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Q.

Two tuning forks with natural frequencies of 400 Hz each move 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 4 Hz. Find the speed of the tuning fork. (velocity of sound $= 340 m / s$ )

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Detailed Solution

$f' = f [\frac{v}{v - v_{s}}]$ fork $' T_{1}'$ is moving towards the observer
$f " = f [\frac{v}{v + v_{s}}]$ fork $' T_{2}'$ is moving away from the observer
$\Rightarrow$ Beat frequency $f' - f " = 4$
$4 = \frac{v f}{v - v_{s}} - \frac{v f}{v + v_{s}} \Rightarrow 4 = \frac{2 v_{s} \times v f}{(v^{2} - v_{s}^{2})} (∵ v_{s} < < v)$

$4 = \frac{2 v_{s} \times v f}{v^{2}} \Rightarrow v_{s} = \frac{4 v}{2 f} v_{s} = \frac{2 \times 340}{20} = 1.7 m / s$

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