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

Mass per unit length   =
Tension = T = 200 N

Velocity =
Time taken .


A stone dropped from the top of the tower 300m high splashes in to the water of a pond near the base of the tower. When is the splash heard at the top, given that speed of sound in air is 340m/s?

Height of the tower, H = 300 m
Speed of the sound in air, V = 340 m/s
Acceleration due to gravity, g = 9.8 m/s2
Initial velocity, u = 0

Let t2 be the time taken to cover 300 m, .
Total time for splash to be heard, T = t1 + t2 = (7.82 + 0.88) s = 8.70 s.


A steel wire has a length of 12 m and a mass of 2.10 kg. What should be the tension in the wire so that the speed of the transverse wave in the wire equals the speed of sound in dry air at 20°C (i.e. 343 m/s)?

Mass per unit length M
Velocity v = 343 ms-1

We know that v
Therefore v2
T = Mv2


Use the formula V= to explain why the speed of sound in air
i. is independent of pressure
ii. increases with temperature
iii. increases with humidity.

i. The pressure P and volume V of a gas of mass m at given temperature are related as PV = constant
or = constant, where
ρ is the density of the gas.
or = constant
Hence the velocity of sound in a gas is independent of pressure.

ii. Let Vt and V0 be the velocity of sound in gas at t°C or 0°C respectively. From the given equation , we have Vt =
and V0 =
or    -----(i)

From the gas equation,
Substituting in eq. (1), we have

Thus, the velocity of sound in a gas is proportional to the square root of its absolute temperature. For small changes in temperature and we can use the binomial expansion to write
terms of order and higher are neglected. Thus V = V + 0.61t, since for air V = 332 ms-1
Thus, the velocity of sound in air increases by about 0.61 ms-1 for each degree Celsius rise in temperature.

iii. The presence of water vapour in a gas lowers its density. Consequently, sound travels faster in humid gas than in dry gas under the same conditions. An increase in humidity increases the speed of sound.


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-vt or x+vt (i.e ) y = f (x ±vt ). Is the converse true? In other words, does every function of x-vt or x+vt represent a travelling wave? Examine if the following functions for y can possibly represent a traveling wave:
(i) (x - vt )2
(ii) log(x + vt)
(iii) (iv)

The converse is not true because every function of (x - vt) or (x + vt) is not necessarily a wave function. The requirement for an acceptable function for a travelling wave is that it should be finite everywhere and at all times, only function (iii) satisfies this requirement. The remaining functions cannot possibly represent a wave.


A bat emits ultrasonic sound of 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 = 340m/s, in water = 1486 m/s )?

(i) Reflected sound λair
(ii) Transmitted sound


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

Speed = 1.7 × 103 ms-1
Frequency = 4.2
× 106 Hz



A transverse harmonic wave on a string is described by
y(x, t ) = 3 sin (36 t + 0.018x +
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 its frequency?

iii. What is the initial phase at the origin?

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

(i) The given equation can be written as
y = 3.0 sin (0.018x + 36t +
This represents a wave travelling in the negative x-direction. Comparing this with the standard equation,


This is the least distance between two successive crests in the wave. Comparing the terms of t,

Velocity =
                  = 2000 cm s-1  

(ii) Amplitude = A = 3.0 cm

(iii) Frequency = 

(iv) Initial phase   .


For a wave described as y(x, t ) = 3 sin (36 t + 0.018x + π/4) plot the displacement (y) versus (t) graphs for x = 0, 2 and 4 cm. what are the shape of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?

The given equation is
             y = 3.0 sin [36t + 0.018x +

At x = 0 cm,
            y = 3.0 sin                            -------(1)

The value of y at t = 0 is given by
            y = 3.0 sin [
               = 2.1 cm

At x = 2 cm
            y = 3.0 sin [36t + 0.036x +
π/4]           -------(2)
The value of y at t = 0 is given by
            y = 3.0 sin [0.036x +
               = 2.2 cm

At x = 4 cm
            y = 3.0 sin [36t + 0.072x +
π/4]           -------(3)
The value of y at t = 0 is given by
           y = 3.0 sin [0.072x +
             = 2.6 cm

Equation (1), (2) and (3) are the sinusoidal. They all have equal amplitude frequency. All the graph in the below figure are sinusoidal, but they differ in initial phase.


For the travelling wave
y = 2 cos2
π(10t - 0.0080x + 0.35)
where x &y are in cm and t in s. What is the phase difference between the oscillatory motion at the two points separated by the distance of

(i) 4m,

(ii) 0.5 m,


(i) Phase difference
                                 = 2
π ×0.008 ×100 ×4            (since 1/λ = 0.0080 cm)
                                 = 20.1 rad

(ii) 2.5 rad
π rad.


The transverse displacement of a string (clamped at it's two ends) is given by
y(x, t)= 0.060  
where x and y are in metres and t is in seconds. The length of the string is 1.5m and its mass is 3
×10-2 kg. Answer the following. 

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

(b) Interpret the wave as a superposition of two waves travelling in opposite direction. What are the wavelength, frequency and speed of propagation of each wave?

(c) Determine the tension in the string?

(a) The function does not have the characteristic form (vt + x) of a travelling wave. Hence the function does not represent a travelling wave; it represents a stationary wave.

(b) The stationary wave in the given question may be written as y = 2A sin
α cos βWhere A = 0.03 m, and β = 120πt.
Now 2A sin
α cos β = A sin (α + β) + A sin (α - β)
                            = 0.03 sin
i.e. y = y0 + y2
Hence the two component waves are

λbe the wavelength, ν the frequency and v the speed of each wave. Then
= coefficient of x in the argument of the sine function =
λ = 3 m

ω = 2πν = coefficient of t in the argument of the sine function
                   = 120
πwhich gives ν = 60 Hz.
Hence v =
νλ = 60 × 3 = 180 m/s

(c) Mass per unit length (m) =
We know that
ν = where T is tension in the string.
T = mν2 = 2.0 × 10-2 × (180)2 = 648 N.


For the wave on a string described in the previous problem, do all the points on the string oscillate with the same frequency, phase and amplitude?

In a stationary wave on a string, all points on the string vibrate with the same frequency and the same phase, but the amplitude is zero at nodes and maximum at antinodes.


Given below are some functions of x and t to represent the displacement of an elastic wave. State which of these represent (1) A travelling wave (2) A stationary wave (3) None. 

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


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

(d) y = cos x sin t + cos 2x sin 2t

(a) Stationary
(b) Unacceptable function for any wave.
(c) Travelling wave
(d) Superposition of 2 stationary waves.


A wire stretched between 2 rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10-2 kg and its linear density is 4 × 10-2 kg/m. What is

(a) the speed of the transverse wave on the string and

(b) tension on the string?

Frequency ν = 45 Hz
Linear density
μ  = Mass of wire/Length of wire
                        = = 0.875 m

For fundamental mode, 
λ       = 2L = 2 × 0.875

(a) Velocity of wave v =
ν × λ
Velocity v          = 45 × 2 × 0.875
                           = 78.8 ms-1

We know           v =
Therefore           v2 =          
(b) T =
μv2 = 4 × 10-2 × (78.8)2 = 248.4 N.


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 340Hz) When the tube length is 25.5cm and 79.3cm. Estimate the speed of sound in air at the temperature of the experiment and determine the end-correction. What is this correction for?

In our discussion we have taken the open end of a closed pipe to be an antinode. This is not strictly true. In fact, the particles of air just at the open end are not perfectly free because of the restriction imposed by the pipe. The true antinode is slightly away from the open tube as shown in the figure below.


The distance e is called end-correction. The effective length of the pipe = (L + e).
The tuning fork is in resonance at two lengths is about three times the first, 25.5 and 79.3 cm. Since the second resonance length is about three times the first, it is clear that the fork is in resonance with the first harmonic (i.e. the fundamental mode) when L1 = 25.5 cm and with the third harmonic when L2 = 79.3 cm. Thus we have,


where N is the frequency of the tuning fork. Dividing (ii) by (i) we get
3(25.5 + e) = 79.3 + e or e = 1.4 cm
Then the speed of sound as obtained from (i) is
V = 4N(L1 + e)
   = 4
× 340 × (25.5 + 1.4)
   = 36584 cms-1 = 365.84 ms-1
   ~ 366 ms-1.


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

For a rod fixed at its middle l            
λ                                      = 2 × 100
                                                         = 200 cm = 2m
Therefore the speed of the sound in steel v = n
λ= 2.53 × 103 × 2= 5060 ms-1.


A pipe 20 cm long is closed at 1 end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will this same source be in resonance with the pipe if both ends are open?

Let N be the frequency of the source and np that of the pth harmonic of a closed pipe. The source will resonantly excite that harmonic mode of the pipe for which N = νπ for any value of p = 1,3,5………
Now for a closed pipe we know that

Therefore for resonance,               
Putting N = 430 Hz, v = 340 m/s and L = 20 cm = 0.2 m, we have
which gives p = 1.01 ~ 1
Hence the source of frequency 430 Hz will resonantly excite the first harmonic (i.e. fundamental mode).
In a open pipe, the condition of resonance with the same source will be

which is not an integer. Hence the source will not be in resonance with any harmonic of the open pipe.


Two sitar strings A and B playing the note 'Ga' are slightly 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 3 Hz. If the original frequency of A is 324Hz, what is the frequency of B?

Beat of frequency produced by the strings A and B = 6 Hz
Original frequency of the string A = nA = 324 Hz
Therefore frequency of the string B = nB = 324
±  6 Hz
                                                         = 330 or 318 Hz
Now, the frequency of a string is proportional to the square root of tension. Hence, if the tension in A is slightly reduced, its frequency will be slightly decreased, i.e. it will become less than 324 Hz. If the frequency of strong B is 330 Hz, the beat frequency would increase to a value greater than 6 Hz if the tension in A is reduced. But the beat frequency is found to decrease to 3 Hz. Hence, the frequency of B cannot be 330 Hz; it is, therefore 318 Hz. When the tension in A is reduced, its frequency becomes 324 - 321 Hz which will produces beats of frequency 3 Hz with string B of frequency 318 Hz.


Explain why
(a) in a sound wave, displacement node is pressure antinode and vice versa.

(b) bats can ascertain distances, directions, nature and sizes of the obstacles without any 'eyes'.

(c) the reverberation time is larger for an empty hall than a crowded hall.

(d) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes. 

(e) solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and

(f) the shape of a pulse gets distorted during propagation in a dispersive medium.

(a) At the point, where a compression and a rarefaction intersect, the displacement is minimum and is called displacement node. On the other hand, at the mid-points of a compression or rarefaction, the displacement variation is maximum, such a point is called displacement antinode. It is also know as pressure node as the pressure variation is minimum at that point.

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

(c) Audience are also absorbers of sound. The total absorption in case of a crowded hall is more, therefore, as per sabine formula, reverberation time of a crowded hall is smaller than that of an empty hall.

(d) The fundamental note produced by an instrument is accompanied by a number of overtones, the number differs from one instrument to the other. This fact determines the quality of the sound produced by the instrument. Thus, a violin note and a sitar note may have the same frequency. But the two notes appear different because of the different number of overtones accompanying them.

(e) The gases possess only volume elasticity. A such, only longitudinal waves can propagate in gases. On the other hand, the solid possess both volume and shear elasticity and likewise both the longitudinal and transverse waves can propagate through them.

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


A train standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air.

(i) What is the frequency of the whistle for a platform observer when the train

(a) approaches the platform with a speed of 10 m/s

(b) recedes form the platform with a speed of 10 m/s

(ii) What is the speed of sound in each case?

Frequency of the whistle n = 400 Hz
Velocity of the sound V = 340 ms-1
Velocity of listener Vl = 0
Velocity of source Vs = 10ms-1


The direction in which the sound travels to reach the observer is taken as +ve
(a) The train is approaching the listener. So Vs is +ve,
     i.e. Vs = + 10 ms-1.

(b)  When the train recedes from the listener, Vs is -ve    


A train standing in a station yard blows a whistle of frequency 400Hz in still air.

(i) A wind starts blowing in the direction from the yard to the station with the speed of 10 m/s. What are the frequency, wavelength and speed of sound for an observer standing on the stations platform?

(ii) Is the situation exactly equivalent to the case when the air is still and the observer runs towards the yard at a speed of 10m/s?

ν = 400 Hz, v = 340 m/s.
(i) Speed of wind um = 10 m/s. Since the wind is blowing in the direction in which the sound travels, the effective   
Speed of sound = v + um = 340 + 10 = 350 m/s.
Since the observer and the source are both stationary, the frequency of sound heard by the observer = frequency of the source = v = 500 Hz. The wavelength of sound heard by the observer is
= 0.875m
(ii) Speed of observer u0 = 10 m/s. Since the observer is approaching the source of sound, the speed of sound relative to the observer is (u + u0) = 340 + 10 = 350 m/s. 
Since the source is stationary, the wavelength of sound remains unchanged and is given by

The frequency of sound heard by the observer is
The two situations are not equivalent as the frequency and the wavelength of sound heard by the observer are different, although the effective speed of sound is the same. The reason is that in case
(i), with respect to the medium (which is moving with the wind), both the source and observer are in motion. In case
(ii) only the observer is in motion; the medium and the source remaining stationary.


A travelling harmonic wave on the string is described by
(i) What are the displacement and velocity of oscillation of a point at x = 1 cm and t = 1s? Is this velocity equal to the velocity of wave propagation? 

(ii) Locate the points on the string which have the same transverse displacements and velocity as the x = 1cm point and t = 2s, 5s, 11 s.

The particle displacement are given by
                        y(x,t) = 7.5 sin
Let us first find the wavelength and the wave velocity. Now
= coefficient of x in the argument of sine function = 0.005

Also,  = coefficient of t in the argument of the sine function = 12
  = 24 m s-1

Since the wave is travelling in the -x direction at a speed of 24 m s-1. The wave velocity is -24 ms-1.
(a) Putting x = 1 cm and t = 1 s in eq (i), we have
Displacement at x = 1 cm and t = 1 s
                                            = 7.5 sin (12.005 + 0.785) 
                                            = 7.5 sin (12.79) rad
                                            = 7.5 sin (732.7°) (since
π rad = 180°)
                                            = 7.5 sin (4
π + 12.7)
                                            = 7.5 sin (12.7°)
                                            = 7.5
× 0.2198
                                            = 1.65 cm
Particle velocity is obtained by differentiating eq (i) with respect to t.
We get, V(x, t) = = 7.5
× 12 × cos     -----(ii)
Putting x = 1 cm and t = 1s in eq (ii) we get, as before
Velocity at (x = 1 cm and t = 1s) = 7.5
× 12 cos 12.7°= 87.8 cm/s
which is not equal to the wave velocity of -24 m/s

(b) By definition of wavelength (l), all points on the string which are
λ, 2λ, ……….. = nλ away from x = 1 cm (here n is an integer and λ = 12.67 m) have the same displacement and velocity as those at x = 1 cm for all t values.


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) wavelength
(ii) frequency
(iii) speed of propagation? 

(b) If the pulse rate is 1 after 20s, (i.e. the whistle is blown for a split second after every 20s), is the frequency of the note produced by the whistle equal to 1/20 = 0.05 Hz.

(a) The pulse need not have a wavelength or frequency. But has a definite speed of propagation.

(b) The frequency of production of pulse is 1/20 Hz. This cannot be equal to the frequency of sound wave.


One end of a long string of linear mass density = 8.0 ×10-3 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 end absorbs all the incoming energy so that reflected waves at this 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 a function of x and t that describes this wave on the string.

Tension in the string, T = 90 ×9.8 = 882 N
Mass per unit length of the string, m = 8.0
×10-3 kg/m
Frequency of the wave,
ν = 256 Hz
Amplitude of the wave a = 5.0 cm = 0.05 m
As the wave propagating along the string is a transverse travelling wave, the velocity of the wave is given by
v = = 3.2
×102 m/s
ω = 2πn = 2 ×3.14 ×256
                      = 16.1
×102 rad/s
We know that
ν = ω/K
K = = 4.84 rad/m

As the wave propagates along positive x-direction, the equation of the wave is
y = a sin (wt - Kx)
   = 0.05 sin (16.1
×102 t - 484x).


A SONAR system fixed in a submarine operates at a frequency of 40 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km/h. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 m/s.

(a) Frequency of sound received by submarine
V = 40
× 103 Hz
Sonar is the source i.e Vs = 0
Submarine is the listener Vl = 360 kmh-1 = 100 ms-1
Speed of sound in water V = 1450ms-1

The listener moves towards the source V1 is -ve

= 42.75
× 103 Hz

(b) Frequency of the reflected sound received by sonar.
Now sonar is the listener , i. e, V1 = 0
Submarine is the source, i.e., Vs = 100ms-1. it travels in the direction of sound. So Vs is positive. The reflected sound has a frequency 42.75
×103 Hz

= = 45.91
× 103 Hz.


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 and that of P wave is 8.0 km/s. A seismograph records P and S waves from an earthquake. The first P waves arrives 4 min before the first S wave. Assuming the waves travel in straight line, how far away does the earthquake occur?

Time taken t = 4 minutes
                   = 4
× 60 = 240 seconds
Velocity of the transverse wave v1 = 4.0 km/s = 4000 m/s
Velocity of the longitudinal wave v2 = 8 km/s = 8000 m/s
T =
240 =

S = = 192000 m/s = 1920 km/s.


A bat is flitting about in a cave, navigating via ultrasonic bleeps. 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?

Frequency of the sound emitted by bat ν0 = 40 kHz
Speed of the air v = 340 m/s
Speed of the bat = 0.03v
            = kHz
            = 42.27 kHz.

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