Question

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

• A. 504
• B. 520
• C. 260
• D. 252

Answer 1

The correct option is C 260

Suppose two tuning forks are named A and B with frequencies $n_A=256Hz$ (known), $n_B=?$
(unknown), and beat frequency x = 4 bps.


Frequency of unknown tuning for k may be
$n_B$ = 256 + 4 = 260 Hz
or = 256 - 4 = 252 Hz
It is given that on sounding waxed fork A (fork of frequency 256 Hz) and fork B,number of beats (beat frequency) increases.It means that with decrease in frequency of A,the difference in new frequency of A and the frequency of B has increased.This is possible only when the frequency of A while decreasing is moving away from the frequency of B.
This is possible only $n_B=260$ Hz.

Alternate method:It is given $n_A=256Hz,n_B=?$ and x=4 bps
Also after loading A(i.e. $n_A\downarrow$), beat frequency (i.e. x) increases $\uparrow )$.
Apply these informations in two possibilities to know the frequency of unknown
tuning fork.
$n_A\downarrow -n_B=x\uparrow$ ...(i)
$n_B-n_A\downarrow =x\uparrow$ ...(ii)
It is obvious that equation (i) is wrong (ii) is correct so
$n_B=n_A+x=256+4=260Hz.$​