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

A

Explanation

Beat frequency between `P` and `Q` is `(7)/(2)` or `3.5 H_(Z)`.
On loading `P` with wax, its frequency will decrease and beat frequency is increasing.
`Q_(rarr)5 m//s| 5m//s_(larr)Q'`
`:. f_(Q) gt f_(p)`
or `f_(Q) - f_(P) = 3.5`…(i)
Given, `f'_(Q) - f_(Q) = 5`
or `f_(Q) ((332 + 5)/(332 - 5)) - f_(Q) = 5`
Solving this equation, we get
`f_(Q) = 163.5 H_(Z)`
Now, from Eq. (i) we have
`f_(b) = 160 H_(z)`