JEE Main 2022 · 27 June · morning · Single correct · +4 / −1

An observer moves towards a stationary source of sound with a velocity equal to one-fifth of the velocity of sound. The percentage change in the frequency will be :

  1. (a) 20%
  2. (b) 10%
  3. (c) 5%
  4. (d) 0%
Answer and solution

Answer (a) 20%

Solution

Physics sir · 3 steps

The observer runs into the waves faster, so add his speed to the speed of sound in the numerator.

  1. Moving observer. The source is at rest, so the wavelength in air does not change. The observer moving towards the source meets wave crests at a relative speed of v + vₒ instead of v.

    f' = f (v + vₒ)/v

  2. Put in numbers. With vₒ = v/5, the factor is (v + v/5)/v = 6/5 = 1.2, so f' = 1.2f.
  3. So. The percentage change is (f' - f)/f × 100 = 0.2 × 100 = 20%, an increase. The answer is (a).

The idea to keep. For a moving observer and a still source, f'/f = 1 + vₒ/v exactly. So the percentage change in frequency equals the observer's speed as a percentage of the speed of sound.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Sound Waves

Asked about 40 times in JEE Main since 2020. This question is from 2022.