There are three sources of sound of equal intensity with frequencies 400, 401 and 402 vib/sec. The number of beats heard per second is

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 $\mathrm{y}=\mathrm{asin}2\mathrm{\pi }(\mathrm{n}-1)\mathrm{t},$ $\mathrm{y}=\mathrm{asin}2\mathrm{\pi nt}$ and ${\mathrm{y}}_3=\mathrm{asin}2\mathrm{\pi }(\mathrm{n}+1)\mathrm{t}$
Resultant displacement due to all three waves is $\mathrm{y}={\mathrm{y}}_1+{\mathrm{y}}_2+{\mathrm{y}}_3$
$=\mathrm{asin}2\mathrm{\pi nt}+\mathrm{a}\left[\sin 2\mathrm{\pi }\right(\mathrm{n}-1)\mathrm{t}+\sin 2\mathrm{\pi }(\mathrm{n}+1\left)\mathrm{t}\right]$
$=\mathrm{asin}2\mathrm{\pi nt}+\mathrm{a}\left[2\sin 2\mathrm{\pi ntcos}2\mathrm{\pi t}\right]$
$\left[\text{Using }\mathrm{sinC}+\mathrm{sinD}=2\sin \frac{\mathrm{C}+\mathrm{D}}{2}\cos \frac{\mathrm{C}-\mathrm{D}}{2}\right]$
$\text{⇒y=a\hspace{0.17em}(1+2cos2πt)sin2πnt}$
This is the resultant wave having amplitude $=(1+2\cos 2\mathrm{\pi t})$
For maximum amplitude cos 2$\mathrm{\pi }$t = 1 $\Rightarrow$ 2$\mathrm{\pi }$t = 2m$\mathrm{\pi }$ where m = 0, 1, 2, 3, ...
$\Rightarrow$ t = 0, 1, 2, 3 ...
Hence time interval between two successive maximum is 1 sec. So beat frequency = 1
Also for minimum amplitude (1+2cos 2$\mathrm{\pi }$t) = 0 
$\Rightarrow \cos 2\mathrm{\pi t}=-\frac{1}{2}$  
$\Rightarrow 2\mathrm{\pi t}=2\mathrm{m\pi }+\frac{2\mathrm{\pi }}{3}\Rightarrow \mathrm{t}=\mathrm{m}+\frac{1}{3}$
$\Rightarrow \mathrm{t}=\frac{1}{3},\frac{4}{3},\frac{7}{3},\frac{10}{3},....$       (for m = 0, 1, 2, ..) 
Hence time interval between two successive minima is 1 sec so, number of beats per second = 1
Note : PET/PMT Aspirants can remember result only.