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.

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now