A stationary observer receives sound from two identical tuning forks… | JEE Main 2020 (7 Jan, Shift 2)

Q51JEE Main 2020 · 7 JAN · SHIFT 2Difficult

Doppler Effect

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 observer hears 2 beats/sec. The oscillation frequency of each tuning fork is v0 = 1400 Hz and the velocity of sound in air is 350 m/s. The speed of each tuning fork is close to:

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Answer: (c)

Approaching fork: $f_1 = \nu_0\frac{v}{v - u}$ ; receding fork: $f_2 = \nu_0\frac{v}{v + u}$
For $u \ll v$ : beats $= f_1 - f_2 \approx \frac{2\nu_0 u}{v}$
$2 = \frac{2 \times 1400 \times u}{350} = 8u \Rightarrow u = \frac{1}{4}$ m/s
So the answer is (c) $\frac{1}{4}$ m/s.

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Doppler Effect question from JEE Main 2020 (7 Jan, Shift 2)

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