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)

$\frac{n}{2\sqrt{2}}$

Answer 2

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 ${{n}_{2}}=\frac{n}{2\sqrt{2}}$