When a tuning fork of frequency 341 Hz is so | vedclass

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Question

(D) Let the frequency of the first tuning fork be $n_A = 341 \ Hz$ and the frequency of the second tuning fork be $n_B$.The beat frequency is given by

Vedclass · Accepted Answer

$347$

Vedclass · Answer

(D) Let the frequency of the first tuning fork be $n_A = 341 \ Hz$ and the frequency of the second tuning fork be $n_B$.The beat frequency is given by $|n_A - n_B| = 6 \ Hz$.This implies $n_B = 341 \pm 6$,so $n_B$ could be $347 \ Hz$ or $335 \ Hz$.When the second tuning fork is loaded with wax,its frequency $n_B$ decreases $(n_B \downarrow)$.After loading,the new beat frequency is $2 \ Hz$.Case $1$: If $n_B = 335 \ Hz$,then $n_A - n_B = 6 \ Hz$. Loading $n_B$ makes it smaller,so $(n_A - n_B)$ increases. This contradicts the observation that the beat frequency decreased to $2 \ Hz$.Case $2$: If $n_B = 347 \ Hz$,then $n_B - n_A = 6 \ Hz$. Loading $n_B$ makes it smaller,so $(n_B - n_A)$ decreases. As $n_B$ decreases towards $n_A$,the beat frequency decreases. This matches the observation that the beat frequency decreased to $2 \ Hz$.Therefore,the natural frequency of the second tuning fork is $347 \ Hz$.

When a tuning fork of frequency $341 \ Hz$ is sounded with another tuning fork,six beats per second are heard. When the second tuning fork is loaded with wax and sounded with the first tuning fork,the number of beats is two per second. The natural frequency of the second tuning fork is: (in $Hz$)

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