1 Marks Questions

1.Explosions on other planets are not heard on earth. Why?

Ans.This is because no material medium is present over a long distance between earth and planets and is absence of material medium for propagation, sound waves cannot travel.

2.Why longitudinal waves are called pressure waves?

Ans.Because propagation of longitudinal waves through a medium, involves changes in pressure and volume of air, when compressions and rarefactions are formed.

3.Why do tuning forks have two prongs?

Ans.The two prongs of a tuning fork set each other is resonant vibrations and help to maintain the vibrations for a longer time.

4.Velocity of sound increases on a cloudy day. Why?

Ans.Since on a cloudy day, the air is wet i.e. it contains a lot of moisture, As a result of which the density of air is less and since velocity is inversely proportioned to density, hence velocity increases.

5.Sound of maximum intensity is heard successively at an interval of 0.2 second on sounding two tuning fork to gather. What is the difference of frequencies of two tuning forks?

Ans.The beat period is 0.2 second so that the beat frequency is fb = = 5HZ. Therefore, the difference of frequencies of the two tuning forks is 5HZ.

6.If two sound waves has a phase difference of 60⁰, then find out the path difference between the two waves?

Ans.Phase difference, rad

Now, in general for any phase difference, , the path difference (x) :→

7.If the displacement of two waves at a point is given by:-

Calculate the resultant amplitude?

Ans . If a₁ = amplitude of first wave

a₂ = amplitude of second wave

a_r = resultant amplitude

8.   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? The operating frequency of the scanner is 4.2 MHz.

Ans. Speed of sound in the tissue, v = 1.7 km/s =

Operating frequency of the scanner, v = 4.2 MHz =

The wavelength of sound in the tissue is given as:

9. 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)

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

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

Ans.(a) The given equation represents a stationary wave because the harmonic terms kx and t 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 t are in the combination of kx - t.

(d) The given equation represents a stationary wave because the harmonic terms kx and t appear separately in the equation. This equation actually represents the superposition of two stationary waves.

10.  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 or 0.05 Hz?

Ans.(a) (i)No

(ii)No

(iii)Yes

(b) No

Explanation:

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

(b) The short pip produced after every 20 s does not mean that the frequency of the whistle is or 0.05 Hz. It means that 0.05 Hz is the frequency of the repetition of the pip of the whistle.

2 Marks Questions

1.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 this source be in resonance with the pipe if both the ends are open?

Ans.Length of pipe = L = 20cm = 0.2m

Frequency of n^th node =  _n = 430 Hz

Velocity of sound = v = 340m|s

Now,  of closed pipe is :→

Hence, it will be the first normal mode of vibration, In a pipe, open at both ends we, have

As n has to be an integer, open organ pipe cannot be in resonance with the source.

2.Can beats be produced in two light sources of nearly equal frequencies?

Ans .No, because the emission of light is a random and rapid phenomenon and instead of beats we get uniform intensity.

3.A person deep inside water cannot hear sound waves produces in air. Why?

Ans .Because as speed of sound in water is roughly four times the sound in air, hence refractive

index u =

For, refraction r _max = 90⁰, i_max=14⁰. Since i_max ≠ r_max hence, sounds gets reflected in air only and person deep inside the water cannot hear the sound.

4.If the splash is heard 4.23 seconds after a stone is dropped into a well. 78.4 metes deep, find the velocity of sound in air?

Ans .Here, depth of well = S = 78.4m

Total time after which splash is heard = 4.23s

If t₁ = time taken by store to hit the water surface is the well

t₂ = time taken be splash of sound to reach the top of well.

then t₁ + t₂ = 4.23 sec.

Now, for downward journey of stone;

u = O, a = 9.8 m|s², S = 78.4m,

t = t1 = ?

As, s = ut + at²

If V = velocity of sound in air,

V = 

5.How roar of a lion can be differentiated from bucking of a mosquito? 

Ans. Roaring of a lion produces a sound of low pitch and high intensity whereas buzzing of mosquitoes produces a sound of high pitch and low intensity and hence the two sounds can be differentiated.

6.The length of a sonometer wire between two fixed ends is 110cm. Where the two bridges should be placed so as to divide the wire into three segments whose fundamental frequencies are in the ratio of 1:2:3?

Ans .Let l₁, l₂ and l₃ be the length of the three parts of the wire and f₁, f₂ and f₃ be their respective frequencies.

Since T and m are fixed quantities, and 2 are constant

So, f₁ l₁ = Constant →(1)

f₂ l₂ = Constant (2)

f₃ l₃ = Constant →(3)

Equating equation 1), 2) & 3)

f₁ l₁ = f₂ l₂ = f₃ l₃

Now, l₂ = 

Also, 

Now, Total length = 110cm

i.e l₁ +l₂+l₃= 110cm

i. e. l₁ = 60cm

Now,

7.If string wires of same material of length l and 2l vibrate with frequencies 100HZ and 150 HZ. Find the ratio of their frequencies?

Ans.Since frequency = f of a vibrating string of mass and Tension = T is given by:→ l = length

Let for first case, f₁ = 100HZ ; l₁ = l ; T₁ = Initial Tension

For second case, f₂ = 150HZ ; l₂ = 2l ; T₂ = Final Tension

So,

Hence, the ratio of tensions is

1 : 9

8.Two similar sonometer wires of the same material produces 2 beats per second. The length of one is 50cm and that of the other is 50.1 cm. Calculate the frequencies of two wives?

Ans.The frequency (f) of a son meter wire of length = l, mass = m and lesion = T is given by

50100 = k

9.Why are all stringed instruments provided with hollow boxes?

Ans .The stringed instruments are provided with a hollow box called sound box. When the strings are set into vibration, forced vibrations are produced in the sound box. Since sound box has a large area, it sets a large volume of air into vibration. This produces a loud sound of the same frequency of that of the string.

10.Two waves have equations:-

If in the resultant wave, the amplitude remains equal to the amplitude of the super posing waves. Calculate the phase difference between X₁ and X₂?

Ans.Given, the first wave:→ X₁ = a Sin

The second : wave

Where X₁ = wave function first wave

X₂ = wave function in second wave

a = amplitude

w = Angular frequency

t = time

₁ = phase difference of first wave

₂= phase difference of second wave.

Let The resultant amplitude = ‘a’ and phase difference

a₁ = amplitude of first wave

a₂ = amplitude of second wave

= phase difference between two waves.

Now, in our case,

So, the phase difference between X₁ and X₂ is 120⁰.

11.A Tuning fork of frequency 300HZ resonates with an air column closed at one end at 27⁰C. How many beats will be heard in the vibration of the fork and the air column at 0⁰C?

Ans .According to the question,

Frequency of air column at 27⁰C is 300 HZ

Let l = length of air column and speed of sound = V₂₇

For a pipe; closed at one end, the frequency of n^th harmonic is:→

n = 1, 3, 5, 7 --------

l = length of air column

v = velocity

 So, Let at 0⁰C, the speed of sound = v_o then,

Vo = (0.954) V₂₇

∴ frequency of air column at 0⁰C is :→

The frequency of tuning fork remains 300HZ Number of beats = f₁ – f₂

= 300-286  = 14 Per second

12.A vehicle with horn of frequency ‘n’ is moving with a velocity of 30m|s in a direction perpendicular to the straight line joining the observer and the vehicle. If the observer perceives the sound to have a frequency of n+n₁. Calculate n₁? 

Ans.By Doppler effect, the apparent change in frequency of wave due to the relative motion between the source of waves and observer.

If v₂ = velocity of the listener

v_s = velocity of the source

v = velocity of sound

= frequency of sound reaching from the source to the listener.

= Apparent frequency (i.e. Changed frequency due to movement of source and listener)

But in our case, the source and observer move at right angles to each other. The Doppler Effect is not observed when the source of the sound and the observer are moving at right argyles to each other.

So, if n = original frequency of sound the observer will perceive the sound with a frequency of n (because of no Doppler effect). Hence the n₁ = charge frequency = 0.

13.We cannot hear echo in a room. Explain?

Ans.We know that, the basic condition for an echo to be heard is that the obstacle should be rigid and of large size. Also the obstacle should be at least at a distance of 17m from the source. Since the length of the room is generally less than 17m, the conditions for the production of Echo are not satisfied. Hence no echo is heard in a room.

14.Why do the stages of large auditoriums gave curved backs? 

Ans.The stages of large auditorium have curved backs because when speaker stands at or near the focus of curved surface his voice is rendered parallel after reflection from the concave or parabolic seer face. Hence the voice can be heard at larger distances.

15.Show that Doppler effect in sound is asymmetric?

Ans.The apparent frequency of sound when source is approaching the stationary listener (with velocity v¹) is not the same as the apparent frequency of sound when the listener is approaching the stationary source with a velocity v¹. This shows that Doppler Effect in sound is asymmetric.

Apparent frequency = f¹ =   

This is when source approaches station any listener

V = Velocity of sound in air

u_s = Velocity of sound source = V₁

f¹ = Apparent frequency

f = Original frequency of sound  

Apparent frequency 

→ This is when listener approaching stationary source

Since  ( Hence Doppler effect is asymmetric)

16.An organ pipe P₁ closed at one end vibrating in its first overtone and another pipe P₂ open at both the ends vibrating in its third overtone are in resonance with a given tuning fork. Find the ratio of length of P₁ and P₂?  

Ans.Length of pipe closed at one end for first overtone, l₁ =  

Length of pipe closed at both ends for third overtone; l₂ =

π= wavelength

17.A simple Romanic  wave has the equation

Y = 0.30 Sin (314 t – 1.57x)

t = sec, x = meters, y = cm. Find the frequency and wavelength of this wave.

Another wave has the equation-

Y₁ = 0.1 Sin (314 t – 1.57x + 1.57)

Deduce the phase difference and ratio of intensities of the above two waves? 

Ans.If y is in meters, then equation becomes:→

The standard equation of plane progressive wave is

Comparing equation 1) & 2)

On inspection of the equations of the given two waves,

Phase difference,

18.The component waves producing a stationary wave have amplitude, Frequency and velocity of 8 cm, 30HZ and 180 cm/s respected. Write the equation of the stationary wave?

Ans. Since the ware equation of a travelling wave:-

a = Amplitude

t = time

T = Time Period

x = Path difference

π = wavelength  

Let y₁ = a Sin

Y₂ = a Sin

(It is travelling in opposite direction)

By principle of superposition, wave equation for the resultant wave = y = y₁+y₂

Here a = 8cm; f = 30Hz, V = 180 cm/s

19.A wine of density a g/cm³ is stretched between two clamps 1 m apart while subjected to an extension of 0.05 cm. What is the lowest frequency of transverse vibration in the wire? Let young’s Modulus = y = 9?

Ans.The lowest frequency of transverse vibrations is given by:→

Area = A

Density = P

Here m = mass Per unit length = area  Density

because Density =

m =

 F=35.3  vib/sec

20.Give two cases in which there is no Doppler effect in sound?

Ans.The following are the two cases in which there is no Doppler effect in sound (i.e no change in frequency):-

1) When the source of sound as well as the listener moves in the same direction with the same speed.

2) When one of source | listener is at the centre of the circle and the other is moving on the circle with uniform speed.

21. 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

Mass per unit length,

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

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

22.  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 = 343 m?

Ans. 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

Mass per unit length,

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

∴T =

=   0.175 = 20588.575  N

23.  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.

Ans. (a) Frequency of the ultrasonic sound, v = 1000 kHz = 106 Hz

Speed of sound in air, = 340 m/s

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

(b) Frequency of the ultrasonic sound, v = 1000 kHz = 106 Hz

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

The wavelength of the transmitted sound is given as:

24. (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. (i)

(a) Yes, except at the nodes

(b) Yes, except at the nodes

(c) No

(ii) 0.042 m

Explanation:

(i)

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

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

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

(ii) The given equation is:

For x = 0.375 m and t = 0

Amplitude = Displacement =

25.  A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is  kg and its linear mass density is kg m-1. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?

Ans.(a) Mass of the wire, m = kg

Linear mass density,

Frequency of vibration, ν = 45 Hz

∴Length of the wire,

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

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 = v 45  1.75 = 78.75 m/s

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

T =

=

26. 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?

Ans. Frequency of string A, = 324 Hz

Frequency of string B =

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:

Hence, the beat frequency cannot be 330 Hz

3 Marks Questions

1.Explain briefly the analytical method of formation of beats?

Ans.Let us consider two wares trains of equal amplitude ‘a’ and with different frequencies in same direction.

Let displacements are y₁ and y₂ in time ‘t’.

w₁ = Angular vel. of first ware

w₁ = 

Acc. to superposition principle, the resultant displacement ‘y’ at the same time ‘t’ is:-

y = y1 + y2

 y = a

Using Sin C + Sin D = 2 Cos

We get,

Where, A = 2a Cos

Now, amplitude A is maximum when,

i.e resultant intensity of sound will be maximum at times,

Time interval between 2 successive Maximaxs = 

Similarly, A will be minimum,

i.e. resultant intensity of sound will be minimum at times

Hence time interval between 2 successive minim as are

=

Combining 1) & 2) frequency of beats =

Difference in frequencies of two sources of sound.

2.Show that the frequency of nth harmonic mode in a vibrating string which is closed at both the end is ‘n’ times the frequency of the first harmonic mode?

Ans .When a sting under tension is set into vibration, transverse harmonic wares propagate along its length when length of sting is fixed, reflected waves will also exist. The incident and reflected waves will superimpose on each short to produce transverse stationary waves in sting. Let a harmonic wave be set up in a sting of length =  which is fixed at 2 ends: → x = 0 and x = L

Let the incident wave travels from left to right direction, the wave equation is:→

The wave equation of reflected wave, will have the same amplitude, wavelength, velocity, time but the only difference between the incident and reflected waves will be in their direction of propagation So, :→

Reflection, the wave will suffer a phase change of π. So,

According to the principle of superposition, the wave equation of resultant stationary wave will be:-

1) Let n = 1 (i.e first harmonic mode or fundamental frequency

If n = 2 (second harmonic Mode or first overtone)

i.e frequency of second harmonic Mode is twice the frequency of first harmonic Mode Similarly,   and frequency of n^th harmonic Mode is n times the frequency of first harmonic Mode.

3.Differentiate between the types of vibration in closed and open organ pipes?

Ans.1) In closed pipe, the wavelength of nth mode =  where n = odd integer whereas in open pipe, and n = all integer

2) The fundamental frequency of open pipe is twice that of closed pipe of same length.

3) A closed pipe of length produces the same fundamental frequency as an open pipe of length L.

4) For an open pipe, harmonics are present for all integers and for a closed pipe, harmonics are present for only odd integers hence, open pipe gives richer note.

Put the value of n in equation for frequency

Now, young’s Modulus = y =

Or Stress = y Strain

Stress =

Put the value of Stress in equation 2)

Since fundamental frequency of a stretched spring

f₁=249.5HZ

And f₂ – f₁ = 1

f₂ = 1 + f₁

= 1 + 249.5

f₂ = 250.5 HZ

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?

Ans.Ultrasonic beep frequency emitted by the bat,v = 40 kHz

Velocity of the bat, = 0.03 v

Where, v = velocity of sound in air

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

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

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

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

Ans. Height of the tower, s = 300 m

Initial velocity of the stone, u = 0

Acceleration, a = g = 9.8

Speed of sound in air = 340 m/s

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

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

Therefore, the time after which the splash is heard,

= 7.82 + 0.88 = 8.7 s

6.   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:

(a) (b) (c)

Ans. No;

(a) Does not represent a wave

(b) Represents a wave

(c) Does not represent a wave

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

Explanation:

(a) For x = 0 and t = 0, the function  becomes 0.

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

(b) For x = 0 and t = 0, the function

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

(c) For x = 0 and t = 0, the function

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

7.  For the travelling harmonic wave

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

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

(a) 4 m, (b) 0.5 m, (c),  (d)

Ans.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, = 20 π rad/s

Phase difference is given by the relation:

(a) For x = 4 m = 400 cm

= 0.016 π × 400 = 6.4 π rad

(b) For 0.5 m = 50 cm

= 0.016 π × 50 = 0.8 π rad

(c) For

(d) For

8.  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.

Ans.Frequency of the turning fork, v = 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:

Where,

Length of the pipe,

The speed of sound is given by the relation:

 = 346.8 m/s

9.  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?

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

Fundamental frequency of vibration, v = 2.53 kHz = Hz

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

The distance between two successive nodes is.

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

v = v

=

=  m/s

= 5.06 km/s

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

10.  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, (b) recedes from the platform with a speed of 10 m s-1? (ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 m s-1.

Ans.(i) (a)Frequency of the whistle, v = 400 Hz

Speed of the train, = 10 m/s

Speed of sound, v = 340 m/s

The apparent frequency of the whistle as the train approaches the platform is given by the relation:

(b) The apparent frequency of the whistle as the train recedes from 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.

11.  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. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 m.

Ans.Operating frequency of the SONAR system, v = 40 kHz

Speed of the enemy submarine, = 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 () received and reflected by the submarine is given by the relation:

The frequency () received by the enemy submarine is given by the relation:

Where,

5 Marks Questions

1.  Use the formulato explain why the speed of sound in air

(a) is independent of pressure,

(b) increases with temperature,

(c) increases with humidity.

Ans. (a) Take the relation:

 ………..(i)

Where,

Density,

M= molecular weight of the gas

V= Volume of the gas

Hence, equation(i) reduces to:

……(ii)

Now from the ideal gas equation for n = 1:

PV = RT

For constant T, PV = Constant

Since both M and Y 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:

……(i)

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

PV = RT

P = … (ii)

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

 . ……..(iv)

Where,

 M = pv is a constant

Y and R are also constants

We conclude from equation (iv) that .

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

(c) Letbe the speeds of sound in moist air and dry air respectively.

Letbe the densities of moist air and dry air respectively.

Take the relation:

Hence, the speed of sound in moist air is:

       ……..(i)

And the speed of sound in dry air is:

  …………(ii)

On dividing equations (i) and (ii), we get:

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

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

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

……..(i)

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

(b) 3 cm; 5.73 Hz

(c)

(d) 3.49 m

Explanation:

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

y (x, t) = a sin () … (i)

The given equation is:

….(ii)

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

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

 = 36 rad/s and k = 0.018

We know that:

and

Also,

v = vh

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:

(c) On comparing equations (i) and (ii), we find that the initial phase angle,

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

Wavelength is given by the relation:

3. For the wave described in Exercise 15.8, 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:

.(i)

For x = 0, the equation reduces to:

Also,

Now, plotting y vs. t graphs using the different values of t, as listed in the given table.

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.

4. 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  kg.

Answer the following:

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

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

(c) 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 °t

This equation is similar to the given equation:

Hence, the given function represents a stationary wave.

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

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

The superposition of these two waves yields:

……….(i)

The transverse displacement of the string is given as:

…..(ii)

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

∴Wavelength, h = 3 m

It is given that:

Frequency, v = 60 Hz

Wave speed, v = vh

= 60  3 = 180 m/s

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

 ……….(i)

Where,

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

Mass of the string, m =

Length of the string, l = 1.5 m

Mass per unit length of the string,

Tension in the string = T

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

T =

=

= 648 N

5. A travelling harmonic wave on a string is described by

(a) 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?

(b) 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.

Ans.(a) The given harmonic wave is:

For x = 1 cm and t = 1s,

Where,

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

A =1 cm and t=l s:

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

Where,

And

Speed,

Where,

 =12 rad/s

K =

∴

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

(b) Propagation constant is related to wavelength as:

=1253 cm = 12.56 m

Therefore, all the points at distances n , 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.