 # JEE Main Sound Waves Previous Year Questions with Solutions

Sound is a form of energy that requires a medium for propagation, therefore it is called mechanical waves. Sound waves are longitudinal waves that travel through solids, liquids or gases. Longitudinal waves move through a medium by alternating compression and rarefaction.

This compression and rarefaction create a minute pressure difference that we perceive as sound.

Factors affecting the velocity of sound in the air and gases

• Density: Velocity of sound waves in gas is inversely proportional to the square root of the density of the gas
• Moisture or humidity: The presence of moisture decreases the density of air because the density of water vapour is less than that of air. Hence, sound travels faster in moist air than in dry air.
• Temperature: Sound travels faster on a hot day than on a cold day.
• Pressure: The change in pressure has no effect on the velocity of sound.
• Wind: If u is the velocity of the wind that makes an angle θ with the direction of the velocity of sound v. Then the net velocity of sound is given by v + ucosθ.

## JEE Main Previous Year Solved Questions on Sound Waves

Q1: A musician using an open flute of length 50 cm produces second harmonic sound waves. A person runs towards the musician from another end of a hall at a speed of 10 km/h. If the wave speed is 330 m/s, the frequency heard by the running person shall be close to

(a) 666 Hz

(b) 753 Hz

(c) 500 Hz

(d) 333 Hz

Solution

Frequency of the sound produced by the open flute

f=2(v/2l) = (2 x 330)/(2 x 0.5) = 660 Hz

Velocity of observer, v0 = 10 x (5/18) = (25/9) m/s

As the source is moving towards the observer, according to the Doppler effect.

Frequency detected by observer

f’ = {(v + v0)/v}f = {((25/9) + 330)/330}600

= {2995/(9 x 330)} x 660

f’ = 665.55 ≈ 666 Hz

Q2: A source of sound S is moving with a velocity of 50 m/s towards a stationary observer. The observer measures the frequency of the source when it is moving away from the observer after crossing him? (Take the velocity of sound in air as 350 m/s)

(a) 750 Hz

(b) 857 Hz

(c) 1143 Hz

(d) 807 Hz

Solution

When source is moving towards a stationary observer,

fapp = fsource {(V – 0)/(V – 50)}

1000 = fsource (350/300)

When source is moving away from observer

f’ = fsource{350/(350 + 50)}

f’ = {(1000 x 300)/350} x (350/400)

f’ = 750 Hz

Q3: Two sources of sound S1 and S2 produce sound waves of the same frequency 660 Hz. A listener is moving from source S1 towards S2 with constant speed u m/s and he hears 10 beats. The velocity of sound is 330 m/s. Then u is equals

(a)5.5 m/s

(b) 15.0 m/s

(c)2.5 m/s

(d) 10.0 m/s

Solution

f1 = f[(v – v0)/v]

f2 = f[(v + v0)/v]

Frequency f2 – f1= f x (2v0/v)

10 = 660 x (2u/330)

u = 2.5 m/s

Q4: A resonance tube is old and has a jagged end. It is still used in the laboratory to determine the velocity of sound in air. A tuning fork of frequency 512 Hz produces first resonance when the tube is filled with water to a mark 11 cm below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency 256 Hz which produces first resonance when water reaches a mark 27 cm below the reference mark. The velocity of sound in air, obtained in the experiment, is close to

(a) 322 ms-1

(b) 341 ms-1

(c)335 ms-1

(d) 328 ms-1

Solution

(λ/4) = 0.11 + e

{V/(512)4} = 0.11 + e ——-(1)

{V/(256)4} = 0.27 + e ——-(2)

After solving (1) and (2) we get

V = 328 ms-1

Q5: A small speaker delivers 2W of audio output. At what distance from the speaker will one detect 120 dB intensity sound? (Given reference intensity of sound is 10-12 W/m2)

(a) 40 cm

(b) 20 cm

(c) 10 cm

(d) 30 cm

Solution

Using β = 10 log10(I/I0)

Or 120 = 10 log 10(I/ 10-12) ——(1)

Also I = P/4πr2 = 2/4πr2 ——(2)

On solving the above equations, we get

r = 40 cm

Q6: Two cars A and B are moving away from each other in opposite directions. Both the cars are moving with a speed of 20 m/s with respect to the ground. If an observer in car A detects a frequency 2000 Hz of the sound coming from car B, what is the natural frequency of the sound source in car B? (speed of sound in air = 340 m/s)

(a)2250 Hz

(b)200 Hz

(c)2300 Hz

(d)2150 Hz

Solution

f’ = f{(v – v0)/(v + vs)}

2000 = f{(340 – 20)/(340 + 20)}

f = 2250 Hz

Q7: A person standing on an open ground hears the sound of a jet aeroplane, coming from the north at an angle 600 with ground level. But he finds the aeroplane right vertically above his position. If v is the speed of sound, the speed of the plane is

(a)(√3/2) v

(b)2v/√3

(c) V

(d) v/2

Solution Distance, PQ = vp x t (Distance = speed x time)

Distance , QR = V.t

Cos 600 = PQ/QR

½ = (vp x t)/( V.t)

Vp = V/t

Q8: Three sound waves of equal amplitudes have frequencies(f-1, f, f+1). They superpose to give beats. The number of beats produced per second is

(a) 2

(b) 1

(c)4

(d) 3

Solution

Beat produced between(f-1) and f is 1.

Beat produced between f and f+1 is 1.

Beat produced between(f-1) and (f+1) is 2

∴ No of beats produced per second will be 2

Q9: A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of the air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.

(a) 6

(b) 4

(c) 12

(d) 8

Solution

Fundamental frequency of the closed organ pipe is

f = v/4L

f = 340/(4 x 0.85) = 100 Hz

The natural frequencies of the closed organ pipe is

fn= (2n – 1)f = f, 3f, 5f, 7f, 9f, 11f, 13f …

So the possible frequencies below 1250 Hz are

fn= 100 Hz, 300 Hz, 500 Hz , 700 Hz, 900 Hz, 1100 Hz

Number of frequencies = 6

Q10: An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency10 GHz. What is the frequency of the microwave measured by the observer?

(speed of light = 3 x 108 m/s)

(a) 17.3 GHz

(b) 15.3 GHz

(c) 10.1 GHz

(d) 12.1 GHz

Solution

Doppler effect in light (speed of observer is not very small compared to speed of light)

f’ = f .[(1 + v/c)/(1 – v/c)]½

Here, frequency (v/c) = ½

So, f’ = f .[(3/2)/(1/2)]½

f’ = 10 x √3 = 17.3 GHz