Two uniform strings A and B made of steel are | vedclass

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

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$1:3$

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(B) The frequency of the $n$-th harmonic for a string is given by $f_n = \frac{n}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$.Thus,$f_n = \frac{n}{2lr} \sqrt{\frac{T}{\pi \rho}}$.The first overtone of $A$ is the second harmonic $(n=2)$,so $f_{A,2} = \frac{2}{2l_A r_A} \sqrt{\frac{T}{\pi \rho}} = \frac{1}{l_A r_A} \sqrt{\frac{T}{\pi \rho}}$.The second overtone of $B$ is the third harmonic $(n=3)$,so $f_{B,3} = \frac{3}{2l_B r_B} \sqrt{\frac{T}{\pi \rho}}$.Given $f_{A,2} = f_{B,3}$,we have $\frac{1}{l_A r_A} = \frac{3}{2l_B r_B}$.Given $r_A = 2r_B$,we substitute this into the equation:$\frac{1}{l_A (2r_B)} = \frac{3}{2l_B r_B} \Rightarrow \frac{1}{2l_A} = \frac{3}{2l_B}$.Therefore,$\frac{l_A}{l_B} = \frac{1}{3}$.

Two uniform strings $A$ and $B$ made of steel are made to vibrate 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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