Question

A piano wire A vibrates at a fundamental frequency of 600 Hz. A second identical wire B produces 6 beats per second with it when the tension in A is slightly increased. Find the ratio of the tension in A to the tension in B.

Answer 1

Here given, $n_B=\frac{1}{2I}\sqrt{\frac{T_B}{M}}$

As the tension increases frequency increases.

It is given that 6 beats are peoduced when tension in A is increased.

So, $n_A=606=\frac{1}{2I}\sqrt{\frac{T_A}{M}}$

$\Rightarrow \frac{n_A}{n_B}=\frac{600}{606}$

$=\frac{\left(\frac{1}{2I}\right)\sqrt{\frac{T_B}{M}}}{\left(\frac{1}{2I}\right)\sqrt{\left(\frac{T_A}{M}\right)}}$

$=\frac{\sqrt{T_B}}{\sqrt{T_A}}$

$\Rightarrow \frac{\sqrt{T_A}}{\sqrt{T_B}}=\frac{606}{600}=1.01$

$\Rightarrow \frac{T_A}{T_B}=1.02$