When two tuning forks A and B are sounded tog | vedclass

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(B) Given: Beat frequency $= 4 \,Hz$, frequency of fork $B$ $(f_B)$ $= 384 \,Hz$.The possible frequencies of fork $A$ $(f_A)$ are $f_B \pm 4$, which a

Vedclass · Accepted Answer

$388$

Vedclass · Answer

(B) Given: Beat frequency $= 4 \,Hz$, frequency of fork $B$ $(f_B)$ $= 384 \,Hz$.The possible frequencies of fork $A$ $(f_A)$ are $f_B \pm 4$, which are $388 \,Hz$ or $380 \,Hz$.When a prong of a tuning fork is filed, its mass decreases, which increases its frequency ($f_A$ increases).Case $1$: If $f_A = 380 \,Hz$, filing increases $f_A$ towards $384 \,Hz$, so the beat frequency $(|f_A - f_B|)$ would decrease.Case $2$: If $f_A = 388 \,Hz$, filing increases $f_A$ away from $384 \,Hz$ (e.g., to $389 \,Hz$), so the beat frequency $(|f_A - f_B|)$ increases.Since the problem states that the beat frequency increases, the initial frequency of fork $A$ must be $388 \,Hz$.

When two tuning forks $A$ and $B$ are sounded together, $4$ beats per second are heard. The frequency of the fork $B$ is $384 \,Hz$. When one of the prongs of the fork $A$ is filed and sounded with $B$, the beat frequency increases. Then the frequency of the fork $A$ is: (in $\,Hz$)

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