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

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$1/3$

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(A) The fundamental frequency of a stretched wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$.Thus,$f = \frac{1}{2Lr} \sqrt{\frac{T}{\pi \rho}}$.The first overtone of the first wire is $f_1 = 2f_1 = \frac{2}{2L_1 r_1} \sqrt{\frac{T}{\pi \rho}} = \frac{1}{L_1 r_1} \sqrt{\frac{T}{\pi \rho}}$.The second overtone of the second wire is $f_2 = 3f_2 = \frac{3}{2L_2 r_2} \sqrt{\frac{T}{\pi \rho}}$.Given $f_1 = f_2$,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 $r_1 = 2r_2$,we substitute: $\frac{L_1}{L_2} = \frac{2r_2}{3(2r_2)} = \frac{2}{6} = \frac{1}{3}$.

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