A tuning fork of frequency 392 , Hz resonates | vedclass

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(C) The frequency of a vibrating string is given by $n = \frac{1}{2\ell} \sqrt{\frac{T}{\mu}}$.Since $T$ and $\mu$ are constant,we have $n \propto \fr

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$8$

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(C) The frequency of a vibrating string is given by $n = \frac{1}{2\ell} \sqrt{\frac{T}{\mu}}$.Since $T$ and $\mu$ are constant,we have $n \propto \frac{1}{\ell}$.Taking the derivative,we get $\frac{\Delta n}{n} = -\frac{\Delta \ell}{\ell}$.Given that the length is decreased by $2 \%$,we have $\frac{\Delta \ell}{\ell} = -0.02$.Substituting this,$\frac{\Delta n}{n} = -(-0.02) = 0.02$.This means the frequency increases by $2 \%$.The new frequency $n_2 = n_1 + \Delta n = n_1 + 0.02 	imes n_1 = 1.02 	imes 392$.$n_2 = 399.84 \, Hz$.The number of beats per second is the difference between the frequencies: $|n_2 - n_1| = |399.84 - 392| = 7.84 \, Hz$.Rounding to the nearest integer,the number of beats is $8$.

$A$ tuning fork of frequency $392 \, Hz$ resonates with $50 \, cm$ length of a string under tension $T$. If the length of the string is decreased by $2 \%$,keeping the tension constant,the number of beats heard when the string and the tuning fork are made to vibrate simultaneously is

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