Physics - Wave on String Question with Solution | TestHub

The fundamental frequency of a stretched wire is given by $n=\frac{1}{2L}\sqrt{\frac{T}{m}}$, where $L$ is the length, $T$ is the tension, and $m$ is the linear mass density.

The linear mass density $m$ can be expressed in terms of diameter $d$ and density $\rho$ as:

$m=\frac{\text{mass}}{\text{length}}=\frac{\text{volume}\times \rho }{L}=\frac{(\pi r^{2}L)\rho }{L}=\pi {\left(\frac{d}{2}\right)}^2\rho =\frac{\pi d^2\rho }{4}$

Substituting $m$ in the formula for $n$:

$n=\frac{1}{2L}\sqrt{\frac{4T}{\pi d^2\rho }}=\frac{1}{Ld}\sqrt{\frac{T}{\pi \rho }}$

For the first wire:

$n_A=\frac{1}{Ld}\sqrt{\frac{T}{\pi \rho }}$

For the second wire, given $L^{\prime }=2L$, $d^{\prime }=3d$, ${\rho }^{\prime }=2\rho$, and $T^{\prime }=2T$:

$n_B=\frac{1}{(2L)(3d)}\sqrt{\frac{2T}{\pi (2\rho )}}$

$n_B=\frac{1}{6Ld}\sqrt{\frac{T}{\pi \rho }}$

Taking the ratio of $n_B$ to $n_A$:

$\frac{n_B}{n_A}=\frac{\frac{1}{6Ld}\sqrt{\frac{T}{\pi \rho }}}{\frac{1}{Ld}\sqrt{\frac{T}{\pi \rho }}}=\frac{1}{6}$

Thus, $n_B:n_A=1:6$.