Q 11-14-102JEE MainJEE Main 2020 (7 Jan, Shift 2)Medium
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 $f_0 = 1400$ Hz and the velocity of sound in air is $350$ m/s. The speed of each tuning fork is close to:
Answer: (C) $\dfrac14$ m/s
For $u \ll v$:
$$f_1 = \frac{f_0v}{v - u} \approx f_0\left(1 + \frac uv\right), \qquad f_2 = \frac{f_0v}{v + u} \approx f_0\left(1 - \frac uv\right)$$
$$f_1 - f_2 \approx \frac{2f_0u}{v} = 2 \Rightarrow u = \frac{v}{f_0} = \frac{350}{1400} = \frac14\ \text{m/s}$$
Solution by Sreeraj P, M.Sc Physics