3π2 2
m/s

Case (c) When y = 0
A = -(10π)2 × 0 = 0

Q. 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
centre with a velocity v0 at time t = 0. Determine the amplitude of the resulting
oscillations in terms of the parameters ω, x0 and v0. [Hint: Start with the
equation x = a cos (ωt+θ) and note that the initial velocity is negative.]

Answer
The displacement rquation for an oscillating mass is given by:
== Acos(ωt + θ) where,
A is the amplitude
=is the displacement θ is the phase constant
Velcoity, v = dx/dt = -Aωsin(ωt + θ) At t = 0, x = x0
x0 = Acos θ = x0 ....(i) and, dx/dt = -v0 = Aωsinθ Asinθ = v0/ω ...(ii)
LSquaring and adding equations (i) and (ii), we get:

PDF generated automatically by the HTML to PDF API of PDFmyURL

15.WAVES

Q. 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?

Ans.

Mass of the string, M = 2.50 kg Tension in the string, T = 200 N Length of the string, l =

20.0 m

-1
Mass per unit length, µ = M/l = 2.50/20 = 0.125 kg m
The velocity (v) of the transverse wave in the string is given by the relation:

v = √t/µ
= √200/0.125 = √1600 = 40 m/s

∴ Time taken by the disturbance to reach the other end, t = l/v = 20/40 = 0.5 s.

Q. 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

–1 –2
340 m s ? (g= 9.8 m s )

Answer

Height of the tower, s = 300 m Initial velocity of the stone, u = 0
2

Acceleration, a = g = 9.8 m/s Speed of sound in air = 340 m/s

The time (t1) taken by the stone to strike the water in the pond can be calculated using the second equation of
motion, as:

2
s = ut1 + 1/2 gt1

2
300 = 0 + 1/2 × 9.8 × t1
∴ t1 = √300 × 2/9.8 = 7.82 s
Time taken by the sound to reach the top of the tower, t2 =300/340 = 0.88 s

Therefore, the time after which splash is heard, t = t1 + t2 = 7.82 + 0.88 = 8.7 s.

Q. 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 air at 20 °C
–1

= 343 m s .

Answer

Length of the steel wire, l = 12 m Mass of the steel wire, m = 2.10 kg

Velocity of the transverse wave, v = 343 m/s
-1

Mass per unit length, µ = m/l = 2.10/12 = 0.175 kg m

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

v = √T/µ ∴ T = vu

= 2 × 0.175 = 20588.575 ≈ 2.06 × 104 N.
(343)

15.4. Use the formula v = √γP/ρ to explain why the speed of sound in air
(i) is independent of pressure,
(j) increases with temperature,
(k) increases with humidity.
Ans.
(a) Take the relation:
v = √ γP/ρ ....(i)
where,
Density, ρ = Mass/Volume = M/V
M = Molecular weight of the gas V = Volume of the gas
Hence, equation (i) reduces to: v = √γPV/M ....(ii)
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.
(b) Take the relation: v = √γP/ρ ....(i)
For one mole of any ideal gas, the equation can be written as:

PV = RT
P = RT/V ....(ii)
Substituting equation (ii) in equation (i), we get: v = √γRT/Vρ = √γRT/M .....(iii)
where,

mass, M = ρV is a constant γ and R are also constants

We conclude from equation (iii) that v ∝ √T
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.

(c) Let vm and vd be the speed of sound in moist air and dry air

Respectively Let ρm and ρd be the densities of the moist air and dry air respectively

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

d< m

∴ vm > vd
Hence, the speed of sound in mois air is greater than it is in dry air. Thus, in gaseous medium, the speed

of sound increases with humidity.

Q. 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:
(o) (x – vt)2
(p) log [(x + vt) / x0]
(q) 1 / (x + vt)

Answer

No, the converse is not true. The basic requirements for a wave function to represent a travelling wave is
that for all values of x and t, wave function must have finite value.
Out of the given functions for y, no one satisfies this condition. Therefore, none can represent a travelling

wave.

Q. 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–1 and in water 1486 m s–1.

Answer

(a) Frequency of the ultrasonic sound, ν = 1000 kHz = 6 Hz Speed of sound in air, va = 340 m/s
10

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

relation: λr = v/v
6 -4

= 340/10 = 3.4 × 10 m.

(b) Frequency of the ultrasonic sound, ν = 1000 kHz = 6 Hz Speed of sound in water, vw = 1486 m/s
10

The wavelength of the transmitted sound is given as: λr = 1486 × 6 = 1.49 × -3 m.
10 10

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 s–1? The operating frequency of the scanner is 4.2
MHz.

Ans.

3
Speed of sound in the tissue, v = 1.7 km/s = 1.7 × 10 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:
λ = v/v

36 -4
= 1.7 × 10 / 4.2 × 10 = 4.1 × 10 m.

15.8. A transverse harmonic wave on a string is described by

Where x and y are in cm and t in s. The positive direction of x is from left to right.
(a) Is this a travelling wave or a stationary wave?
If it is travelling, what are the speed and direction of its propagation?
(b) What are its amplitude and frequency?
(c) What is the initial phase at the origin?
(d) What is the least distance between two successive crests in the wave?

Ans.(a) The equation of 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 compaaring both the equations, we find that equation (ii) represents a travelling wave, propgating
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:

v = ω/2π and λ = 2π/k Also,
v = vλ
∴ v = (ω/2π) / (2π/k) = ω/k
= 36/0.018 = 2000 cm/s = 20 m/s
Hence, the speed of the given travelling wave is 20 m/s.

(b) Amplitude of the given wave, a = 3 cm Frequency of the given wave:

v = ω/2π = 36 / 2 × 3.14 = 573 Hz

(c) On comparing equations (i) and (ii), we find that the intial phase angle, Φ = π/4

(d) The distance between two successive crests or troughs is equal to the wavelength of the wave.
Wavelength is given by the relation: k = 2π/λ

∴ λ = 2π/k = 2 × 3.14 / 0.018 = 348.89 cm =3.49 m.

Q. For the wave described in last question, plot the displacement (y) versus (t) 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, frequency or phase?
Ans.
All the waves have different phases.
The given transverse harmonic wave is:

For x = 0, the equation reduces to:

Also,

ω = 2π/t = -1 ∴ t = π/18 s
36rad/s

Now, plotting y vs. t graphs using the different values of t, as listed in the given table
t (s) 0 T/8 2T/7 3T/8 4T/8 5T/8 6T/8 7T/8

y 3 3/√2 -- 0
3/√2 0 3/√2 –3 3/√2

(cm)

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 x. The y-t plots of the three waves are shown in the given
figure.

Q. 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

=======

= 4m
= 0.5 m,
= λ/2
= 3λ/4

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: Φ = kx =
2π/λ

(a) For x = 4 m = 400 cm
Φ = 0.016 π × 400 = 6.4 π rad

(b) For 0.5 m = 50 cm
Φ = 0.016 π × 50 = 0.8 π rad

(c) For x = λ/2
Φ = 2π/λ × λ/2 = π rad

(d) For x = 3λ/4
Φ = 2π/λ × 3λ/4 = 1.5π rad.

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 in s. The length of the string is 1.5
m and its mass is 3.0 ×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 each wave?
= Determine the tension in the string.

Ans.(a) The general equation representing a stationary wave is given by the displacement function:
y (x, t) = 2a sin kx cos ωt
This equation is similar to the given equation:

Hence, the given equation represents a stationary wave.
(b) A wave travelling along the positive x-direction is given as: y1 = a sin (ωt - kx)
The wave travelling along the positive x-direction is given as: y2 = a sin (ωt + kx)
The supposition of these two waves yields: y = y1 + y2 = a sin (ωt - kx) - a sin (ωt + kx)
=a sin (ωt) cos (kx) - a sin (kx) cos (ωt) - a sin (ωt) cos (kx) - a sin (kx) cos (ωt)
= 2 a sin (kx) cos (ωt)

∴Wavelength, λ = 3 m It is given that:

120π = 2πν Frequency, ν = 60 Hz Wave speed, v = νλ = 60 × 3 = 180 m/s

(c) The velocity of a transverse wave travelling in a string is given by the relation:
v = √T/µ ....(i) where,

Velocity of the transverse wave, v = 180 m/s
–2

Mass of the string, m = 3.0 × 10 kg Length of the string, l = 1.5 m

Mass per unit length of the string, µ = m/l
-2

= 3.0 × 1.5 = 10
-2 -1

= 2 × 10 kg m Tension in the string = T

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

T = v2μ –2
2
= (180) × 2 × 10

= 648 N

Q. (i) 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?

Ans.All the points on the string
= have the same frequency except at the nodes (where frequency is zero)
= have thse same phase everywhere in one loop except at the nodes.
However, the amplitude of vibration at different points is different.

Q. 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
stationary wave or (iii) none at all:
(a) y = 2 cos (3x) sin (10t)
(b) y = 2 √x - vt

(c) y = 3 sin (5x – 0.5t) + 4 cos (5x – 0.5t)

(d) y = cos x sin t + cos 2x sin 2t
Answer
(a) The given equation represents a stationary wave because the harmonic terms kx and ωt appear

separately in the equation.

(b) The given equation does not contain any harmonic term. Therefore, it does not represent either a

travelling wave or a stationary wave.
(c) The given equation represents a travelling wave as the harmonic terms kx and ωt are in the combination of kx
– ωt.
(d) The given equation represents a stationary wave because the harmonic terms kx and ωt appear

separately in the equation. This equation actually represents the superposition of two stationary waves.

Q.. A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45

Hz. The mass of
–2 –2

the wire is 3.5 × 10 kg and its linear mass density is 4.0 × 10

–1
kg m . What is (a) the speed of a transverse wave on the string, and (b) the tension in the
string?

Ans. –2

(a) Mass of the wire, m = 3.5 × 10 kg
-2 -1
Linear mass density, μ = m/l = 4.0 × 10 kg m wire, l = m/μ = 3.5 × –2 / 4.0 × -2 = 0.875 m
Frequency of vibration, v = 45 Hz∴ length of the 10 10

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

= = 2l/m where,

n = Number of nodes in the wire For fundamental node, n = 1:

= = 2l

= = 2 × 0.875 = 1.75 m

The speed of the transverse wave in the string is given as:
v = νλ= 45 × 1.75 = 78.75 m/s

(b) The tension produced in the string is given by the relation:

2
T=v µ

2 –2
= (78.75) × 4.0 × 10 = 248.06 N

Q. 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

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 given by the relation:

l1 = π/4 where,

length of pipe, l1 = 25.5 cm = 0.255 m
∴ λ = 4l1 = 4 × 0.255 = 1.02 mThe speed of the sound is given by the relation: v = vλ = 340 × 1.02 = 346.8
m/s.

Q. 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
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 node is λ/2 ∴ l = λ/2
λ = 2l = 2 × l = 2 m
The speed of sound in steel is given by the relation: v = νλ

3
= 2.53 × 10 × 2

3
= 5.06 × 10 m/s
= 5.06 km/s

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

–1
open? (Speed of sound in air is 340 m s ).

Answer

First (Fundamental); No

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

Source frequency = th normal mode of frequency, νn = 430 Hz Speed of sound, v = 340 m/s
n

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

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

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

vn = nv/2l n = 2lvn/v

15.18. 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 frequency is found to
reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?

Answer

Frequency of string A, fA = 324 Hz
Frequency of string B = fB
Beat’s frequency, n = 6 Hz
Beat's Frequency is given as:

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:
v ∝ √T
Hence, the beat frequency cannot be 330 Hz ∴ fB = 318 Hz.

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 only longitudinal waves can propagate
in gases, and
= The shape of a pulse gets distorted during propagation in a dispersive medium.

Answer

= A node (N) is a point where the amplitude of vibration is the minimum and pressure is the maximum.
An antinode (A) is a point where the amplitude of vibration is the maximum and pressure is the
minimum.
Therefore, a displacement node is nothing but a pressure antinode, and vice versa.

= Bats emit ultrasonic waves of large frequencies. When these waves are reflected from the obstacles in their
path, they give
them the idea about the distance, direction, size and nature of the obstacle.

= 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.

= This is because solids have both, the elasticity of volume and elasticity of shape, whereas gases
have only the volume elasticity.

= A sound pulse is a combination of waves of different wavelength. As waves of different λ travel in a
dispersive medium with different velocities, therefore, the shape of the pulse gets distorted.

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 frequency of the whistle for a platform observer when the train

–1
(a) approaches the platform with a speed of 10 m s , (b)

–1
recedes from the platform with a speed of 10 m s ? (ii) What is the speed of sound in each case?

–1
The speed of sound in still air can be taken as 340 m s .

Answer
(i) (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 (v') of the whistle as the train approaches the platform is given by the
relation
:

(ii) 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.

15.21. A train, standing in a station-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
1. 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

–1
observer runs towards the yard at a speed of 10 m s ? The speed of sound in still air can be taken as

–1
340 m s .
Ans.
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:
λ = ve/v = 350/400 = 0.875 m
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 (v').
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.

15.22. 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 t = 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 point at t
= 2 s, 5 s and 11 s.

Answer

(a) The given harmonic wave is

= 90 coss (732.81°) = 90 cos (90 × 8 + 12.81°)

= 90 cos (12.81°)

= 90 × 0.975 =87.75 cm/s
Now, the equation of a propagating wave is given by: y (x, t) = asin (kx + wt + Φ)

where, k = 2π/λ

∴ λ = 2π/k and, ω = 2πv ∴ v = ω/2π

Speed, v = vλ = ω/k where,

ω = 12rad/s k = 0.0050 -1
m

∴ v = 12/0.0050 = 2400 cm/s

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

propagation.

(b) Propagation constant is related to wavelength as: k = 2π/λ
∴ λ = 2π/k = 2 × 3.14 / 0.0050 = 1256 cm = 12.56 m

Therefore, all the points at distance nλ (n = ±1, ±2, .... and so on), 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.
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) 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 1/20 or 0.05 Hz?
Answer

(a) A short pip be a whistle has neither a definite wavelength nor a definite frequency. However, its speed of
propagation is fized, being equal to speed of sound in air.
(b) No, frequency of the note produced by a whistle is not 1/20 = 0.05 Hz. Rather 0.05 Hz is the
frequency of repetition of the short pip of the whistle.

–3
15.24. One end of a long string of linear mass density 8.0 × 10

–1
kg m 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 (fork end) of
the string x = 0 has zero transverse displacement (y = 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, μ = 8.0 × 10-3 kg m-1 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 × 3 – 4.84 x) m.
10 t

15.25. A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. An enemy submarine moves
towards the

–1
SONAR with a speed of 360 km h . What is the frequency of sound reflected by the submarine?

–1
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, ve = 360 km/h = 100 m/s
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 (v')

received and reflected by the submarine is given by the relation:

Q. Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both
–1

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 . 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.

15.27. 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 bat hear reflected off the wall?

Answer

Ultrasonic beep frequency emitted by the bat, ν = 40 kHz Velocity of the bat, vb = 0.03 v
where, v = velocity of sound in air
The apparent frequency of the sound striking the wall is given as: