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(A) The fundamental frequency of a stretched wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$.Thus,$n = \frac{1}{2l

Vedclass · Accepted Answer

$1: 3$

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(A) The fundamental frequency of a stretched wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 ho$.Thus,$n = \frac{1}{2lr} \sqrt{\frac{T}{\pi ho}}$.The first overtone of the first wire is $n_1 = 2 	imes n_{1, 	ext{fund}} = 2 	imes \frac{1}{2l_1 r_1} \sqrt{\frac{T}{\pi ho}} = \frac{1}{l_1 r_1} \sqrt{\frac{T}{\pi ho}}$.The second overtone of the second wire is $n_2 = 3 	imes n_{2, 	ext{fund}} = 3 	imes \frac{1}{2l_2 r_2} \sqrt{\frac{T}{\pi ho}} = \frac{3}{2l_2 r_2} \sqrt{\frac{T}{\pi ho}}$.Given $n_1 = n_2$,we have $\frac{1}{l_1 r_1} = \frac{3}{2l_2 r_2}$.Rearranging for the ratio of lengths: $\frac{l_1}{l_2} = \frac{2r_2}{3r_1}$.Since $r_1 = 2r_2$,substituting this gives $\frac{l_1}{l_2} = \frac{2r_2}{3(2r_2)} = \frac{1}{3}$.

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