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Two tuning forks with natural frequencie...

Two tuning forks with natural frequencies of `340 Hz` each move relative to a stationary observer. One fork moves away form the observer, while the other moves towards him at the same speed. The observer hears beats of frequency `3 Hz`. Find the speed of the tuning fork.

Text Solution

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As observer is at rest.
`n.=n(V/(Vpm V_s)), n_1=n (V/(V-U))` and `n_2=n(V/(V+U))`
Now as beat frequency is 3 Hz so `n_1` and `n_2` are very close which is possible only when :
U < V . So using Binomial theorem -
`n_1=n(1-U/V)^(-1) =n(1+U/V)` and `n_2=n(1+U/V)^(-1)=n(1-U/V)`
So beat frequency `Deltan=n_1=n_2=n ((2U)/V)`
`rArr U=V/2 ((Deltan)/n)=330/2(3/340)` =1.45 m/s
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