Physics - Wave on String Question with Solution | TestHub

The fundamental frequency of a sonometer wire is given by $n=\frac{1}{2L}\sqrt{\frac{T}{m}}$

Since linear mass density $m=\pi r^2\rho =\frac{\pi D^2\rho }{4}$, we have $n=\frac{1}{LD}\sqrt{\frac{T}{\pi \rho }}$

This implies $n\propto \frac{\sqrt{T}}{LD}$

Given $T^{\prime }=3T$, $L^{\prime }=3L$, and $D^{\prime }=2D$

The new frequency is $n^{\prime }=\frac{1}{L^{\prime }{D}^{\prime }}\sqrt{\frac{T^{\prime }}{\pi \rho }}$

Substituting the given values:

$n^{\prime }=\frac{1}{(3L)(2D)}\sqrt{\frac{3T}{\pi \rho }}$

$n^{\prime }=\frac{\sqrt{3}}{6}\left(\frac{1}{LD}\sqrt{\frac{T}{\pi \rho }}\right)$

$n^{\prime }=\frac{\sqrt{3}}{6}n=\frac{n}{2\sqrt{3}}$

Answer: $\frac{\mathrm{n}}{2\sqrt{3}}$