Question

The fundamental frequency of a sonometer wire is n. If its radius is doubled and its tension becomes half, the material of the wire remains same, the new fundamental frequency will be:

• A. $ n $
• B. $ \frac{n}{\sqrt{2}} $
• C. $ \frac{n}{2} $
• D. $ \frac{n}{2\sqrt{2}} $

Answer 1

Answer: (D)

Frequency of sonometer wire is given by
$ n=\frac{1}{2l}\sqrt{\frac{T}{m}} $
where m is mass of string per unit length, and is tension in the string.
Also, $ m=\pi {{r}^{2}}d $
r being radius of string and d is the density of material of string.
So, $ n=\frac{1}{2l}\sqrt{\frac{T}{\pi {{r}^{2}}d}} $
or $ n\propto \frac{\sqrt{T}}{r} $
or $ \frac{{{n}_{1}}}{{{n}_{2}}}=\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\times \left( \frac{{{r}_{2}}}{{{r}_{1}}} \right) $
Given, $ {{r}_{2}}=2{{r}_{1}},\,{{T}_{2}}=\frac{{{T}_{1}}}{2}, $
$ {{n}_{1}}=n $
Hence, $ \frac{n}{{{n}_{2}}}=\sqrt{2}\times 2 $
or $ {{n}_{2}}=\frac{n}{2\sqrt{2}} $