Two strings X and Y of a sitar produce a beat | vedclass

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(A) Given: Frequency of string $X$ $(n_X)$ = $300 \ Hz$. Initial beat frequency = $|n_X - n_Y| = 4 \ Hz$. This implies the initial frequency of $Y$ $(

Vedclass · Accepted Answer

$296$

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(A) Given: Frequency of string $X$ $(n_X)$ = $300 \ Hz$. Initial beat frequency = $|n_X - n_Y| = 4 \ Hz$. This implies the initial frequency of $Y$ $(n_Y)$ is either $300 - 4 = 296 \ Hz$ or $300 + 4 = 304 \ Hz$. When the tension of string $Y$ is increased,its frequency $n_Y$ increases because $n \propto \sqrt{T}$. Case $1$: If $n_Y = 296 \ Hz$,increasing the tension makes $n_Y$ approach $300 \ Hz$ (e.g.,$298 \ Hz$),so the beat frequency $|300 - 298| = 2 \ Hz$. This matches the given condition. Case $2$: If $n_Y = 304 \ Hz$,increasing the tension makes $n_Y$ move further away from $300 \ Hz$ (e.g.,$306 \ Hz$),so the beat frequency $|300 - 306| = 6 \ Hz$. This does not match the condition. Therefore,the original frequency of $Y$ was $296 \ Hz$.

Two strings $X$ and $Y$ of a sitar produce a beat frequency of $4 \ Hz$. When the tension of the string $Y$ is slightly increased,the beat frequency is found to be $2 \ Hz$. If the frequency of $X$ is $300 \ Hz$,then the original frequency of $Y$ was .... $Hz$.

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