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Waves and Sound — JEE physics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceJEE physicsWaves and Sound1 Mark
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.
$260$
Suppose two tuning forks are named $A$ and $B$ with frequencies $n_A=256 Hz$ (known), $n_{\text {～}}=$ ? (unknown), and beat frequency $x=4$ bps.

$n_B  =256+4=260\  Hz$
$\text { or }  =256-4=252 Hz$
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 $n_1=260 Hz$.
Alternate method : It is given $n_A=256 Hz, n_B=$ ? and $x=4$ bps
Also after loading $A($ i.e. $n, \downarrow$ ), beat frequency (i.e. $x$ ) increases $(\uparrow)$.
Apply these informations in two possibilities to known the frequency of unknown tuning fork.
$n \downarrow-n=x \uparrow \dots(i)$
$n-n \downarrow=x \uparrow \dots(ii)$
It is obvious that equation (i) is wrong (ii) is correct so $n_n=n+x=256+4=260\ Hz$

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