Explanation

(b) Given,
$y_{1} = 4\sin{500\pi t}\ldots..(i)$
$y_{2} = 2\sin{506\ \pi t}\ \ \ldots..(ii)$
Comparing Eqs. (i) and (ii)
Y=a sin ωt……………(iii)
We have,
$\omega_{1} = 500\pi\ $
$\Longrightarrow f_{1}\ \frac{500\pi}{2\pi} = 250beats/s$
And
$\omega_{2} = 506\pi$
$\Longrightarrow \ \ \ f_{22} = \frac{506\pi}{2\pi} = 253\ beats/s$
Thus, number of beats produced
$= f_{2} - f_{1} = 253 - 250$
=3 beats/s
=3×60 beats/min
=180 beats/min