Two tuning folks with natural frequencies $340$ 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 $3$ Hz. Find the speed of the tuning fork. (Velocity of sound in air $= 330$ m/s).

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Solution

As observer is at rest-
$n^{'} = n (\frac{V}{V \mp V s}), n_{1} = n (\frac{V}{V - U})$ and $n_{2} = n (\frac{V}{V + U})$
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-
$n_{1} = n {(1 - \frac{U}{V})}^{- 1} = n (1 + \frac{U}{V})$ and $n_{2} = n {(1 + \frac{U}{V})}^{- 1} = n (1 - \frac{U}{V})$
So beat frequency $Δ n = n_{1} - n_{2} = n (\frac{2 U}{V})$
$\Rightarrow U = \frac{V}{2} (\frac{Δ n}{n}) = \frac{330}{2} (\frac{3}{340}) = 1.45$ m/s.

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