Two tuning forks when sounded together produced 4 beats/sec. The … - Vidyadip

MCQ

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
$520$
✓
$260$
D
$252$

✓

Answer

Correct option: C.

$260$

c
(c) Suppose two tuning forks are named $A$ and $B$ with frequencies ${n_A} = 256\,Hz$ (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

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

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