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Question

(A) The frequency of the $n^{	ext{th}}$ harmonic for a stretched string is given by $f = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 ho

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$1: 3$

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(A) The frequency of the $n^{	ext{th}}$ harmonic for a stretched string is given by $f = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 ho$. Thus,$f = \frac{n}{2L r} \sqrt{\frac{T}{\pi ho}}$. For the $1^{	ext{st}}$ wire,the $1^{	ext{st}}$ overtone is the $2^{	ext{nd}}$ harmonic $(n_1 = 2)$. So,$f_1 = \frac{2}{2L_1 r_1} \sqrt{\frac{T}{\pi ho}} = \frac{1}{L_1 r_1} \sqrt{\frac{T}{\pi ho}}$. For the $2^{	ext{nd}}$ wire,the $2^{	ext{nd}}$ overtone is the $3^{	ext{rd}}$ harmonic $(n_2 = 3)$. So,$f_2 = \frac{3}{2L_2 r_2} \sqrt{\frac{T}{\pi ho}}$. Given $f_1 = f_2$ and $r_1 = 2r_2$: $\frac{1}{L_1 (2r_2)} = \frac{3}{2L_2 r_2}$. $\frac{1}{2L_1} = \frac{3}{2L_2} \implies \frac{L_1}{L_2} = \frac{1}{3}$. Therefore,the ratio is $1: 3$.

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