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(B) The frequency of a vibrating wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$. Since tension $T$ and mass per unit length $\mu$ are consta

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$1608$

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(B) The frequency of a vibrating wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$. Since tension $T$ and mass per unit length $\mu$ are constant, we have $n \propto \frac{1}{l}$, which implies $n l = 	ext{constant}$.Let $n$ be the frequency of the tuning fork.At $l_1 = 100 \, cm$, the frequency of the wire is $n_1 = n + 8$ (since the wire is shorter, its frequency is higher).At $l_2 = 101 \, cm$, the frequency of the wire is $n_2 = n - 8$.Using $n_1 l_1 = n_2 l_2$:$(n + 8) 	imes 100 = (n - 8) 	imes 101$$100n + 800 = 101n - 808$$101n - 100n = 800 + 808$$n = 1608 \, Hz$.

When a vibrating tuning fork is placed on a sound box of a sonometer, $8$ beats per second are heard when the length of the sonometer wire is kept at $101 \, cm$ or $100 \, cm$. Then the frequency of the tuning fork is (Consider that the tension in the wire is kept constant). (in $ \, Hz$)

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