Physics - Wave on String Question with Solution | TestHub

The frequency of the $p$-th harmonic of a stretched string is given by $f=\frac{p}{2L}\sqrt{\frac{T}{\mu }}$

The linear mass density is $\mu =\pi r^2\rho$, where $r$ is the radius and $\rho$ is the density of the material.

Substituting $\mu$, we get $f=\frac{p}{2Lr}\sqrt{\frac{T}{\pi \rho }}$

For string X, the second overtone corresponds to the third harmonic, so $p_X=3$.

For string Y, the third overtone corresponds to the fourth harmonic, so $p_Y=4$.

Since both strings are made of the same material and are under the same tension, $\rho$ and $T$ are constant.

Given that the frequencies are equal ($f_X=f_Y$):

$\frac{3}{2L_X{r}_X}\sqrt{\frac{T}{\pi \rho }}=\frac{4}{2L_Y{r}_Y}\sqrt{\frac{T}{\pi \rho }}$

$\frac{3}{L_X{r}_X}=\frac{4}{L_Y{r}_Y}$

$\frac{L_X}{L_Y}=\frac{3}{4}\times \frac{r_Y}{r_X}$

It is given that the radius of string X is thrice that of string Y, so $r_X=3r_Y$, which means $\frac{r_Y}{r_X}=\frac{1}{3}$.

$\frac{L_X}{L_Y}=\frac{3}{4}\times \frac{1}{3}=\frac{1}{4}$

The ratio of the length of string X to that of Y is $1:4$.

Answer: $1:4$