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$(A)$ The fundamental frequency of a stretched wire is given by the formula: $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$, where $\mu = \pi r^2 \rho$ is th

Vedclass · Accepted Answer

$200$

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$(A)$ The fundamental frequency of a stretched wire is given by the formula: $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$, where $\mu = \pi r^2 ho$ is the linear mass density.Substituting $\mu$, we get: $f = \frac{1}{2lr} \sqrt{\frac{T}{\pi ho}}$.Given the initial conditions: $f_1 = 600 \,Hz$, length $l_1 = l$, radius $r_1 = r$, and tension $T_1 = T$.New conditions: $l_2 = 2l$, $r_2 = r/2$, and $T_2 = T/9$.The ratio of frequencies is: $\frac{f_2}{f_1} = \frac{l_1}{l_2} 	imes \frac{r_1}{r_2} 	imes \sqrt{\frac{T_2}{T_1}}$.Substituting the values: $\frac{f_2}{600} = \frac{l}{2l} 	imes \frac{r}{r/2} 	imes \sqrt{\frac{T/9}{T}}$.$\frac{f_2}{600} = \frac{1}{2} 	imes 2 	imes \sqrt{\frac{1}{9}} = 1 	imes \frac{1}{3} = \frac{1}{3}$.Therefore, $f_2 = \frac{600}{3} = 200 \,Hz$.

$A$ wire under tension vibrates with a fundamental frequency of $600 \,Hz$. If the length of the wire is doubled, the radius is halved and the wire is made to vibrate under one-ninth the tension. Then the fundamental frequency will become (in $\,Hz$)

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