Explanation:
A Let, for wire 1
Tension $=\mathrm{T}_{1}$, Length $=l_{1}$, Radius $=\mathrm{r}_{1}$, Density $=\rho_{1}$,
Fundamental frequency $=\mathrm{f}_{1}$
Similarly for wire 2-
$\mathrm{T}_{2}, l_{2}, \mathrm{r}_{2}, \rho_{2}, \mathrm{f}_{2}$
Given, $r_{1}=2 r_{2}, \rho_{1}=\rho_{2}=\rho$ (for same material of wire) Since, fundamental frequency-
$\mathrm{f}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}$
Where,
$\mu=\frac{\text { mass of wire }}{\text { length of wire }}=\frac{(\text { volume }) \times \text { density }}{\text { length of wire }}$
$\mu=\frac{\left(\pi \mathrm{r}^{2} l\right)(\rho)}{l}=\pi \mathrm{r}^{2} \rho$
Form equation (i)
$\mathrm{f}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\pi \mathrm{r}^{2} \rho}}=\frac{1}{2 l r} \sqrt{\frac{\mathrm{T}}{\pi \rho}}$
for the first overtone,
$\mathrm{f}_{1}=2\left[\frac{1}{2 l_{1} \mathrm{r}_{1}} \sqrt{\frac{\mathrm{T}_{1}}{\pi \rho_{1}}}\right]$
For second overtone, $f_{2}=3\left[\frac{1}{2 l_{2} r_{2}} \sqrt{\frac{T_{2}}{\pi \rho_{2}}}\right]$
According to question,
$\therefore 2\left[\frac{1}{2 l_{1}\left(2 \mathrm{r}_{2}\right)} \sqrt{\frac{\mathrm{T}}{\pi \rho}}\right]=3\left[\frac{1}{2 l_{2} \mathrm{r}_{2}} \sqrt{\frac{\mathrm{T}}{\pi \rho}}\right]\left(\because \rho_{1}=\rho_{2}=\rho\right)$
$\frac{1}{2 l_{1}}=\frac{3}{2 l_{2}}$
$\frac{l_{1}}{l_{2}}=\frac{1}{3}$
$l_{1}: l_{2}=1: 3$
