In a guitar,two strings A and B made of the s | vedclass

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(C) The frequency of a stretched string is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$. Thus,$f \propto \sqrt{T}$.When the tension $T$ in string

Vedclass · Accepted Answer

$524$

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(C) The frequency of a stretched string is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$. Thus,$f \propto \sqrt{T}$.When the tension $T$ in string $B$ is decreased,its frequency $f_B$ decreases.The initial beat frequency is $|f_A - f_B| = 6 \, Hz$. Given $f_A = 530 \, Hz$,the possible values for $f_B$ are $530 + 6 = 536 \, Hz$ or $530 - 6 = 524 \, Hz$.Case $1$: If $f_B = 536 \, Hz$,decreasing the tension decreases $f_B$. As $f_B$ moves towards $530 \, Hz$,the beat frequency $|530 - f_B|$ decreases (e.g.,$536 	o 535 \implies$ beats $6 	o 5$). This contradicts the problem statement.Case $2$: If $f_B = 524 \, Hz$,decreasing the tension decreases $f_B$. As $f_B$ moves away from $530 \, Hz$,the beat frequency $|530 - f_B|$ increases (e.g.,$524 	o 523 \implies$ beats $6 	o 7$). This matches the problem statement.Therefore,the original frequency of $B$ is $524 \, Hz$.

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