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

[CPMT 1976; MP PMT 1993]

• 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 (known), $n_B$ = ? (unknown), and beat frequency x = 4 bps.

Frequency of unknown tuning fork 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 if $n_B$ = 260 Hz.

Alternate method : It is given $n_A$ = 256 Hz,$n_B$ = ? 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 known 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

$n_B=n_A+x=256+4=260Hz$.