CLASS – 11

WORKSHEET- WAVES

A. INTRODUCTION TO WAVE

(1 Mark Questions)

1. With propagation of longitudinal waves through a medium, the quantity transmitted is

(a) matter (b) energy
(c) matter and energy (d) energy, matter and momentum
Sol. (b)
Only energy is transmitted from one point to another and during propagation of any
longitudinal waves n a medium transmission of energy through the medium without
matter being transmitted.

2. Why are the longitudinal waves also called pressure waves?

Sol. Since propagation of longitudinal waves through a medium creates pressure disturbances
in the medium, hence these waves are called pressure waves.

3. What is the direction of oscillations of the particles of the medium through which (i) a
transverse and (ii) a longitudinal wave is propagating?
Sol. In transverse waves, particles of the medium oscillate in a direction perpendicular to the
direction of propagation of waves. In longitudinal waves particles of the medium
oscillates in the direction of propagation of waves.

4. Ocean waves hitting a beach are always found to be nearly normal to the shore. Why?
Sol. Ocean waves are transverse waves travelling in concentric circles of the ever-increasing
radius. When they hit the shore, their radius of curvature is so large that they can be
treated as plane waves. Hence they hit the shore nearly normal to the shore.

5. Which of the following wave functions does not represent a travelling wave?
(a) y = (x – vt)2 (b) y = log(x +vt) (c) y = 1/x +vt (d) all of these
Sol. (d) The basic requirement for a wave function to represent a travelling wave is that for all
values of x and t wave function must have a finite value. Out of the given functions for y
no one satisfied the given condition. Therefore none can represent a travelling wave.

(2 Marks Questions)

6. Solids can support both longitudinal and transverse waves, but only longitudinal waves
can propagate in gases. Give reason.
Sol. Solids possess both the volume, elasticity and the shear elasticity. Therefore they can
support both longitudinal and transverse waves. On the other hand, gases have only the
volume elasticity and no shear elasticity, so only longitudinal waves can propagate in
gases.

(3 Marks Questions)

7. You have learnt that travelling wave in one dimension is represented by a fraction y =
f(x, t) where x and t must appear in the combination x – v t or x +vt, i.e. y = f(x±vt). Is
the converse true? Examine if the following functions for y can possibly represent a
travelling wave: (a) (x – vt)2 (b) log[(x+vt)x0] (c) exp[-(x+vt)/x0] (d) 1/(x+vt).
Sol. If y = f(x±vt) represents a travelling wave, then the converse may not be true i.e. every
function of x – vt or x +vt may not always a travelling wave. The basic requirement for a
function to represent travelling wave is that it must be finite for all value of x and t.
The functions (i), (ii) and (iv) are not finite for all values of x and t, hence they cannot
represent a travelling wave. Only fraction (iii) satisfies the condition to represent a
travelling wave.

B. PLANE PROGRESSIVE WAVE OR HARMONIC WAVE

(1 Mark Questions)

1. Two astronauts on the surface of the moon cannot talk to each other, why?
Sol. Two astronauts cannot talk on Moon like they do on Earth because sound needs medium
to travel and there is no air on Moon.

2. What is the evidence that (i) sound is a wave, (ii) sound is a mechanical wave and (iii)
sound waves are longitudinal waves?
Sol. (i) Sound waves show the phenomenon of diffraction. (ii) Sound waves require material
for propagation. (iii) Sound waves cannot be polarized.

3. Do displacement, particle velocity and pressure variation in a longitudinal wave vary

with the same phase?
Sol. No particle velocity is out of phase by /2 with the displacement and the pressure
variation is out by phase by it with the displacement.

4. A progressive wave is represented by y = 5sin (100t – 2x) where x and y are in m and t
is in s. The maximum particle velocity is
(a) 100 m/s (b) 200 m/s (c) 400 m/s (d) 500 m/s
Sol. (d)
 The given progressive wave is
y=5sin(100πt−2πx)
particle velocity, νp=dy/dt = 500πsin(100πt−2x)
ν(pmax)=500ms−1

5. The propagation constant of a wave is also called its

(a) wavelength (b) frequency (c) wave number
(d) angular wave number
Ans. (d)

6. A wave of wavelength 2m propagate through a medium. What is the phase difference

between two particles on the line of propagation? Given that the distance between the
particles is 75m.
Sol. The phase difference at any instant of time t, between two particles separated by distance
2π
x is given by = Δx where x is path difference and  is wavelength.
λ
2π
 = ×75;  = 75.
2

7. Newton assumed that sound propagation in a gas takes under

(a) isothermal condition (b) adiabatic condition
(c) isobaric condition (d) isentropic condition
Sol. (a)
Newton assumed that sound propagation in a gas takes under isothermal condition.

8. For vrms is the rms speed of molecules in a gas and v is the speed of sound waves in the
gas, then the ratio vrms/v is
3  3
(a) (b) (c) 3 (d)
 3 
Sol. (a)
3P
rms speed of gas molecule is vrms =√ ρ …(i)

γP
Speed of sound in the gas is v = √ ρ …(ii)

vrms 3
Divide (i) by (ii) we get, = √γ
v

9. What kind of thermodynamical process occur in air, when a sound wave propagates
through it?
Sol. When the sound wave travel through air adiabatic changes take place in the medium.
10. State the factors on which the speed of a wave travelling along a stretched ideal string
depends.
Sol. The speed of a wave travelling along a stretched ideal string depends on the tension on
T
the string (T) and mass per unit length (m).  = √μ

11. What is the effect of pressure on the speed of sound in air? Justify your answer.
γP
Sol. The speed of sound in a gas is given by v = √
ρ

At constant temperature, PV = constant. Pm/ = constant

Since m is constant, so P/ = constant i.e when the pressure changes, density also
changes in the same ratio so that the factor P/ remains unchanged. Hence, the pressure
has no effect on the speed of sound in a gas for a given temperature.

(2 Marks Questions)

12. If the phase difference between two sound waves of wavelength  is 60°, what is the
corresponding path difference?
λ
Sol. Path difference of a given phase difference  is given by, x = 2π δ
Given  = 60° = /3
λ π λ
Therefore x = 2π × 3 = 6

13. Define wave number and angular wave number and give their SI units.
Sol. Wave number is the number of waves present in a unit distance of medium (υ ̅ = 1/λ). SI
unit of υ̅ is m-1. Angular wave number of propagation constant difference is 2. It
represents phase change per unit path difference and denoted by k = 2. SI unit of k is
rad m-1.

14. A steel wire has a length of 12.0m and a mass of 2.10kg. What should be the tension in
the wire so that speed of a transverse wave on the wire equals to the speed of sound in
dry air at 20°C = 343m/s.
Sol. Here l = 12.0m, M = 2.10kg, T = ?, v = 34.3 ms -1
Mass per unit length,  = M/l = 2.10/12.0 = 0.175 kgm-1
T
As v = √μ.

Therefore T = v2 = (343)2 ×0.175 = 2.06×104N

15. Discuss the effect of the following factors on the speed of sound: (a) pressure (b) density
(c) humidity (d) temperature.
Sol. Effect on speed of sound due to:
(a) Pressure: There is no effect of pressure change on speed of sound as long as
temperature remain constant.
(b) Density: The speed of sound is inversely proportional to the square root of density of
medium.
(c) Humidity: Speed of sound increases with increase in humidity.
(d) Temperature: Speed of sound in a gas is directly proportional to square root of its
temperrture.

16. A String of mass 2.50kg is under a tension of 200N. The length of the stretched string is
20.0m. If a transverse jerk is struck at one end of the string, how long does the
disturbance take to reach the other end?
2.50
Sol. Given m = 20.0 kgm-1; T = 200N
T 200×20.0
Speed of transverse jerk is v = √m = √ = √1600 = 40 ms −1
250
Distance 20
Therefore time taken by the jerk to reach the other end= = 40 = 0.5s.
speed

(3 Marks Questions)

 
17. The equations of displacements of two waves are y1  10sin 3t   and
 3
y2  5 sin 3t  3 cos3t  . Find the ratio of their amplitudes.
π
Sol. Here y1 = 10sin[3πt + 3 ]
Therefore amplitude of this wave is A1 = 10 and y2 – 5[sin 3t + √3cos t]
π √3 π π π
= 10[cos 3 sin3πt + cos3πt] = 10 [cos 3 sin3πt + sin 3 cos3πt] = 10sin (3πt + 3 )
w
Therefore amplitude of this wave is A2 = 10
Their corresponding ratio = A1/A2 = 10/10 = 1

18. A mechanical wave travels along a string is described by y(x, t) = 0.005 sin (3.0t – 80x)
in which numerical constants are in SI units. Calculate (a) amplitude of displacement (b)
amplitude of velocity (c) wavelength (d) amplitude of acceleration (e) the time period (f)
frequency of oscillation.
Sol. (a) y(x, t) = 0.005sin(3.0t – 80x)
2πt 2πt
Also, y(x, t) = A sin( − )
T λ
A = 0.005m
2π
(b) Amplitude of velocity A = 3×0.005 = 0.015 ms-1
T
 2π
(c) Wavelength, k = where k = 80
λ
2π π
80 = ; λ = 40
λ
2π 2
(d) Amplitude of acceleration = ( T ) A = (3)2 ×0.005 = 0.045 ms-2
2π 2π
(e) Time period;  = or T = = 120sec
T 3
(f) Frequency,  = 1/T = 0.47Hz.

19. What is the nature of sound waves in air? How is the speed of sound waves in
atmosphere affected by the (i) humidity (ii) temperature?
Sol. The nature of sound wave is longitudinal in air.
γP 1
(i) As v = √ ρ , i. e. v ∝
√ρ
The density of water vapours is less than that of dry air. Since the speed of sound is
inversely proportional to the square root of density, so speed of sound increases with
increase in temperature.
nRT
(ii) We know that, PV = nRT pr P = V
γP γnRT γRT
Also v = √ ρ = √ =√
ρV M

Where M = molecular weight of the gas. As , R and M are constants, so v ∝ √T, i.e.
velocity of sound in a gas is directly proportional to the square root of its temperature
hence we conclude that the velocity of sound in air increases with increase in
temperature.

20. A stone dropped from the top of the tower 300 m high splashes into water of a pond near
the base of the tower. When is the splash heard at the top? Speed of sound in air =
340m/s, g = 9.8 m/s2.
Sol. Let t be the time taken by the stone to reach the water surface.
Here s = 300m, u = 0, a = g = 9.8ms-2
As s = ut + ½ at2
2×300
Therefore 300 = ½ × 9.8 × t 2 or t2 = = 61.2
9.8
Therefore t = √61.2 = 7.82s
distance 300
Time taken by the splash to reach from water surface to the top, t’ = = 340 = 0.88s
speed
Therefore time taken by the splash to be heard on the top = t + t’ = 7.82+0.88 = 8.7s.

21. A steel wire has a length of 12.0 ms and a mass of 2.10kg. What should be the tension in
the wire equals the speed of sound in dry air at 20°C is 343 ms -1?
T T
Sol. Speed of transverse wave in the steel wire is given by v = √ or v 2 = or T = v 2 m
m m
 Given v = 343 ms-1. M = 2.10/12.0 kg m-1
Therefore T = (343)2 × (2.10/12.0) = 2.06×104N

22. Use the formula v = √P/r to explain why the speed of sound in air (a) is independent of
pressure, (b) increases with temperature (c) increases with humidity.
γP
Sol. (a) Take the relation v = √ ρ …(i)
Where, Density, ρ = Mass Volime = M V ρ=Mass/Volume=M/V
M=Molecular weight of the gas V= Volume of the gas
γPV
Hence, equation (i) reduces to : v = √ ... (ii)
M
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.
γP
(b) Take the relation: v = √ ρ ... (i)
For one mole of an ideal gas, the gas equation can be written as: PV = RT
p = RT/V ... (ii)
Substituting equation (ii) in equation (i), we get
γRT γRT
v=√V =√ ... (iv)
ρ M

Where, Mass , M= ρV is a constant, Y and R are also constants

We conclude from equation (iv) 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 speeds of sound in moist air and dry air respectively
Let ρm and ρd be the densities of moist air and dry air respectively
γP
Take the relation v = √ ρ ... (i)

γPV
And the speed of sound in dry air is: vd = √ (ii)
M
On dividing equations (i) and (ii), we get:
γP ρ ρ
vm/vd = √ρ × γPd = √ρ d
m m

However, the presence of water vapour reduces the density of air, i.e.,
ρd < ρ m
∴ vm > vd
 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.

23. A bat emits ultrasound frequency 100 kHz in air. If this sound meets a water surface,
what is the wavelength of (i) the reflected sound (ii) the transmitted sound? Speed of
sound in air = 340 ms_1 and in water = 1486 ms-1.
Sol. Here v = 100kHz = 105Hz, d = 340 ms-1, w = 1486 ms-1
Frequency of both the reflected and transmitted sound remains unchanged.
υa 340
(i) Wavelength of reflected sound, λa = = 105 = 3.4×10-3 m
v
υw 1486
(ii) Wavelength of transmitted sound, λw = = = 1.49×10-2m
v 105

24. A hospital uses and ultrasonic scanner to locate tumours in a tissue. What is the
wavelength of sound in a tissue in which the speed of sound is 1.7 kms -1 ? The operating
frequency of the scanner is 4.2 MHz.
Sol. Here  = 1.7 kms-1 = 1.7×103 ms-1, v = 4.2 MHz = 4.2×106Hz
υ 1.7×103
Wavelength, λ = v = 4.2×106 = 4.047×10-4 m

25. A transverse harmonic wave on a strong is described by:

Y(x, t) = 3.0 sin(36t + 0.081x + /4)
Where x, y are in cm and t in s. The positive direction of x is from left to right.
(i) Is this a travelling or a stationary wave? If it is travelling, what are the speed and
direction of its propagation
(ii) What are its amplitude and frequency?
(iii) What is the initial phase at the origin?
(iv) Hat is the least distance between two successive crests in the wave?
Sol. Given y(x, t) = 3.0sin(36t + 0.018x + /4) …(i)
2π 2π
The standard equation for a harmonic wave is y(x, t) = A sin( T t + x + ϕ0 ) …(ii)
λ
Comparing (i) and (ii) we get A = 3.0, 2/T = 36, 2 = 0.018, 0 = /4
(i) The given equation represents travelling wave propagating from right to left (as x term
is +ve)
λ λ/2π 1/0.018 36
 = T = T/2π = = 0.018 = 2000 cms-1 = 20ms-1
1/36
(ii) Amplitude, A = 3.0cm
1 36 18
Frequency, v = T = 2π = 3.14 = 5.73s-1
(iii) Initial phase at the origin,  = /4 rad
(iv) Least distance between two successive crests is equal to wavelength,
 = 2/0.018 = 349.0cm = 3.49m.
26. 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 travelling wave, (ii) a
stationary wave or (iii) none at all
(a) y = 2cos (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 + 2x sin 2t
Sol. (a) As the function is the product of two separate harmonic functions of x and t, so it
represents a stationary wave.
(b) It cannot represent any type of wave.
(c) Here y = 3 sin (5x – 0.5t) + 4 cos (5x – 0.5t) = 3sin + 4cos  [ = 5x – 0.5t]
If we put A cos  = 3 and A sin  = t then y = a sin()
It represents a simple harmonic travelling wave of amplitude, A = √32 + 42 = 5 and  =
= tan-1(4/3).

(5 marks Questions)

27. For the wave described by y(x, t) = 3.0 sin (36t + 0.018x + /4). Plot the displacement (u)
versus (t) graphs for x = 0, 2 and 4cm. What are the shaped of these graphs? In which
aspects does the oscillatory motion in travelling wave differ from one point to another:
amplitude, frequency or phase?
Sol. All the waves have different phases.
The given transverse harmonic wave is:
y (x,t) = 3.0sin(36t + 0.018x + π/4) ...(i)
For x = 0, the equation reduces to:
y (0,t)=3.0 sin (36t + π/4)
Also, ω=2π/t = 36rad/s−1
∴ t = π/18s
Now, plotting y vs. t graphs using the different values of t, as listed in the given table
t 0 T/8 2T/8 3T/8 4T/8 5T/8 6T/8 7T/8 T
y √3 3 √3 0 −√3 - 3 −√3 0 √3
2 2 2 2 2
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.
28. For the travelling harmonic wave y(x, t) = 2.0 cos 2(10t – 0.0080x + 0.35s), where x
and y are in cm and t is in s. Calculate the phase difference between oscillatory motion of
two points by a distance of (a) 4m (b) 0.5m (c) /2 (d) 3/4.
Sol. Given equation of a travelling harmonic wave is
y(x, t) = 2.0 cos 2p(10t – 0.0080x + 0.35) …(i)
The standard equation of a travelling harmonic wave is
2π 2π
y(x, t) = A cos [ T t − x + ϕ0 ] …(ii)
λ
2π
Comparing eqn (i) and (ii) we get = 2π × 0.0080cm−1 …(iii)
λ
2π
= λ = 2π × 10 and ϕ0 = 0.35
2π
We know that phase difference = × path difference …(iv)
λ
(a) When path difference = 4m = 400cm, then from (iv)
2π
Phase difference = ×400 = 2 × 0.0080 × 400 [by using (iii)]
λ
= 6.4 rad.
(b) When path difference = 0.5m = 50cm then phase difference = 2 ×0.0080×50 = 0.8
rad
2π λ
(c) When path difference 2, then phase difference = × 2 =  rad
λ
2π
(d) When path difference = λ
2π 2π 3π π
Phase difference = × = rad = (π + 2 )
λ λ 2
∵ cos () = - cos 
π
∴ Effective phase difference = 2 rad

29. The equation of a plane progressive wave is given by equation: y = 10 sin 2(t – 0.005x),
where x and y are in cm and t in seconds. Calculate (i) amplitude (ii) frequency (iii)
wavelength (iv) velocity of wave.
Sol. Here y = 10 sin 2(t – 0.005x)
2π
y = 10sin200 (200t − x) …(i)
The equation of a travelling wave is given by
2π
y = a sin λ (vt –x) …(ii)
Comparing the equations (i) and (ii) we have a = 10cm, A = 200cm and v = 200 cms -1
v 200
Now frequency, υ = λ = 200 = 1 Hz.

30. A transverse harmonic wave on a string is described by y(x, t) = 3.0sibn (36t + 0.018x +
/4) 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?
2π
Sol. The equation of the form y(x,t) = A sin( λ (υt + x) + ϕ) …(i)
represents a harmonic wave of amplitude A, wavelength l and traveling from right to left
with a velocity v.
Now the give equation for the transverse harmonic wave is
36 π
y(x, t) = 3.0sin(36t + 0.018x + /4) = 3.0sin [0.018 (0.018 t + x) + 4 ]
= 3.0sin [0.018(2000t +x) + /4) …(ii)
(a) Since the equation (i) and (ii) are of the same form, the gien equation also represents a
travelling wave propagating from right to left. Further the coefficient of t gives the speed
of the wave. Therefore v = 2000cms1 = 20ms-1
(b) Obviously amplitude, A = 3.0cm
2π 2π
Further = 0.018 or  = 0.018cm
λ
v 2000
Therefore υ = λ = × 0.018 = 5.73 s-1
2π
(c) Initial phase at origin, /4 rad
(d)Least distance between two successive crests in the wave is equal to wavelength.
2π
Therefore  = 0.018 = 349.0cm = 3.49m
 x 
31. A standing wave set up in a medium is given by y  4 cos   where x and y are in cm
 3 
and t is in seconds. (i) Write the equation of the two component waves and give
amplitude and velocity of each wave. (ii) What is the distance between the adjacent
nodes? (iii)What is the velocity of the particle of the medium at x = 3cm and time t =
1/8s?
Sol. (i) The standing wave is formed by superposition of the waves y1 = 2sin(40t = x/3) and
y2 = 2sin(40t + x/3) as y = y1 + y2 and their amplitude = 2cm,  = 40 rads-1 and k =
/3 rad cm-1
ω 40π
Velocity, v = = = 120 cm/s
k π/3
2π 2π 2π
(ii) ∵ = k or λ = = π/3 = 6cm
λ k
Distance between adjacent nodes = /2 = 3cm
(iii) Particle velocity at v = 3cm at time t = 1/8s
dy1 πx 40π π×3
v= = 80πcos (40πt − ) = 80πcos ( − ) = 80π m/s
dt 3 8 3
32. The transverse displacement of a string (clamped at its two ends) is given by y(x, t) =
0.06 sin (2/3) × cos 120t where x, y are in m and t in s. The length of the string is 1.5m
and its mass is 3.0×10-2kg. 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 in opposite directions.
What are the wavelength frequency and speed of propagation of each wave?
(c) Determine the tension in the string.
2π
Sol. y(x, t) = 0.06 sin 3 × cos 120 t …(i)
(a) The displacement which involves harmonic functions of x and t separately represents
a stationary wave and the displacement, which is harmonic functuion of the form (ct +x)
represents a travelling wave. Hence the equation given above represents a stationary
wave.
2π
(b) When a wave pulse y1 = a sin λ (ct – x_ travelling along x axis is superimposed by the
reflected pulse.
2π
y2 = - a sin λ (ct +x) from the other end, a stationary wave is formed and is given by
2π 2π
y = y1 + y2 = −asin × cos vt …(ii)
λ λ
On comparing (i) and (ii) we have
2π 2π
= or λ = 3m
λ 3
2π
v = 120π or v = 60λ = 60 × 3 = 180ms−1
λ
v 180
Now frequency υ = λ = = 60 Hz
3
T
(c) Velocity of transverse wave in a string is given by v = √
μ
3×10−2
Here  = = 2 × 10−2 kgm−1
1.5
Also v = 180 ms-1
Therefore T = v2 = (180)2 ×2×10-2 = 648N.

C. REFLECTION AND REFRACTION OF WAVES

(1 Mark Questions)

1. The phenomenon of echo is an example of

(a) reflection (b) refraction (c) beat (d) resonance
Sol. (a)
Like all waves, sound waves can be reflected. Sound waves suffer reflection from the large
obstacles. As a result of reflection of sound wave from a large obstacle, the sound is heard which
is named as an echo. Ordinarily echo is not heard as the reflected sound gets merged with the
original sound. Certain conditions have to be satisfied to hear an echo distinctly (as a separate
 sound).
Hence, the phenomenon of echo of sound waves is due to reflection.

2. When you shout in front of a hill, your own shout is repeated. Explain.
Sol. The sound is heard more than once because of the time difference between the initial
production of the sound waves and their return from the reflecting surface. For example,
when we shout or clap near a suitable reflecting object such as a tall building or a
mountain, we will hear the same sound again a little later.

(2 Marks Questions)

3. Explain why we cannot hear an echo in a small room?

Sol. For an echo of a simple sound to be heard, the minimum distance between the speaker
and the walls should be 17m. so in any room having length less than 17m, our ears cannot
distinguish between sound received directly and sound received after reflection due to
persistence of hearing. The sensation of hearing of any sound persists in our brain for
0.1s.

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

Sol. The phenomenon of persistence or prolongation of sound after the source has stoped
emitting sound is called reverberation. The time for which the sound persists until it
becomes inaudible is called the reverberation time.

D. SUPERPOSITION OF WAVE

E. STANDING WAVES

(1 Mark Questions)

1. Why do stationary waves not transport energy?

Sol. There is no energy transport in a standing wave because the two waves that make them
up carry equal energy in opposite directions.

2. When you shout in front of an open organ pipe, what happens to the wavelength of the
fundamental note?
Sol. In an open organ pipe, the distance between any two consecutive nodes or antinodes is
equal to the half of the wavelength. Thus, for an organ pipe, the wavelength of the
fundamental note is twice the length of the organ pipe.
3. When are stationary waves produced?
Sol. Stationary waves are produced when two exactly identical progressive waves (having the
same amplitude, wavelength & same speed) traveling through a medium along the same
path in exactly opposite directions, interfere with each other.

(2 Marks Questions)

4. What are the differences between stationary waves and progressive waves?

Sol.

Stationary waves Progressive waves

(i) The disturbance remains confined to a The disturbance travels forward, being
particular region, and there is no onward handed over from one particle to the
motion. neighbouring particles.
(ii) There is no transfer of energy in the Energy is transferred in the medium along
medium. the waves.
(iii) The amplitude of vibration of particles The amplitude of vibration of each particle
varies from zero at nodes to maximum at is same.
antinodes.

5. Differentiate between harmonics and overtones.

Sol. A harmonic is made of vibration that is a whole number multiple of the fundamental
mode. The first harmonic is the fundamental frequency. The second harmonic is twice its
frequency etc. Many instruments, especially bells, oscillate in modes that are not whole
number multiples of the fundamental frequency. These higher modes are called
overtones. Overtones include harmonic, but harmonic do not include overtones. The first
overtone is not the fundamental. The second harmonic is the first overtone.

(3 Marks Questions)

6. Give any three differences between progressive wave and stationary wave. A stationary
wave is y = 12 sin 300t cos 2x. What is the distance between two nearest modes?
Sol. Same as Q 4
Node: Point of wave where velocity of particle is maximum e.g., mean position of a wave
is known as node.
y = 12 sin 300t.cos2x [use trigonometric formula, 2sinA.cosB = sin(A+B_) + sin(A – B)]
Now y = 6[sin(300t + 2x) + sin(300t – 2x)]
= 6sin (300t + 2x) + 6sin(300t – 2x)
At mean postionnodwe will be formed so y = 0
sin(300t + 2x) = - sin(300t – 2x) = sin(300t + 2x) = sin( + 300t – 2x)
 4x = -  or x = - /4
Hence node formed at x = /4 and – /4. So distance between node = /4 + /4 = /2.

7. An open pipe has a fundamental frequency of 240 Hz. The first overtone of a closed
organ pipe has the same frequency as the first overtone of the open pipe. How long is
each pipe? Velocity of sound at room temperature is 350m/s.
Sol. The fundamental frequency of the open organ pipe is 240Hz.
The first overtone of an open organ pipe is also the second harmonic.
So its frequency is 240×2 = 480Hz.
If c be the fundamental frequency of a closed organ pipe, the frequency of its first
overtone which is also the third harmonic is 3c.
3c = 480 or c = 160Hz
v v 350
The length of closed organ pipe is given by υC = 4L ⇒ L = 4υ = 4×160 = 54.7cm
C
v
The length of organ open pipe is given by υ0 = 2L
v 350
L = 2υ = 2×240 = 72.9cm.
0

8. The length of a wire between the two ends of a sonometer is 105cm. Where should the
two bridges be placed so that the fundamental frequencies of the three segments are in the
ratio of 1:5:15?
Sol. Total length of the wire, L = 105cm
3 = 1:5:15
Let L1, L2 and L3 be the length of the three parts.
As  ∝ 1/L
Therefore L1:L2:L3 = 15:5:1
Sum of the ratios = 15+5+1=21
Therefore L1 = 15/21 ×105 = 75cm, L2 = 5/21 × 105 = 25cm, L3 = 1/20 × 105 = 5cm
Hence the bridges should be placed at 75cm and (75+25) = 100cm from one end.

9. A metre-long tube open at one end, with a movable piston at the other end, shows
resonance with a fixed frequency source (a turning 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 effect may be neglected.
Sol. The frequency of nth mode of vibration for a closed end pipe is given by
(𝟐𝐧−𝟏)𝛖
vn = where n = 1, 2, 3, …
𝟒𝐋
Suppose the lengths L1 = 25.5cm and L2 = 79.3cm of the resonance columns correspond
to n = n1 and n = n2 respectively. Then
(𝟐𝐧𝟏 −𝟏)𝛖 (𝟐𝐧𝟐 −𝛖)
v= = …(1)
𝟒𝐋𝟏 𝟒𝐋𝟐
 (𝟐𝐧𝟏 −𝟏)𝛖 (𝟐𝐧𝟐−𝟏)𝛖 𝟐𝐧 −𝟏 𝟐𝟓.𝟓 𝟏
𝐨𝐫 𝟑𝟒𝟎 = = or 𝟐𝐧𝟏 −𝟏 = 𝟕𝟗.𝟑 = 𝟑
𝟒×𝟐𝟓.𝟓 𝟒×𝟕𝟗.𝟑 𝟐
This is possible if n1 =1 and n2 = 2. Thus the resonance length 25.5cm corresponds to the
fundamental note and 79.3cm corresponds to first overtone or third harmonic. From
equation (1) we get
𝟒𝐋 𝐯 𝟒×𝟐𝟓.𝟓×𝟑𝟒𝟎
𝛖 = 𝟐𝐧 𝟏−𝟏 = = 34680 cms-1 = 346.8 ms-1.
𝟏 𝟐×𝟏−𝟏

(5 Marks Questions)
10. What are stationary waves? Explain the formation of stationary waves graphically.
Sol. Stationary Waves are defined as a combination of two waves having equal amplitude and
frequency but moving in opposite directions. A standing wave is formed due to
interference. Specifically, a standing wave is a wave that oscillates in time but its peak
altitude profile does not move in space.
Explanation: Consider two harmonic progressive waves (formed by strings) that have the
same amplitude and same velocity but move in opposite directions. Then the
displacement of the first wave (incident wave) is
y1 = A sin (kx – ωt) (waves move toward right) …(1)
and the displacement of the second wave (reflected wave) is
y2 = A sin (kx + ωt) (waves move toward left) …(2)
both will interfere with each other by the principle of superposition, the net displacement
is y = y + y2 …… (3)
Substituting equation (1) and equation (2) in equation (3), we get
y = A sin (kx – ωt) + A sin (kx + ωt) …(4)
Using trigonometric identity, we rewrite equation (4) as
y(x, t) = 2A cos (ωt) sin (kx) …(5)
This represents a stationary wave or standing wave, which means that this wave does not
move either forward or backward, whereas progressive or travelling waves will move
forward or backward. Further, the displacement of the particle in equation (5) can be
written in more compact form,
y(x, t) = A’ cos (ωt) where, A’ = 2A sin (foe), implying that the particular element of the
string executes simple harmonic motion with amplitude equals to A’. The maximum of
this amplitude occurs at positions for which

where m takes half integer or half integral values. The position of maximum amplitude is
known as antinode.
Expressing wave number in terms of wavelength, we can represent the anti-nodal
positions as
The distance between two successive antinodes can be computed by

Similarly, the minimum of the amplitude A’ also occurs at some points in the space, and
these points can be determined by setting sin (kx) = 0 ⇒ kx = 0, π, 2π, 3π, … = nπ where
n takes integer or integral values. Note that the elements at these points do not vibrate
(not move), and the points are called nodes. The nth nodal positions is given by,
F. BEATS

(1 Mark Questions)

1. When two waves of almost equal frequencies v1 and v2 reach at a point simultaneously,
the time interval between successive maxima is
(a) v1 + v2 (b) v1 – v2 (c) 1/ v1 + v2 (d) 1/ v1 – v2
Sol. (d)
When two waves of almost equal frequencies v1 and v2 reach at a point simultaneously,
beats are produced.
Beat frequency, vbeat = v1 – v2
Time interval between successive maxima = 1/vbeat = 1/v1 – v2.

2. Which of the following phenomenon is used by the musicians to tune their musical
instruments?
(a) interference (b) diffraction (c) beats (d) polarization
Ans. (c)

3. Why do we not hear beats due to sound waves emitted by the violin section of an
orchestra?
Sol. All the violin section of an orchestra are tuned to the same frequency. Since there is no
difference in the frequencies of these violins, no bears are heard. (For beats to be heard,
the frequencies should be slightly different).

4. Two sound source produce 12 beats in 4 seconds. By how much do their frequencies
differ?
Sol. Beat frequency is equal to the beats produced per second. Therefore v = 12/4 = 3Hz.

(2 Marks Questions)

5. If two sound waves of frequencies 480Hz and 536Hz superpose, will they produce beats?
Would you hear the beats?
Sol. Yes, the sound waves will produce 56 beats every second. But due to persistence of
hearing, we would not be able to hear these beats.

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

Sol. When the stem f the tuning fork gently pressed against the top of sonometer box, the air
enclosed in box also vibrates and increases the intensity of sound. The holes bring the
inside air in contact with the outside air and check the effect of elastic fatigue.
7. How does the frequency of a tuning fork change, when the temperature is increased?
Sol. As the temperature increases, the length of the prong of the tuning fork increases and
Young’s modulus changes. So frequency of the tuning fork decreases.

(3 Marks Questions)

8. Calculate the speed of sound in a gas in which two waves of lengths 100cm and 101cm
produce 24 beats in 6 seconds.
Sol. Here 1 = 100cm = 1m, 2 = 101cm = 1.01m
v v v v
Let v be the velocity of the sound in the gas. Then υ1 = λ or 1 and υ2 = λ = 1.01
1 2

Beat frequency  2 = 24/6 = 4Hz

v = v/1.01 = 4 of v = 404m/s.

9. Two sitar strings A and B playing the note ‘Ga’ are slight out of tune and produce beats
of frequency 6Hz. The tension in the string A is slightly reduced and the beat frequency
is found to reduce to 3Hz. If the original frequency of A I s324Hz, what is the frequency
of B?
Sol. We know that  ∝ √T where  = frequency, T = tension
The decrease in the tension of a string decreases its frequency. So let us assume that
original frequency A of A is more than B of B.
Thus A – B = ± 6Hz (given) and A = 324Hz.
Therefore 324 – B = ±6 or B = 324±6 = 318Hz or 330Hz
On reducing tension of A,  = 3Hz.
If B = 330Hz and on decreasing tension in A, A will be reduced i.e. no. of beats will
increase, but this is not so because no. of beats becomes 3. Therefore B must be 318Hz
because on reducing the tension in string A, its frequency may be reduced to 321Hz, thus
giving 3 beats with B = 318Hz.

10. What is beat phenomenon?

Sol. When two or more waves superimpose each other with slightly different frequencies,
then a sound of periodically varying amplitude at a point is observed. This phenomenon
is known as beats. The number of amplitude maxima per second is called the frequency.
If we have two sources, then their difference in frequency gives the beat frequency
Number of beats per second, n = |f1 – f2| per second.
(5 Marks Questions)
11. What are beats? Explain their formation analytically? Prove that the beat frequency is
equal to the difference in frequencies of the two superposing waves.
Sol. The periodic variation in the intensity of the sound is called the beats.
Explanation:
When two waves of nearly equal frequencies, traveling in the same direction
superimpose, the intensity of sound at a point changes with time periodically. The
periodic variation in the intensity of sound is called beats.
One loud sound followed by a faint sound or one faint sound followed by one loud sound
constitute a beat.
The time-lapse between two successive loud sounds or two faint sounds is called the beat
period and the reciprocal of the beat period is called beat frequency.
Analytical treatment of beats:
Let the two waves be represented by
(1)
(2)
If y is the resultant displacement due to superimposition of the waves , then

Where A=2rcosπ(ν₁ -ν₂)t is the amplitude of the resultant wave. As intensity is directly
proportional to the square of the amplitude, therefore, intensity is maximum when A is
maximum. i.e.

or
or
i.e. At times ....... intensity will be maximum.
Therefore, the interval between two successive loud sounds is,

Therefore, beat frequency,

i.e. The beat frequency is equal to the difference of frequencies of two superimposing
harmonic waves.
Hence, proved.
G. INTERFERENCE

(1 Mark Questions)

1. Intensities of two waves, which produce interference are 9:4. The ratio of maximum and
minimum intensity is
(a)9:4 (b) 3:2 (c) 25:1 (d) 5:1
Ans. (c)

(3 Marks Questions)

2. Two periodic waves of intensities I1 and I2 pas through a region at the same time in the
same direction. What is the sum of the maximum and minimum intensities?
I
Sol. Other factors such as  and v remaining the same, I = A2 × constant (K) or A = √K
On superposition Amax = A1 + A2 and Amin = A1 – A2
Therefore Amax2 = A12 + A22 + 2 A1A2
I I I 2√I I
⇒ max = K1 + K2 + K1 2
K
Amin2 = A12 – A22 – 2A1A2
Imin I1 I 2√I1 I2
⇒ = − K2 −
K K K

Therefore Imax + Imin = 2I1+ 2I2

H. DOPPLER’S EFFECT

(1 Mark Questions)

1. Doppler effect is applicable for

(a) sound waves only (b) light waves only
(c) both sound and light waves (d) none of these
Ans. (c)

2. A train approaching a railway platform with a speed of 20m/s starts blowing the whistle.
Speed of sound in air is 340 m/s. If the frequency of the emitted sound from the whistle is
640Hz, the frequency of sound as heard by the person standing on the platform is
(a) 600 Hz (b) 640 Hz (c) 680Hz (d) 720 Hz
Sol. (c)
The apparent frequency of sound, observed by a stationary listener , when a source of sound is
approaching to listener , is given by
𝐯
v’ = ( )
𝐯−𝐯𝐬
Given  =640Hz, v = 340m/s, vs = 20m/s
 𝟑𝟒𝟎 𝟔𝟒𝟎×𝟑𝟒𝟎
Hence v’ = 𝟔𝟒𝟎 (𝟑𝟒𝟎−𝟐𝟎) = =680Hz.
𝟑𝟐𝟎

3. What is Doppler effect?

Sol. An increase (or decrease) in the frequency of sound, light, or other waves as the source
and observer move towards (or away from) each other. The effect causes the sudden
change in pitch noticeable in a passing siren, as well as the red shift seen by astronomers.

4. An observer moves towards a stationary source of sound with a velocity one-fifth of the
velocity of sound. The percentage change in the apparent frequency is
(a) zero (b) 5% (c) 10% (d) 20%
Ans. (d)

(2 Marks Questions)

5. What is the speed of the observer for whom a note is 10 percent lower than the emitted
frequency?
Sol. As the apparent frequency () is less than the emitted frequency (), the observer must
move away from the source. If v is the speed of sound and v0 that of the observer, then v’
v−v0
=( )υ
v
As apparent frequency is 10% lower than the emitted frequency,
10 90
∴ υ′ = υ − 100 υ = 100 υ = 0.9υ
v−v0 v−v0
Or 0.9 = ( ) υor 0.9 = ( )
v v
Or 0.9v = v – v0 or v0 – v – 0.9v = 0.1v
Thus the speed of the observer is (1/10)th of the speed of sound.

6. 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 360km/h. What is the frequency of
sound reflected by the submarine. Take the speed of sound in water to be 1450m/s.
Sol. Frequency,  = 45 kHz, speed of sound = 1450 ms-1, speed of enemy submarine = 360
km h-1 = 100 ms-1.
Firstly sound is observed by enemy submarine. Here observer (enemy submarine) is
moving towards the source (SONAR), so va = 100 ms-1, vs = 0. Frequency of sonar waves
received by the enemy submarine
v−v 1450+100 1550
’ = v−v0 × υ = × 40 = 1450 × 40 = 42.76 kHz
s 1450−0
After the sound is reflected, enemy submarine acts as a source of sound of frequency ’.
This source moves with a speed of 100 ms_1 towards the observer (SONAR) so vs = +100
ms-1, v0 = 0. Frequency of sound reflected by the enemy submarine,
v−v0 1450−0
” = × υ′ = 1450−100 × 42.76 = 45.93 kHz.
v−vs
(3 Marks Questions)

7. A whistle revolve in a circle with angular velocity  = 20 rad/s. If the frequency of the
sound is 385Hz and speed is 340 m/s, then find the frequency heard by the observer when
the whistle is at B.

Sol. Speed of source (whistle), vs = r = 0.5×20 = 10m/s

Actual frequency,  = 385 Hz, speed of sound, v = 340 m/s
When the whistle (at point B) is moving towards the observer, then frequency
v 340
υ′ = v−v × υ = 340−10 × 385 = 396.7Hz
s

8. A railway engine and a car are moving parallel but in opposite direction with velocities
144 km/h and 72 km/h respectively. The frequency of engine’s whistle is 500Hz and the
velocity of sound in 340 m/s. Calculate the frequency of sound heard in the car when (i)
the car and engine are approaching each other (ii) both are moving away from each other.
v+v
Sol. Apparent frequency, f ′ = f0 (v−v0 )
0
340+144×5/18 380
(i)f ′ = 500 ( ) = 500 × = 593.75Hz
340−72×5/18 320

v−v 340−144×5/8 300

(ii) f ′ = f0 (v+v0 ) = 500 × ( 340+72×5/8 ) = 500 × 360 = 416.67 Hz
s

(5 marks Questions)

9. 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 a speed of 10m/s.
What are frequency, wavelength and speed of sound for an observer standing on the
station platform? Is the situation exactly identical to the case when the air is still and the
observer runs toward the yard at a speed of 10m/s? The speed of sound in still air can be
taken as 340m/s.
Sol. Here v = 340m/s.  = 400Hz
(a) Speed of wind, vw = 10m/s
As the direction of blow of wind (yard to station) is the same as the direction of sound,
therefore, for the observer standing on the platform;
Velocity of sound, V’ = v +vw = 340+10 = 250m/s
 As there is no relative motion between source of the sound and the observer; the
frequency of sound will remain unchanged.
Thus frequency of sound = 400Hz
v′ 350
Wavelength of sound, λ′ = = 400 = 0.875 m
υ
(b) Speed of the observer, v0 = 10m/s (towards yard). When the observer moves towards
the source of sound, the apparent frequency,
v+v0 340+10
υ′ = υ= × 400 = 411.5 Hz
v 340
The wavelength of sound waves is not affected due to the motion of the observer and
hence the wavelength of the sound waves will remain unchanged.
Speed of sound relative to the observer = 340+10 = 350m/s.
Therefore, situations (a) and (b) are not equivalent.

10. A train standing at the outer signal of a railway station blows a whistle of frequency
400Hz 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 10m/s. (b) recedes from the platform with a
speed of 10m/s.
(ii) What is the speed of sound in each case if the speed of sound in still air is 340 m/s.
Sol. (i) Here  = 400Hz, v = 340m/s
(a) Train approaches the platform, vs = 10m/s
v 340×400
υ′ = v−v × υ = = 412.12 Hz
s 340−10
(b) Train recedes from the platform, vs =10m/s
v 340×400
υ′ = v+v × υ = = 388.6Hz
s 340+10
(ii) The speed of speed in each case remains same i.e. 340m/s.

11. (a) What is Doppler effect?

(b) Derive an expression for the apparent frequency when a source moves towards a
stationary observer.
(c) A policeman on duty detects a drop of 15% in the pitch of the horn of a motor car as it
crosses him. Calculate the speed of car, if the velocity of sound is 330 m/s.
Sol. (a) A Doppler effect: The Doppler effect or Doppler shift (or simply Doppler, when in
context) is the change in frequency of a wave in relation to an observer who is moving
relative to the wave source.
(b) Let n = actual frequency of the source, n0 = apparent frequency of the source, v =
velocity of sound in air, vs = velocity of the source, v1 = velocity of the observer.
If the source is moving towards observer then apparent frequency is given by
v
n = n0 ( ) i.e. apparent frequency increases.
v−vs
 f c c
(c) f1 = [c−v / c+v]
2
100 c+v
= = c−v
90
100c – 100v = 90c – 90v
10c = 190v
v = 10/190 c = 1/19 ×330m/s
v = 17.3 m/s.

12. Explain why (or how):

(a) in a sound wave, a displacement node is a pressure antinode and vice versa.
(b) bats can ascertain distances, directions, nature and sizes of the obstacles without any
“eyes”.
(c) a violin note and sitar note may have the same frequency, yet we can distinguish
between the two notes.
(d) solids can support both longitudinal and transverse waves, but only longitudinal
waves can propagate in gases, and
(e) the shape of a pulse gets distorted during propagation in a dispersive medium.
Sol. (a) At the displacement node (the point of zero displacement), the variation of pressure is
maximum. Hence the displacement node is the pressure antinode and vice-versa.
(b) Bats can produce and detect ultrasonic waves (sound waves of frequencies above
20KHz). (i) From the interval of time between their producing the waves and receiving
the echo after the reflection from an object, they can estimate the distance of the object
from them. (ii) From the intensity of the echo, they can estimate the nature and size of the
object. (iii) Also, from the small time interval between the reception of the echo by their
two ears, they can determine the direction of the object.
(c) The instruments produce different overtones (integral multiples of fundamental
frequency). Hence the quality of sound produced by the two instruments of even same
fundamental frequency is different.
(d) Solids have both volume and shear elasticity. So both longitudinal and transverse
waves can propagate through them. On the other hand, gases have only volume elasticity
and not shear elasticity. So only longitudinal waves can propagate through them.
(e) When the pulse passes through a dispersive medium, the wavelength of the wave
changes. Consequently, the shape of the pulse changes i.e. it gets distorted.

13. 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 ms~1. (b) recedes from the platform
with a speed of 10 ms-1(ii) What is the speed of sound in each case? The speed of sound
in still air can be taken as 340 ms-1 .
Sol. Given, f = 400 Hz, v = 340 ms-1,vs = 10 ms-1
 (i) speed of train .
(a) train approaches the platform

(b) train recedes from the platform

(ii). The speed of sound does not change, i.e., it is 340 ms -1 for both cases.

I. ASSERTION REASON TYPE QUESTIONS:

(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of
assertion.
(c) If assertion is true but reason is false (d) If both assertion and reason are false
(e) If assertion is false but reason is true
1. Assertion: Transverse waves are not produced in liquids and gases.
Reason: Light waves are transverse waves.
Ans. (e) Assertion is false, but reason is true.
2. Assertion: Sound passes through air in the form of longitudinal waves.
Reason: Longitudinal waves are easier to propagate.
Ans. (c) Assertion is true but reason is false
A longitudinal wave motion travel in the form of compressions and rarefaction which
involve changes in volume and density of the medium. As air possesses volume
elasticity, therefore sound comes to us from the source in the form of longitudinal waves
only.
3. Assertion: Particle velocity and wave velocity both are independent of time.
Reason: For the propagation of wave motion, the medium must have the properties of
elasticity inertia.
Ans. (e) Assertion is false but reason is true
The velocity of every oscillating particle of the medium is different of its different
positions in one oscillation but the velocity of wave motion is always constant i.e. particle
velocity vary with respect to time, while the wave velocity is independent of time.
4. Assertion: Wave produced by a motor boat sailing in water are both longitudinal and
transverse waves.
Reason: The longitudinal and transverse waves cannot be produced simultaneously.
Ans. (c) Assertion is true but reason is false
The propeller of a motor boat cuts the water surface laterally and also pushes it in
backward direction. Hence it will result in both longitudinal and transverse waves.
5. Assertion: Violet shift indicates decrease in apparent wavelength of light.
Reason: Violet shift indicates decrease in apparent wavelength of light.
Ans. (b) Both assertion and reason are true but reason is not the correct explanation of
assertion.
As c < r. therefore violet shift means apparent wavelength of light from a star
decreases. Obviously apparent frequency increases. This would happen when the star is
approaching the earth.

J. CHALLENGING PROBLEMS:
1. One end of a long string of linear mass density 8.0 x 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
(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.
Sol. Tension in the string, T = 90×9.8 = 882N; Mass per unit length of the string, m= 8.0×10 -3
kg m-1; Frequency of the wave, v = 250Hz; Amplitude of the wave, A = 5.0cm = 0.05m
The velocity of the transverse wave along the string is
T 882
υ = √m = √8×10−3 = 3.32 × 102 ms −1
Angular frequency,  = 2v = 2×3.142×256 = 16.1×102 rad s-1
ω ω 16.1×102
As υ = ∴k= = 3.31×102 = 4.84 rad m−1
k υ
As the wave propagates along the positive X axis, so the displacement equation is
y = A sin (t – kx) or y = 0.05sin(16.1×102t – 4.84x), x and y are in m.

2. 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 ms -1.
Sol. Same as Q 6 Sec H.

3. Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can
experience both transverse (S) and longitudinal (P) sound waves. Typically the speed of S
wave is about 4.0 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?
Sol. Suppose the earthquake occurs at a distance of x km from the seismograph.
 Speed of S wave = 4.0 km s-1
Time taken by the S wave to reach the seismograph = x/4 s
Speed of P wave = 8.0 kms-1
Time taken by the P wave to each the seismograph = x/8 s
x x
But 4 − 8 = 4×60s
Therefore x = 1920 km.

4. 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?
Sol. As the bat approaches the stationary flat wall surface, the apparent frequency is
υ
v ′ = υ−υ v
s
The stationary flat surface (source) reflects the sound of frequency v’ to the bat
(observer) moving towards the flat surface. So the apparent frequency is
υ+υ0 υ+υ0 υ υ+υ0
v" = × v′ = × υ−υ v = ×v
υ υ s υ−υs
υ+0.03υ 1.03
= υ−0.03υ × v = 0.97 × 40kHz = 42.47 kHz.