Question

Two tuning forks when sounded together produced $4$beats/sec. The frequency of one fork is $256Hz$. The number of beats heard increases when the fork of frequency $256Hz$ is loaded with wax. The frequency (in $Hz$) of the other fork is

Answer 1

Given, known frequency of tuning fork $A$ is $N_A=256Hz$.

Unknown frequency of tuning fork $B$ is $N_B=?$

Two tuning forks when sounded together produced $4$beats/sec,
$x=4bps$

The number of beats heard increases when the fork of frequency $256Hz$ is loaded with wax. The frequency of tuning fork $B$ is,

$N_B-N_A=x$
$N_B=N_A+x$
$N_B=256+(\pm 4)$
$N_B=260HzOr{N}_B=252Hz$

The frequency of sounding waxed tuning fork $A$ decreases, the difference in new frequency of $A,B$ has increased.

This is possible only when the frequency of $A$ while decreasing is moving away from the frequency of $B$.
So, the possible frequency, $N_B=260Hz$.

Final Answer: $N_B=260Hz$