JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Waves & Sound

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A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observers hears 2 beats/sec. The oscillation frequency of each tuning fork is

v_{0} = 1400 Hz

and the velocity of sound in air is

350 m / s

. The speed of each tuning fork is close to:

(A) $\frac{1}{4} m / s$ ✓ Correct
(B) $1 m / s$
(C) $\frac{1}{2} m / s$
(D) $\frac{1}{8} m / s$

Answer: (A)

Solution

The approaching fork is heard at

f_{0} \frac{v}{v - v _{s}}

and the receding one at

f_{0} \frac{v}{v + v _{s}}

, and their difference is the

2

Hz beat. Because the fork speed is tiny compared with the speed of sound, this difference is well approximated by

\frac{2 f _{0} v _{s}}{v} = 2

, giving

v_{s} = \frac{v}{f _{0}} = \frac{350}{1400} = \frac{1}{4}

m/s.

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Waves & Sound

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