A tuning fork of known frequency 256,Hz produ | vedclass

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(D) Let the frequency of the piano string be $n_P$. The frequency of the tuning fork is $n_f = 256\,Hz$.The beat frequency is given by $|n_f - n_P| =

Vedclass · Accepted Answer

$256 - 5\,Hz$

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(D) Let the frequency of the piano string be $n_P$. The frequency of the tuning fork is $n_f = 256\,Hz$.The beat frequency is given by $|n_f - n_P| = 5\,Hz$. This implies $n_P = 256 \pm 5\,Hz$,so $n_P$ is either $251\,Hz$ or $261\,Hz$.When the tension $T$ in the piano string is increased,the frequency $n_P$ increases because $n_P \propto \sqrt{T}$.If $n_P$ was $251\,Hz$,increasing the tension makes $n_P$ approach $256\,Hz$,which would decrease the beat frequency $|256 - n_P|$.If $n_P$ was $261\,Hz$,increasing the tension makes $n_P$ move further away from $256\,Hz$,which would increase the beat frequency.Since the beat frequency decreases from $5\,Hz$ to $2\,Hz$,the initial frequency must have been $251\,Hz$.Therefore,the frequency of the piano string before increasing the tension was $256 - 5 = 251\,Hz$.

$A$ tuning fork of known frequency $256\,Hz$ produces $5$ beats per second with the vibrating string of a piano. The beat frequency decreases to $2$ beats per second when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was

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