Discussion Forum

A wire having tension 225 N  produces six beats  per second when it is tuned with a fork .When tension changes to 256 n , it is tuned with the same fork, the number of beats remains unchanged. The frequency of the fork will be 

Answer:

Explanation:

The fundamental frequency of wire vibrating  under tension T is given by

$f_{1}=\frac{1}{2}\sqrt{\frac{T}{\mu}}$

Here, $\mu$= mass of string per unit length

Let the frequency of tunning fork be x which beat 6 beats per second

Let (x-f)=$\pm$ 6

$\therefore$     $f_{1}=\frac{1}{2L}\sqrt{\frac{225}{\mu}}$ and $f_{2}=\frac{1}{2L}\sqrt{\frac{256}{\mu}}$

$\therefore$   $\frac{f_{1}}{f_{2}}=\frac{15}{16}$

$\therefore$    $f_{2}=\frac{16}{15}\times f_{1}$

$\Rightarrow$    $f_{2}=\frac{16}{15} (6+x)$

Taking two cases of f₁ and equating both , we have

$(x+6)=\frac{16}{15} (x-6)$

 $\therefore$      15x+90=16x-96

 $\therefore$     x=186 Hz