A tuning fork of unknown frequency makes $3\,beats/sec$ with a standard fork of frequency $384\,Hz$ . The beat frequency decreases when a small piece of wax is put on the prong of the first. The frequency of the fork is .... $Hz$
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answer $: 387 H z$
beats $=\left|f_{1}-f_{2}\right|$
since on adding wax to unknown tuning fork the frequency $f_{1}$ reduced
and also beats reduced this means that $f_{1}>f_{2}$
$f_{1}-384=3$
$f_{1}=387 H z$
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