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 $f_1 = 341 \ Hz$ and the frequency of the second tuning fork be $f_2$.The beat frequency is given by

Vedclass · Accepted Answer

$347$

Vedclass · Answer

(D) Let the frequency of the first tuning fork be $f_1 = 341 \ Hz$ and the frequency of the second tuning fork be $f_2$.The beat frequency is given by $|f_1 - f_2| = 6 \ Hz$.This implies $f_2 = 341 \pm 6$,so $f_2 = 335 \ Hz$ or $f_2 = 347 \ Hz$.When the second tuning fork is loaded with wax,its frequency $f_2$ decreases to $f_2'$.Given that the new beat frequency is $|f_1 - f_2'| = 2 \ Hz$.Case $1$: If $f_2 = 335 \ Hz$,loading with wax decreases the frequency further (i.e.,$f_2' Case $2$: If $f_2 = 347 \ Hz$,loading with wax decreases the frequency (i.e.,$f_2' Thus,the original 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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