Question

When a tuning fork of frequency 341 Hz is sounded with another tuning fork, six beats per second are heard. When the second tuning fork is loaded with wax and sounded with the first fork, the number of beats is two per second. The natural frequency (in $\text{Hz}$)of the second tuning fork is

Answer 1

Given beat frequency is $x=6bps$, which decreases (i.e, $x\downarrow$) after second fork is loaded with wax.

Hence,
$|n_B\downarrow -n_A|=|x\downarrow |$
$⟹n_b=n_A+x=341+6=347Hz.$