A certain fork is found to give ` 2 beats//s` when sounded in conjuction with a stretched string vibrating transversely under a tension of either `10.2 or 9.9 kg weight`. Calculate the frequency of fork.

Answer

268 Hz`

Explanation

`n//sqrt(T) = constant`. (i)
If `f` is frequency of fork , the frequency of
string `= f +- 4` (ii)
From Eqs.(i) and (ii), we get
`( f + 4)/(sqrt(10.2)) = ( f - 4)/(sqrt(9.9))`
Solving for `f` , we get ` f = 532 Hz`.