Two uniform wires of same material are vibrating under the same tension. If the first overtone of first wire…
Two uniform wires of same material are vibrating under the same tension. If the first overtone of first wire is equal to the $2^{\text {nd }}$ overtone of $2^{\text {nd }}$ wire and radius of the first wire is twice the radius of the $2^{\text {nd }}$ wire then the ratio of length of first wire to $2^{\text {nd }}$ wire is
$1: 3$
$3: 1$
$1: 9$
$9: 1$
Solution
Fundamental frequency of the first wire is
$\mathrm{n}=\frac{1}{2 l_1} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}=\frac{1}{2 l_1} \sqrt{\frac{\mathrm{T}}{\pi \mathrm{r}_1^2 \rho}}=\frac{1}{2 l_1 \mathrm{r}_1} \sqrt{\frac{\mathrm{T}}{\pi \rho}}$
The first overtone $\mathrm{n}_1=2 \mathrm{n}=\frac{1}{l_1 \mathrm{r}_1} \sqrt{\frac{\mathrm{T}}{\pi \rho}}$
Similarly, the second overtone of the second wire will be,
$\mathrm{n}_2=\frac{3}{2 l_2 \mathrm{r}_2} \sqrt{\frac{\mathrm{T}}{\pi \rho}}$
Given that $\mathrm{n}_1=\mathrm{n}_2$
$\begin{aligned}
& \therefore \quad \frac{1}{l_1 \mathrm{r}_1} \sqrt{\frac{\mathrm{T}}{\pi \rho}}=\frac{3}{2 l_2 \mathrm{r}_2} \sqrt{\frac{\mathrm{T}}{\pi \rho}} \\
& \therefore \quad 3 l_1 \mathrm{r}_1=2 l_2 \mathrm{r}_2 \\
& \\
& \quad \frac{l_1}{l_2}=\frac{2 \mathrm{r}_2}{3 \mathrm{r}_1}
\end{aligned}$
$=\frac{2 r_2}{3\left(2 r_2\right)} \quad \ldots .\left(\because r_1=2 r_2\right)$
$=\frac{1}{3}$