A tuning fork of frequency 392 Hz resonates | vedclass

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(C) The fundamental frequency of a vibrating string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$.Since tension $T$ and mass per unit length $\m

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) The fundamental frequency of a vibrating string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$.Since tension $T$ and mass per unit length $\mu$ are constant,we have $n \propto \frac{1}{l}$.Taking the logarithmic derivative,we get $\frac{\Delta n}{n} = -\frac{\Delta l}{l}$.Given that the length is decreased by $2\%$,we have $\frac{\Delta l}{l} = -0.02$.Therefore,the change in frequency is $\frac{\Delta n}{n} = -(-0.02) = 0.02$.This means the new frequency $n_2$ is $n_2 = n_1(1 + 0.02) = 392 	imes 1.02 = 399.84 \ Hz$.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 heard 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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