The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves, $\mathrm{y}_{1}(\mathrm{x}, \mathrm{t})=4 \sin (\mathrm{kx}-\omega \mathrm{t})$ and $y_{2}(x, t)=2 \sin \left(k x-\omega t+\frac{2 \pi}{3}\right)$, are: (Take the angular frequency of initial waves same as $\omega$)

(D) $\left[2 \sqrt{3}, \frac{\pi}{6}\right]$

(A) $\left[6, \frac{2 \pi}{3}\right]$

(B) $\left[6, \frac{\pi}{3}\right]$

(C) $\left[\sqrt{3}, \frac{\pi}{6}\right]$

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Waves & Sound

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 - ω t)

and

y_{2} (x, t) = 2 sin (k x - ω t + \frac{2 π}{3})

, are:
(Take the angular frequency of initial waves same as

ω

)

Answer: (D)

Adding two waves of amplitudes

4

and

2

with a phase difference

\frac{2 π}{3}

gives a resultant amplitude

A = 4^{2} + 2^{2} + 2 \cdot 4 \cdot 2 cos 12 0^{\circ} = 20 - 8 = 23

. Its phase relative to the first wave satisfies

tan ϕ = \frac{2 sin 12 0 ^{\circ}}{4 + 2 cos 12 0 ^{\circ}} = \frac{3}{3} = \frac{1}{3}

, so

ϕ = \frac{π}{6}

. The resultant wave has amplitude

23

and phase

\frac{π}{6}

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Single correct option