A wire of length L, diameter d, and density o | vedclass

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(C) The fundamental frequency of a stretched wire is given by the formula: $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $\mu$ is the linear mass den

Vedclass · Accepted Answer

$1 : 6$

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(C) The fundamental frequency of a stretched wire is given by the formula: $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $\mu$ is the linear mass density.Linear mass density $\mu = 	ext{Area} 	imes 	ext{Density} = (\pi r^2) ho = \pi (d/2)^2 ho = \frac{\pi d^2 ho}{4}$.Substituting $\mu$ into the frequency formula: $n = \frac{1}{2L} \sqrt{\frac{T}{\frac{\pi d^2 ho}{4}}} = \frac{1}{Ld} \sqrt{\frac{T}{\pi ho}}$.For the first wire: $n_A = \frac{1}{Ld} \sqrt{\frac{T}{\pi ho}}$.For the second wire: $n_B = \frac{1}{(2L)(3d)} \sqrt{\frac{2T}{\pi (2ho)}} = \frac{1}{6Ld} \sqrt{\frac{T}{\pi ho}}$.Taking the ratio $n_B : n_A = \frac{\frac{1}{6Ld} \sqrt{\frac{T}{\pi ho}}}{\frac{1}{Ld} \sqrt{\frac{T}{\pi ho}}} = \frac{1}{6}$.Thus, the ratio $n_B : n_A$ is $1 : 6$.

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