A string A has twice the length, twice the diameter, twice the tension and twice the density of another string B. The overtone of A which will, have the same fundamental frequency as that of B is

Fundamental frequency of string \(B\) is,
\(\begin{aligned}
\mathrm{n}_{\mathrm{B}} & =\frac{1}{2 l_{\mathrm{B}}} \sqrt{\frac{\mathrm{~T}_{\mathrm{B}}}{\mu_{\mathrm{B}}}} \text {. where, } \mu=\text { mass per unit length. } \\
\mathrm{m}_{\mathrm{B}} & =\rho \pi \mathrm{r}^2 l \\
\therefore \quad \mu_{\mathrm{B}} & =\frac{\rho \pi \mathrm{r}^2 l}{l}=\rho \pi \mathrm{r}^2 \\
\therefore \quad \mu_{\mathrm{A}} & =2 \rho \pi(2 \mathrm{r})^2=8 \mu_{\mathrm{B}} \\
\mu_{\mathrm{A}} & =\frac{8 f \pi \mathrm{r}^2}{2 l}=4 \mu_{\mathrm{B}}
\end{aligned}\)
\(\therefore \quad\) Fundamental frequency of string A is,
\(\mathrm{n}_{\mathrm{A}}=\frac{1}{2\left(2 l_{\mathrm{B}}\right)} \sqrt{\frac{2 \mathrm{~T}_{\mathrm{B}}}{8 \mu_{\mathrm{B}}}}=\frac{1}{4}\left(\frac{1}{2 l_{\mathrm{B}}} \sqrt{\frac{\mathrm{~T}_{\mathrm{B}}}{\mu_{\mathrm{B}}}}\right)=\frac{1}{4} \mathrm{n}_{\mathrm{B}}\)
Frequency of \(\mathrm{p}^{\text {th }}\) overtone \(=(p+1) n\) Here, \((p+1)=4 \Rightarrow p=3\)
\(\therefore \quad 3^{\text {rd }}\) Overtone of string A will have same frequency as fundamental frequency of string B .