A tuning fork of unknown frequency produces 4 | vedclass

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Question

(A) Let the unknown frequency be $f$.Given that the beat frequency is $4 \,Hz$ when sounded with a $254 \,Hz$ tuning fork,the possible frequencies are

Vedclass · Accepted Answer

$258 \,Hz$

Vedclass · Answer

(A) Let the unknown frequency be $f$.Given that the beat frequency is $4 \,Hz$ when sounded with a $254 \,Hz$ tuning fork,the possible frequencies are $f = 254 \pm 4$,which means $f = 258 \,Hz$ or $f = 250 \,Hz$.When the tuning fork is loaded with wax,its frequency decreases.Case $1$: If $f = 250 \,Hz$,loading with wax will decrease the frequency further (e.g.,to $249 \,Hz$),resulting in a beat frequency of $|254 - 249| = 5 \,Hz$. This contradicts the problem statement.Case $2$: If $f = 258 \,Hz$,loading with wax will decrease the frequency (e.g.,to $257 \,Hz$ or $256 \,Hz$ or $255 \,Hz$),resulting in a beat frequency of $|254 - 255| = 1 \,Hz$ or $|254 - 256| = 2 \,Hz$ or $|254 - 257| = 3 \,Hz$. However,if the frequency decreases such that the beat frequency remains $4 \,Hz$,it implies the frequency must have been higher than the reference fork. Specifically,if the frequency was $258 \,Hz$ and it decreases,it moves closer to $254 \,Hz$,reducing the beat frequency. The only way to maintain $4$ beats is if the frequency was $258 \,Hz$ and it was loaded such that it becomes $250 \,Hz$ (which is physically unlikely for a small amount of wax) or if the question implies the beat frequency condition holds for the higher value. Given standard physics problems of this type,$f = 258 \,Hz$ is the correct choice.

$A$ tuning fork of unknown frequency produces $4$ beats per second when sounded with another tuning fork of frequency $254 \,Hz$. It gives the same number of beats per second when the unknown tuning fork is loaded with wax. The unknown frequency before loading with wax is ..........

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