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

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(D) Given: Frequency of string $A$,$n_A = 425 \, Hz$. Initial beat frequency $x_1 = 5 \, Hz$.Beat frequency is given by $|n_A - n_B| = 5 \, Hz$. This

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$420$

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(D) Given: Frequency of string $A$,$n_A = 425 \, Hz$. Initial beat frequency $x_1 = 5 \, Hz$.Beat frequency is given by $|n_A - n_B| = 5 \, Hz$. This implies $n_B$ could be $420 \, Hz$ or $430 \, Hz$.When the tension of string $B$ is increased,its frequency $n_B$ increases because $n \propto \sqrt{T}$.Case $1$: If $n_B = 420 \, Hz$,increasing tension increases $n_B$. As $n_B$ approaches $n_A$ $(425 \, Hz)$,the beat frequency $|n_A - n_B|$ decreases. This matches the observation that the beat frequency decreased to $3 \, Hz$.Case $2$: If $n_B = 430 \, Hz$,increasing tension increases $n_B$ further away from $n_A$ $(425 \, Hz)$,which would increase the beat frequency. This contradicts the observation.Therefore,the original frequency of $B$ must be $420 \, Hz$.

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