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(A) The frequency of the $n$-th harmonic for a stretched string is given by $f_n = \frac{n}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$.For t

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$1:3$

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(A) The frequency of the $n$-th harmonic for a stretched string is given by $f_n = \frac{n}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$.For the first overtone of $A$ $(n=2)$: $f_{A} = \frac{2}{2l_A} \sqrt{\frac{T}{\pi r_A^2 \rho}} = \frac{1}{l_A r_A} \sqrt{\frac{T}{\pi \rho}}$.For the second overtone of $B$ $(n=3)$: $f_{B} = \frac{3}{2l_B} \sqrt{\frac{T}{\pi r_B^2 \rho}} = \frac{3}{2l_B r_B} \sqrt{\frac{T}{\pi \rho}}$.Given $f_A = f_B$ and $r_A = 2r_B$,we equate the expressions:$\frac{1}{l_A r_A} = \frac{3}{2l_B r_B} \Rightarrow \frac{1}{l_A (2r_B)} = \frac{3}{2l_B r_B}$.Simplifying,we get $\frac{1}{2l_A} = \frac{3}{2l_B} \Rightarrow \frac{l_A}{l_B} = \frac{1}{3}$.Thus,the ratio of the lengths $l_A : l_B = 1:3$.

Two uniform stretched strings $A$ and $B$,made of steel,are vibrating under the same tension. If the first overtone of $A$ is equal to the second overtone of $B$ and if the radius of $A$ is twice that of $B$,the ratio of the lengths of the strings is

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