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(C) The fundamental frequency of a stretched wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu$ is the mass per unit length.Since $\

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$1 : 6$

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(C) The fundamental frequency of a stretched wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu$ is the mass per unit length.Since $\mu = 	ext{Area} 	imes 	ext{Density} = (\pi r^2) ho = \pi (d/2)^2 ho = \frac{\pi d^2 ho}{4}$,we can write $n = \frac{1}{2L} \sqrt{\frac{T}{\frac{\pi d^2 ho}{4}}} = \frac{1}{Ld} \sqrt{\frac{T}{\pi ho}}$.For wire $A$: $n_A = \frac{1}{Ld} \sqrt{\frac{T}{\pi ho}}$.For wire $B$: $n_B = \frac{1}{(2L)(3d)} \sqrt{\frac{2T}{\pi (2ho)}} = \frac{1}{6Ld} \sqrt{\frac{T}{\pi ho}}$.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 $1 : 6$.

$A$ wire of length $L$,diameter $d$,and density of material $\rho$ is under tension $T$,having a fundamental frequency of vibration $n_A$. Another wire of length $2L$,tension $2T$,density $2\rho$,and diameter $3d$ has a fundamental frequency of vibration $n_B$. The ratio $n_B : n_A$ is

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