When two tuning forks (fork 1 and fork 2) are | vedclass

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(D) Let the frequency of fork $1$ be $n_1 = 200 \, Hz$ and the frequency of fork $2$ be $n_2$.Initially,the beat frequency is $|n_1 - n_2| = 4 \, Hz$.

Vedclass · Accepted Answer

$196$

Vedclass · Answer

(D) Let the frequency of fork $1$ be $n_1 = 200 \, Hz$ and the frequency of fork $2$ be $n_2$.Initially,the beat frequency is $|n_1 - n_2| = 4 \, Hz$. This implies $n_2 = 200 \pm 4$,so $n_2 = 204 \, Hz$ or $n_2 = 196 \, Hz$.When tape is attached to the prong of fork $2$,its frequency $n_2$ decreases.If $n_2$ was $204 \, Hz$,decreasing it would make the beat frequency $|200 - n_2'|$ smaller than $4 \, Hz$.If $n_2$ was $196 \, Hz$,decreasing it further (e.g.,to $194 \, Hz$) would make the beat frequency $|200 - 194| = 6 \, Hz$.Since the beat frequency increased to $6 \, Hz$,the original frequency of fork $2$ must have been $196 \, Hz$.

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