Two tuning forks with natural frequency $340 H z$ each moves relative to a stationary observer. One fork moves away from the observer, while the other moves towards him at the same speed. The observer hears beats of frequency $3 H z$ . Find the speed of the tuning fork.

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Solution

$\Rightarrow f_{1} =$ frequency observed by observer due to fork (1)
$= (\frac{V_{s o u n d}}{V_{s o u n d} - V}) f_{0}$
$\Rightarrow f_{2} =$ frequency observed by observer due to fork (2)
$= (\frac{V_{s o u n d}}{V_{s o u n d} + V}) f_{0}$
$\Rightarrow$ Beat frequency $= | f_{1} - f_{2} | = 3 H z$
$= ∣ ∣ ∣ (\frac{1}{V_{s o u n d} - V}) - (\frac{1}{V_{s o u n d} + V}) ∣ ∣ ∣ (V_{s o u n d}) f_{0} = 3 H z$
$\Rightarrow V_{s o u n d}$ in air $= 330 m / s$
$\Rightarrow [(\frac{1}{(330 - V)} - \frac{1}{(330 + V)})] (330) (340) = 3$
$\Rightarrow [\frac{330 + V - 330 + V}{(330 - V) (330 + V)}] (330) (340) = 3$
$\Rightarrow (2 V) (330) (340) = 3 (330 - V) (330 + V)$
$\Rightarrow 74800 V = 108900 - V^{2}$
$\Rightarrow V^{2} + 74800 V - 108900 = 0$
$\Rightarrow V ≃ 1.45 m / s .$
Hence, the answer is $V ≃ 1.45 m / s .$

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