Two uniform wires of same material are vibrat | vedclass

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(B) The frequency of the $n^{th}$ harmonic for a stretched string is given by $f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$, where $L$ is the length, $T$

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$1/3$

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(B) The frequency of the $n^{th}$ harmonic for a stretched string is given by $f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$, where $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density.Linear mass density $\mu = ho \cdot A = ho \cdot \pi r^2$, where $ho$ is the density of the material and $r$ is the radius.Thus, $f_n = \frac{n}{2L} \sqrt{\frac{T}{ho \pi r^2}} = \frac{n}{2Lr} \sqrt{\frac{T}{ho \pi}}$.For the first wire: $n_1 = 2$ (first overtone is the second harmonic), $r_1 = 2r_2$, $L_1 = L_1$.$f_{1, 	ext{overtone}} = \frac{2}{2L_1 r_1} \sqrt{\frac{T}{ho \pi}} = \frac{1}{L_1 (2r_2)} \sqrt{\frac{T}{ho \pi}}$.For the second wire: $n_2 = 3$ (second overtone is the third harmonic), $r_2 = r_2$, $L_2 = L_2$.$f_{2, 	ext{overtone}} = \frac{3}{2L_2 r_2} \sqrt{\frac{T}{ho \pi}}$.Given $f_{1, 	ext{overtone}} = f_{2, 	ext{overtone}}$:$\frac{1}{2L_1 r_2} = \frac{3}{2L_2 r_2}$.$\frac{1}{L_1} = \frac{3}{L_2} 	ext{ implies } \frac{L_1}{L_2} = \frac{1}{3}$.

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