A tuning fork sounded together with a tuning | vedclass

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(D) Let the frequency of the unknown tuning fork be $n_B$ and the known frequency be $n_A = 256 \ Hz$.The beat frequency is given by $|n_A - n_B| = 2

Vedclass · Accepted Answer

$254$

Vedclass · Answer

(D) Let the frequency of the unknown tuning fork be $n_B$ and the known frequency be $n_A = 256 \ Hz$.The beat frequency is given by $|n_A - n_B| = 2 \ Hz$.This implies $n_B = 256 \pm 2$,so $n_B$ is either $258 \ Hz$ or $254 \ Hz$.When the known tuning fork $(256 \ Hz)$ is loaded with wax,its frequency $n_A$ decreases.Given that the beat frequency decreases from $2 \ Hz$ to $1 \ Hz$ after loading,we analyze the two cases:Case $1$: If $n_B = 258 \ Hz$,then $n_B - n_A = 258 - 256 = 2 \ Hz$. After loading,$n_A$ decreases,so $(n_B - n_A)$ increases. This contradicts the observation.Case $2$: If $n_B = 254 \ Hz$,then $n_A - n_B = 256 - 254 = 2 \ Hz$. After loading,$n_A$ decreases,so $(n_A - n_B)$ decreases. This matches the observation $(1 \ Hz)$.Therefore,the frequency of the unknown tuning fork is $254 \ Hz$.

$A$ tuning fork sounded together with a tuning fork of frequency $256 \ Hz$ emits $2$ beats per second. On loading the tuning fork of frequency $256 \ Hz,$ the number of beats heard is $1$ per second. The frequency of the unknown tuning fork is: (in $Hz$)

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