Two sitar strings A and B playing the note Sa | vedclass

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

Vedclass · Accepted Answer

$260$

Vedclass · Answer

(C) The frequency of a stretched string is given by $v = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$. Thus,$v \propto \sqrt{T}$.When the tension $T$ of string $B$ is decreased,its frequency $v_B$ decreases.Initially,the beat frequency is $|v_B - v_A| = 5 \ Hz$. This means $v_B$ could be $255 + 5 = 260 \ Hz$ or $255 - 5 = 250 \ Hz$.If $v_B = 250 \ Hz$,decreasing the tension would make $v_B$ even smaller (e.g.,$248 \ Hz$),which would increase the beat frequency to $|248 - 255| = 7 \ Hz$.However,the problem states that the beat frequency decreases to $3 \ Hz$.If $v_B = 260 \ Hz$,decreasing the tension makes $v_B$ smaller (e.g.,$258 \ Hz$),which results in a new beat frequency of $|258 - 255| = 3 \ Hz$.This matches the given condition.Therefore,the original frequency of $B$ is $260 \ Hz$.

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