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 $y = a \sin 2 \pi (n - 1) t,$ $y = a \sin 2 \pi nt$ and $y_3 = a \sin 2 \pi (n + 1) t$

Resultant displacement due to all three waves is $y = y_{1} + y_{2} + y_{3}$

$= a \sin 2 \pi n t + a [\sin 2 \pi (n - 1) t + \sin 2 \pi (n + 1) t]$

$= a \sin 2 \pi n t + a [2 \sin 2 \pi n t \cos 2 \pi t]$

$\left[\text{Using } \sin C + \sin D = 2 \sin \frac{C + D}{2} \cos \frac{C - D}{2}\right]$

⇒ ⇒ $y = a(1 + \cos 2 \pi t) \sin 2 \pi n t$

This is the resultant wave having amplitude $= \left(1 + \cos 2 \pi t \right)$

For maximum amplitude cos 2 π π t = 1 ⇒ ⇒ 2 π π t = 2m π π where m = 0, 1, 2, 3, ...

⇒ ⇒ t = 0, 1, 2, 3 ...

Hence time interval between two successive maximum is 1 sec. So beat frequency = 1

Also for minimum amplitude (2cos 2 π π t) = 0

$\Rightarrow \cos 2 \pi t = - \frac{1}{2}$ $\Rightarrow 2 \pi t = 2 m \pi + \frac{2 \pi}{3} \Rightarrow t = + \frac{1}{3}$ $\Rightarrow t = \frac{1}{3}, \frac{4}{3}, \frac{7}{3}, \frac{10}{3}, \ldots \quad (\text{for } m = 0, 1, 2, \ldots)$

Hence time interval between two successive minima is 1 sec so, number of beats per second = 1