Physics - Wave on String Question with Solution | TestHub

The frequency of the $n$-th harmonic of a stretched string is given by:

$f=\frac{n}{2L}\sqrt{\frac{T}{\mu }}$

The linear mass density $\mu$ can be written in terms of the radius $r$ and density $\rho$ of the string as $\mu =\pi r^2\rho$. Substituting this into the frequency formula, we get:

$f=\frac{n}{2Lr}\sqrt{\frac{T}{\pi \rho }}$

For string A, the first overtone corresponds to the second harmonic ($n_A=2$).
For string B, the second overtone corresponds to the third harmonic ($n_B=3$).

Since both strings are made of steel and are under the same tension, $\rho$ and $T$ are constant for both. We are given that the frequencies are equal ($f_A=f_B$) and $r_A=2r_B$.

Equating the frequencies:

$\frac{2}{2L_A{r}_A}\sqrt{\frac{T}{\pi \rho }}=\frac{3}{2L_B{r}_B}\sqrt{\frac{T}{\pi \rho }}$

$\frac{1}{L_A{r}_A}=\frac{3}{2L_B{r}_B}$

Substituting $r_A=2r_B$:

$\frac{1}{L_A(2r_B)}=\frac{3}{2L_B{r}_B}$

$\frac{1}{2L_A}=\frac{3}{2L_B}$

$\frac{L_A}{L_B}=\frac{1}{3}$

The ratio of the lengths of the strings is $1:3$.

Answer: $1:3$