A tuning fork of frequency 512, Hz makes 4 be | vedclass

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(D) Let the frequency of the tuning fork be $v_{1} = 512\, Hz$ and the frequency of the piano string be $v_{2}$.The beat frequency is given by $|v_{1}

Vedclass · Accepted Answer

$508$

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(D) Let the frequency of the tuning fork be $v_{1} = 512\, Hz$ and the frequency of the piano string be $v_{2}$.The beat frequency is given by $|v_{1} - v_{2}| = 4\, Hz$.Therefore,$v_{2} = 512 \pm 4$,which means $v_{2} = 516\, Hz$ or $v_{2} = 508\, Hz$.When the tension in the piano string is increased,its frequency $v_{2}$ increases.Case $1$: If $v_{2} = 516\, Hz$,increasing the tension will increase $v_{2}$ further (e.g.,to $517\, Hz$ or $518\, Hz$),which would increase the beat frequency $(|512 - 517| = 5\, Hz)$. This contradicts the problem statement.Case $2$: If $v_{2} = 508\, Hz$,increasing the tension will increase $v_{2}$ (e.g.,to $510\, Hz$),which would decrease the beat frequency $(|512 - 510| = 2\, Hz)$. This matches the problem statement.Thus,the initial frequency of the piano string was $508\, Hz$.

$A$ tuning fork of frequency $512\, Hz$ makes $4$ 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 .... $Hz$

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