JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Waves & Sound

A source, approaching with speed

u

towards the open end of a stationary pipe of length

L

, is emitting a sound of frequency

f_{s}

. The farther end of the pipe is closed. The speed of sound in air is

v

and

f_{0}

is the fundamental frequency of the pipe. For which of the following combination(s) of

u

and

f_{s}

, will the sound reaching the pipe lead to a resonance ?

Answer: (A), (D)

Solution

The frequency heard at the pipe due to Doppler effect is:

f = f_{s} (\frac{v}{v - u})

For a closed pipe, resonance occurs at odd harmonics:

f_{0}, 3 f_{0}, 5 f_{0}, \dots

(A)

f = f_{0} (\frac{v}{v - 0.8 v}) = f_{0} (\frac{v}{0.2 v}) = 5 f_{0}

✓ (odd harmonic)

(B)

f = 2 f_{0} (\frac{v}{0.2 v}) = 10 f_{0}

✗ (even harmonic)

(C)

f = 0.5 f_{0} (\frac{v}{0.2 v}) = 2.5 f_{0}

✗ (not an integer harmonic)

(D)

f = 1.5 f_{0} (\frac{v}{v - 0.5 v}) = 1.5 f_{0} (\frac{v}{0.5 v}) = 3 f_{0}

✓ (odd harmonic)

Correct answers: (A) and (D)

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · One or more correct options