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# A sitar wire is replaced by another wire

Question:

A sitar wire is replaced by another wire of same length and material but of three times earlier radius. If the tension in the wire remains the same, by what factor will the frequency change?

Solution:

The frequency of the stretched wire is given as:

$v=\frac{n}{2 L} \sqrt{\frac{T}{m}}$

Given that,

No.of harmonic n, length L, and tension T are the same in both the cases.

Therefore,

$v \propto \frac{1}{\sqrt{m}} \Rightarrow \frac{v_{1}}{v_{2}}=\frac{\sqrt{m_{2}}}{\sqrt{m_{1}}}$

Substituting the values of mass per unit length, we get

$\frac{m_{2}}{m_{1}}=\frac{\pi r_{2}^{2} \rho}{\pi r_{1}^{2} \rho}=\frac{(3 r)^{2}}{r^{2}}=\frac{9}{1}$

Solving the above the equation, we get v2 = 1/3 v1

Therefore, the frequency of sitar is reduced by 1/2 of its actual value.