Question

Three sound waves of equal amplitudes have frequencies $(v-1),v,(v+1)$. They superpose to give beats. The number of beats produced per second will be

• A. v
• B. v/2
• C. 2
• D. 1

Answer 1

Answer: (C)

If $(v-1),v,(v+1)$ be the frequencies of the three waves and a be the amplitude of each then $y_1=a\sin 2\pi (v-1)t,y_2=a\sin 2\pi vt$
and $y_3=a\sin 2\pi (v+1)t$
Resultant displacement due to all three waves is
$y=y_1+y_2+y_3$
$=a\sin 2\pi vt+a[\sin 2\pi (v-1)t$
$+\sin 2\pi (v+1)t]$
$=a\sin 2\pi vt+a[2\sin 2\pi vt\cos 2\pi t]$
$=a[2\cos 2\pi t+1]\sin 2\pi vt$
$=a^{\prime }\sin 2\pi$ vt with $d=a[1+2\cos 2\pi t]$
So, $I\propto (d{)}^2\propto a^2(1+2\cos 2\pi t{)}^2$
For $I$ to be max. or min.
$\frac{dI}{dt}=0\Rightarrow \frac{d}{dt}(1+2\cos 2\pi t{)}^2=0$
ie, $2(1+2\cos \pi t)(2\sin 2\pi t)\times 2\pi =0$
$\sin 2\pi t=0$ or $1+2\cos 2\pi t=0$
So, if $1+2\cos 2\pi t=0$
$\Rightarrow 2\pi t=2\pi n\pm \frac{2\pi }{3}$
with $n=0,1,2,\dots$
$t=\frac{1}{3},\frac{2}{3},\frac{4}{3},\frac{5}{3},\dots$ and for these value of $t$
$\cos 2\pi t=-\left(\frac{1}{2}\right),I=0$,
ie, $I$ is minimum and if $\sin 2\pi t=0$
$2\pi t=n\pi ,n=0,1,2,..$
$\Rightarrow t=0,\frac{1}{2},1,\frac{3}{2},2\dots$
$I$ is therefore $9a^2,a^2,9a^2,a^2$
ie, intensity is maximum (with two different values) ie, number of beats per sec is two.