Physics - Wave on String Question with Solution | TestHub

The frequency of the $p$-th overtone of a stretched string is given by $f=\frac{(p+1)v}{2L}$, where $v$ is the wave speed and $L$ is the length of the wire.

For the first wire, the first overtone ($p=1$) frequency is:
$f_1=\frac{2v_1}{2L_1}=\frac{v_1}{L_1}$

For the second wire, the second overtone ($p=2$) frequency is:
$f_2=\frac{3v_2}{2L_2}$

Given that $f_1=f_2$, we have:
$\frac{v_1}{L_1}=\frac{3v_2}{2L_2}$

$\Rightarrow \frac{L_1}{L_2}=\frac{2v_1}{3v_2}$

The velocity of a transverse wave in a wire is $v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{T}{\pi r^2\rho }}$, where $T$ is tension, $r$ is radius, and $\rho$ is density.

Since $T$ and $\rho$ are the same for both wires, $v\propto \frac{1}{r}$.

$\Rightarrow \frac{v_1}{v_2}=\frac{r_2}{r_1}$

Given that the radius of the first wire is twice the radius of the second wire ($r_1=2r_2$), we get:
$\frac{v_1}{v_2}=\frac{r_2}{2r_2}=\frac{1}{2}$

Substituting this into the length ratio equation:
$\frac{L_1}{L_2}=\frac{2}{3}\times \frac{1}{2}=\frac{1}{3}$

Answer: $\frac{1}{3}$