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

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(B) Let the frequency of fork $1$ be $f_1 = 200 \,Hz$ and the frequency of fork $2$ be $f_2$.Initially,the beat frequency is $|f_1 - f_2| = 4 \,Hz$. T

Vedclass · Accepted Answer

$196$

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(B) Let the frequency of fork $1$ be $f_1 = 200 \,Hz$ and the frequency of fork $2$ be $f_2$.Initially,the beat frequency is $|f_1 - f_2| = 4 \,Hz$. This implies $f_2 = 200 \pm 4$,so $f_2 = 204 \,Hz$ or $f_2 = 196 \,Hz$.When tape is attached to the prong of fork $2$,its mass increases,which causes its frequency $f_2$ to decrease.After adding the tape,the new beat frequency is $|f_1 - f_2'| = 6 \,Hz$,where $f_2' Case $1$: If $f_2 = 204 \,Hz$,then $f_2'$ decreases from $204 \,Hz$. The beat frequency $|200 - f_2'|$ would decrease (e.g.,if $f_2' = 202 \,Hz$,beats $= 2 \,Hz$). This contradicts the observation that beats increased to $6 \,Hz$.Case $2$: If $f_2 = 196 \,Hz$,then $f_2'$ decreases from $196 \,Hz$ (e.g.,$f_2' = 194 \,Hz$). The beat frequency $|200 - 194| = 6 \,Hz$. This matches the observation.Therefore,the original frequency of fork $2$ is $196 \,Hz$.

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