Explanation

(c) Let $n - 1( = 400),n( = 401)$ and $n + 1( = 402)$ be the frequencies of the three waves. If a be the amplitude of each then $y_{1} = a\sin{2\pi(n - 1)t,}\ y_{2} = a\sin{2\pi nt}$ and $y_{3} = a\sin{2\pi(n + 1)t}$
Resultant displacement due to all three waves is
$y = y_{1} + y_{2} + y_{3}$
$= a\sin{2\pi nt + a\lbrack\sin{2\pi(n - 1)t + \sin{2\pi(n + 1)t}}\rbrack}$
$= a\sin{2\pi nt + a\lbrack 2\sin{2\pi nt\cos{2\pi t}}\rbrack}$
$= a\left\lbrack 2\cos{2\pi t + 1} \right\rbrack\sin{2\pi nt}$
$= a^{'}\sin{2\pi nt\ with\ a^{'} = a\lbrack 1 + 2\cos{2\pi t}\rbrack}$
So, $I \propto \left( a^{'} \right)^{2} \propto a^{2}\left( 1 + 2\cos{2\ \pi t} \right)^{2}$
For $I$ to be max or min
$\frac{dI}{dt} = 0 \Rightarrow \frac{d}{dt}\left( 1 + 2\cos{2\ \pi t} \right)^{2} = 0$
$i.e.,\ 2\left( 1 + 2\cos{\pi t} \right)\left( 2\sin{2\ \pi t} \right) \times 2\pi = 0$
$\sin{2\pi t = 0\ or\ 1 + 2\cos{2\pi t = 0}}$
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 $t$
$\cos{2\pi t = - \left( \frac{1}{2} \right),\ I = 0,\ i.e.,}\ I$ is min and if $\sin{2\pi t = 0}$
$2\pi t = n\pi,\ n = 0,1,2,\ \ldots \Rightarrow t = 0,\frac{1}{2},1,\frac{3}{2},2\ldots$
$(I)$ from equation (i)
$9a^{2},a^{2},9a^{2},a^{2}$
$i.e.,$ Intensity is max. (with two different values)
$i.e.,$ number of beats per sec is two