Two wires are held tight between the same pair of supports. The...

The fundamental frequency of a stretched wire is given by:

$f = \dfrac{1}{2L} \sqrt{\dfrac{T}{\mu}}$

The linear mass density $\mu$ can be written in terms of the volume density $\rho$ and radius $r$ as $\mu = \pi r^2 \rho$. Substituting this in the frequency formula:

$f = \dfrac{1}{2Lr} \sqrt{\dfrac{T}{\pi \rho}}$

Since both wires are held between the same pair of supports, their lengths $L$ are equal. The ratio of their fundamental frequencies is:

$\dfrac{f_1}{f_2} = \dfrac{r_2}{r_1} \sqrt{\dfrac{T_1}{T_2} \times \dfrac{\rho_2}{\rho_1}}$

Given $\dfrac{T_1}{T_2} = \dfrac{2}{1}$, $\dfrac{r_1}{r_2} = \dfrac{3}{1}$, and $\dfrac{\rho_1}{\rho_2} = \dfrac{1}{2}$, we substitute these values:

$\dfrac{f_1}{f_2} = \dfrac{1}{3} \sqrt{\dfrac{2}{1} \times \dfrac{2}{1}}$

$\dfrac{f_1}{f_2} = \dfrac{1}{3} \sqrt{4} = \dfrac{2}{3}$