A tuning fork arrangement (pair) produces 4 , | vedclass

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(D) Let the frequency of the known tuning fork be $n_A = 288 \, Hz$ and the unknown fork be $n_B$.Initially,the beat frequency is $x = |n_A - n_B| = 4

Vedclass · Accepted Answer

$292$

Vedclass · Answer

(D) Let the frequency of the known tuning fork be $n_A = 288 \, Hz$ and the unknown fork be $n_B$.Initially,the beat frequency is $x = |n_A - n_B| = 4 \, Hz$.This implies $n_B = 288 \pm 4$,so $n_B$ is either $292 \, Hz$ or $284 \, Hz$.When wax is added to the unknown fork,its frequency $n_B$ decreases $(n_B \downarrow)$.After adding wax,the new beat frequency is $x' = 2 \, Hz$.Case $1$: If $n_B = 284 \, Hz$,then $n_B$ decreases further away from $288 \, Hz$,so the beat frequency would increase $(|288 - 283| = 5 \, Hz)$,which contradicts the observation.Case $2$: If $n_B = 292 \, Hz$,then $n_B$ decreases towards $288 \, Hz$,so the beat frequency decreases $(|288 - 290| = 2 \, Hz)$.This matches the given observation.Therefore,the frequency of the unknown fork is $292 \, Hz$.

$A$ tuning fork arrangement (pair) produces $4 \, beats/sec$ with one fork of frequency $288 \, cps$. $A$ little wax is placed on the unknown fork and it then produces $2 \, beats/sec$. The frequency of the unknown fork is .... $cps$.

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