A tuning fork `P` of unknows frequency gives `7` beats in `2` seconds with another tuning fork `Q`. When `Q` runs towards a wall with a speed of `5 m//s` it gives `5` beats per second with its echo. On loading `P` with wax, it gives `5` beats per second with `Q` . What is the frequency of `P`? Assume speed of sound = `332 m//s` .

Answer

`160 Hz`

Explanation

Let frequency of `P` is `n` then possible frequency of `Q` is `n + (7)/(2)` or `n - (7)/(2)` but on loading `P` with wax frequency will decrease and here beat frequency increases so `Q` must have frequency `n_(Q) = n + (7)/(2)`
Now from dopplar effect for `Q` frequency echo received by `Q`.
`n'_(Q) = (V + V_(Q))/(V - V_(Q))n_(Q) , (n_(Q') - n_(Q))/(n_(Q)) = (2V_(Q))/(V - V_(Q)) , (n_(Q)^(') - n_(Q) = 5)`
`(5)/(n + 7//2) = (2 xx 5)/(332 - 5)`
`2n + 7 = 327 rArr n = 160 Hz`