When two tuning forks are sounded together,5 | vedclass

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(D) The frequency of a sonometer wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$.Since the tension $T$ and mass per unit length $\mu$ are con

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$480 \ Hz, 485 \ Hz$

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(D) The frequency of a sonometer wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$.Since the tension $T$ and mass per unit length $\mu$ are constant,$f \propto \frac{1}{l}$.Let $f_1$ and $f_2$ be the frequencies of the two tuning forks corresponding to lengths $l_1 = 0.97 \ m$ and $l_2 = 0.96 \ m$ respectively.Thus,$f_1 = \frac{k}{0.97}$ and $f_2 = \frac{k}{0.96}$,where $k = \frac{1}{2} \sqrt{\frac{T}{\mu}}$.Given the beat frequency is $f_2 - f_1 = 5 \ Hz$.Substituting the values: $\frac{k}{0.96} - \frac{k}{0.97} = 5$.$k \left( \frac{0.97 - 0.96}{0.96 	imes 0.97} ight) = 5$.$k \left( \frac{0.01}{0.9312} ight) = 5 \implies k = 5 	imes 93.12 = 465.6$.Now,$f_1 = \frac{465.6}{0.97} = 480 \ Hz$.And $f_2 = \frac{465.6}{0.96} = 485 \ Hz$.

When two tuning forks are sounded together,$5$ beats per second are heard. One of the forks is in unison with $0.97 \ m$ length of sonometer wire and the other is in unison with $0.96 \ m$ length of the same wire. The frequencies of the two tuning forks are

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