Physics - Wave on String Question with Solution | TestHub

The fundamental frequency of a stretched wire is given by $n=\frac{1}{2L}\sqrt{\frac{T}{m}}$.

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

Substituting $m$, we get $n=\frac{1}{2Lr}\sqrt{\frac{T}{\pi \rho }}$.

For the first wire, the $1^{\text{st}}$ overtone is the second harmonic, which is $2n_1=\frac{2}{2L_1{r}_1}\sqrt{\frac{T}{\pi \rho }}=\frac{1}{L_1{r}_1}\sqrt{\frac{T}{\pi \rho }}$.

For the second wire, the $2^{\text{nd}}$ overtone is the third harmonic, which is $3n_2=\frac{3}{2L_2{r}_2}\sqrt{\frac{T}{\pi \rho }}$.

Given that the $1^{\text{st}}$ overtone of the $1^{\text{st}}$ wire is equal to the $2^{\text{nd}}$ overtone of the $2^{\text{nd}}$ wire:

$\frac{1}{L_1{r}_1}\sqrt{\frac{T}{\pi \rho }}=\frac{3}{2L_2{r}_2}\sqrt{\frac{T}{\pi \rho }}$

$\frac{1}{L_1{r}_1}=\frac{3}{2L_2{r}_2}$

It is also given that the radius of the $1^{\text{st}}$ wire is twice the radius of the $2^{\text{nd}}$ wire, so $r_1=2r_2$. Substituting this into the equation:

$\frac{1}{L_1(2r_2)}=\frac{3}{2L_2{r}_2}$

$\frac{1}{2L_1}=\frac{3}{2L_2}$

$\frac{L_1}{L_2}=\frac{1}{3}$

The ratio of the length of the $1^{\text{st}}$ wire to that of the $2^{\text{nd}}$ wire is $1:3$.

Answer: $1:3$