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 $y_{1}=a \sin 2 \pi(v-1) t, y_{2}=a \sin 2 \pi v t$
and $y_{3}=a \sin 2 \pi(v+1) t$
Resultant displacement due to all three waves is
$y=y_{1}+y_{2}+y_{3}$
$=a \sin 2 \pi v t+ a[\sin 2 \pi(v-1) t$
$+\sin 2 \pi(v+1) t]$
$=a \sin 2 \pi v t+a[2 \sin 2 \pi v t \cos 2 \pi t]$
$=a[2 \cos 2 \pi t+1] \sin 2 \pi v t$
$=a' \sin 2 \pi$ 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.
$\frac{d I}{d t}=0 \Rightarrow \frac{d}{d t}(1+2 \cos 2 \pi t)^{2}=0$
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 $n=0,1,2, \ldots$
$t=\frac{1}{3}, \frac{2}{3}, \frac{4}{3}, \frac{5}{3}, \ldots$ and for these value of $t$
$\cos 2 \pi t=-\left(\frac{1}{2}\right), I=0$,
ie, $I$ is minimum and if $\sin 2 \pi t=0$
$2 \pi t=n \pi, n=0,1,2, . .$
$\Rightarrow t=0, \frac{1}{2}, 1, \frac{3}{2}, 2 \ldots$
$I$ is therefore $9 a^{2}, a^{2}, 9 a^{2}, a^{2}$
ie, intensity is maximum (with two different values) ie, number of beats per sec is two.