Two tuning forks F_1 and F_2 produce 6 beats | vedclass

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Question

(B) The initial beat frequency is $n_b = |n_1 - n_2| = 6 \ Hz$. Given $n_1 = 256 \ Hz$,the possible frequencies for $F_2$ are $256 + 6 = 262 \ Hz$ or

Vedclass · Accepted Answer

$262$

Vedclass · Answer

(B) The initial beat frequency is $n_b = |n_1 - n_2| = 6 \ Hz$. Given $n_1 = 256 \ Hz$,the possible frequencies for $F_2$ are $256 + 6 = 262 \ Hz$ or $256 - 6 = 250 \ Hz$.When wax is applied to a tuning fork,its frequency decreases.If $n_2 = 250 \ Hz$,applying wax would decrease it further (e.g.,to $249 \ Hz$),making the beat frequency $|256 - 249| = 7 \ Hz$. This contradicts the observation that the beat frequency remains $6 \ Hz$.If $n_2 = 262 \ Hz$,applying wax decreases it (e.g.,to $262 - \Delta n$). If the beat frequency remains $6 \ Hz$,it implies the frequency of $F_2$ was higher than $F_1$ and decreased towards $F_1$. However,if the beat frequency remains $6 \ Hz$ after applying wax,it indicates the frequency of $F_2$ was $262 \ Hz$ and it decreased to $262 - 6 = 256 \ Hz$ (which would result in $0$ beats) or the question implies the beat frequency is still $6$ after the change. Given the standard interpretation of such problems,if $n_2 = 262 \ Hz$,applying wax decreases it,and if it remains $6$,it implies the original frequency was $262 \ Hz$.

Two tuning forks $F_1$ and $F_2$ produce $6$ beats per second when sounded together. The frequency of $F_1$ is $256 \ Hz$. When wax is applied to $F_2$,the number of beats per second remains $6$. What is the frequency of $F_2$ in $Hz$?

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