Q 11-14-143JEE MainJEE Main 2025 (7 Apr, Shift 1)Medium
Two harmonic waves moving in the same direction superimpose to form a wave $x = a\cos(1.5t)\cos(50.5t)$, where $t$ is in seconds. Find the period with which they beat (close to the nearest integer).
Answer: (D) $2\ \text{s}$
Using $2\cos A\cos B = \cos(A + B) + \cos(A - B)$:
$$x = \frac a2\cos(52t) + \frac a2\cos(49t)$$
The two component waves have $\omega_1 = 52\ \text{rad/s}$ and $\omega_2 = 49\ \text{rad/s}$. The beat frequency is
$$f_{\text{beat}} = \frac{\omega_1 - \omega_2}{2\pi} = \frac{3}{2\pi}\ \text{Hz},\qquad T_{\text{beat}} = \frac{2\pi}{3} \approx 2.09\ \text{s} \approx 2\ \text{s}$$
Solution by Sreeraj P, M.Sc Physics