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

The Correct Option is 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}} $