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

9

Explanation

Let `f_(1)` and `f_(2)` be the frequency of tuning forks `P` and `Q`, then `I f_(1)-f_(2) I=7//2`
Apparent frequency for `O` corresponding to signal directly coming from `Q=f_(2)((v)/(v+v_(q)))`
Apparent frequency of the echo `= f_(2)((v)/(v-v_(q))):. Delta f_(2)-f_(2)[(2v_(q)v)/(v^(2)-v_(q)^(2))]`
Since, `Delta f_(2) = 5` (given) `:. f_(2) =163.5 Hz` Now, `f_(1) = 163.5 pm 3.5 = 167` or `160 Hz`, when `P` is filled, its frequency will increase, since it si given that filled `P` gives greater number of beats with `Q`. It implies that `f_(1)` must be `167 Hz`.