Two sitar strings A and B playing the note 'G | vedclass

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Question

(D) The frequency of string $A$ is $f_A = 324 \ Hz$.Let the frequency of string $B$ be $f_B$.The beat frequency is given by $n = |f_A - f_B| = 6 \ Hz$

Vedclass · Accepted Answer

$318$

Vedclass · Answer

(D) The frequency of string $A$ is $f_A = 324 \ Hz$.Let the frequency of string $B$ be $f_B$.The beat frequency is given by $n = |f_A - f_B| = 6 \ Hz$.This implies $f_B = 324 \pm 6$,so $f_B = 330 \ Hz$ or $318 \ Hz$.The frequency of a string is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,which means $f \propto \sqrt{T}$.When the tension $T$ in string $A$ is reduced,its frequency $f_A$ decreases.If $f_B = 330 \ Hz$,then initially $f_B - f_A = 330 - 324 = 6 \ Hz$. As $f_A$ decreases,the difference $(f_B - f_A)$ increases,so the beat frequency would increase.If $f_B = 318 \ Hz$,then initially $f_A - f_B = 324 - 318 = 6 \ Hz$. As $f_A$ decreases,the difference $(f_A - f_B)$ decreases. Since the beat frequency reduced to $3 \ Hz$,this matches the condition.Therefore,the frequency of $B$ is $318 \ Hz$.

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