Question

A stretched sonometer wire is in unison with a tuning fork. When the length is increased by 4%, the number of beats heard per second is 6. Find the frequency of the fork.

Answer 1

Let length of wire is $L$

Length is increased by $4\%$

then number of beat heard per second $=6$

$n_1-n_2=6/s=6Hz$..................$\left(1\right)$

$n_1$ and $n_2$ are the frequency of initial and final.

$\begin{array}{c}n=\frac{1}{2l}\sqrt{\frac{T}{m}}\\ Now,n_1{l}_1=n_2{l}_2\\ n_1{l}_1=n_2\times 1.042\\ or,n_1=n_2\times 1.04\\ or,n_1=\frac{26}{25}{n}_2\\ or,25n_1=26n_2\\ or,n_2=\frac{25}{26}{n}_2.........\left(2\right)\\ from\left(1\right)and\left(2\right)weget\\ n_1-\frac{25}{26}{n}_1=6\\ or,n_1=156Hz\\ \therefore Frequencyofturningforkis156Hz\end{array}$