Q 11-14-146JEE MainJEE Main 2025 (8 Apr, Shift 2)Medium
The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves, $y_1(x, t) = 4\sin(kx - \omega t)$ and $y_2(x, t) = 2\sin\left(kx - \omega t + \dfrac{2\pi}{3}\right)$, are: (Take the angular frequency of the initial waves to be the same, $\omega$.)
Answer: (D) $\left[2\sqrt3,\ \dfrac\pi6\right]$
Add the amplitudes as phasors at $120^\circ$:
$$A = \sqrt{4^2 + 2^2 + 2(4)(2)\cos120^\circ} = \sqrt{16 + 4 - 8} = 2\sqrt3$$
$$\tan\phi = \frac{2\sin120^\circ}{4 + 2\cos120^\circ} = \frac{\sqrt3}{3} \Rightarrow \phi = \frac\pi6$$
Solution by Sreeraj P, M.Sc Physics