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 $f_{0}$ = 1400Hz and the velocity of sound in air is 350 m/s. The speed of each tuning fork is close to:

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$\frac{1}{8} m / s$

$\frac{1}{4} m / s$

1m/s

$\frac{1}{2} m / s$

answer is A.

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Detailed Solution

$\begin{array}{l} B e a t = A p p a r e n t f r e q u e n c y d u e t o s o u r c e 1 - A p p a r e n t f r e q u e n c y d u e t o s o u r c e 2 \\ \Rightarrow B e a t = f_{0} (\frac{C}{C - V}) - f_{0} (\frac{C}{C + V}) = 2 [h e r e C = s p e e d o f s o u n d] \\ \Rightarrow f_{0} C (\frac{2 V}{C^{2} - V^{2}}) = 2 \\ S i n c e, V < < C, \\ w e c a n n e g l e c t V^{2} t e r m . \\ \Rightarrow V = \frac{C}{f_{0}} = \frac{350}{1400} \\ \Rightarrow V = \frac{1}{4} m / s \end{array}$

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