The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves,…
(Take the angular frequency of initial waves same as $\omega$)
- $\left[6, \frac{2 \pi}{3}\right]$
- $\left[6, \frac{\pi}{3}\right]$
- $\left[\sqrt{3}, \frac{\pi}{6}\right]$
- $\left[2 \sqrt{3}, \frac{\pi}{6}\right]$
Solution

$\begin{aligned} & \begin{array}{l}\mathrm{A}=\sqrt{2^2+4^2+2 \times 2 \times 4 \times \cos 120^{\circ}} \\ \quad=\sqrt{12}=2 \sqrt{3} \\ \tan \phi=\frac{2 \sin 120^{\circ}}{4+2 \cos 120^{\circ}}=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}} \\ \phi=\frac{\pi}{6}\end{array}\end{aligned}$
Asked in: JEE Main 2025 (08 Apr Shift 2)