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Two vibrating strings of the same material but lengths $L$ and $2L$ have radii $2r$ and $r$ respectively. They are stretched under the same tension . Both the strings vibrate in their fundamental modes, the one of length $L$ with frequency $f_1$ and the other with frequency $f_2$. The ratio $\frac{f_1}{f_2}$ is given by
$2$
$4$
$8$
$1$
Solution
$\frac{\mathrm{f}_{1}}{\mathrm{f}_{2}}=\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}} \times \frac{\mathrm{L}_{2}}{\mathrm{L}_{1}}=\sqrt{\frac{\mathrm{T}_{1}}{\mu_{1}} \frac{\mu_{2}}{\mathrm{T}_{2}}} \times \frac{\mathrm{L}_{2}}{\mathrm{L}_{1}}$
Material is same
$\Rightarrow \mathrm{m}_{1}=\rho$ volume $=\rho \times \mathrm{A}_{1} \mathrm{L}_{1}=\rho \pi \mathrm{r}_{1}^{2} \mathrm{L}_{1}$
$\frac{\mathrm{f}_{1}}{\mathrm{f}_{2}}=\sqrt{\frac{\mathrm{m}_{2}}{\mathrm{L}_{2}} \frac{\mathrm{L}_{1}}{\mathrm{m}_{1}} \frac{\mathrm{L}_{2}^{2}}{\mathrm{L}_{1}^{2}}}=\sqrt{\frac{\mathrm{m}_{2} \mathrm{L}_{2}}{\mathrm{m}_{1} \mathrm{L}_{1}}}=\sqrt{\frac{\mathrm{r}_{2}^{2} \mathrm{L}_{2}}{\mathrm{r}_{1}^{2} \mathrm{L}_{1}} \times \frac{\mathrm{L}_{2}}{\mathrm{L}_{1}}}$
$=\frac{r_{2} L_{2}}{r_{1} L_{1}}=\frac{r(2 L)}{(2 r)(L)}=1$
