The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves, y₁(x, t) = 4sin (kx - ω t) and y₂(x, t) = 2sin (kx - ω t + 2π/3), are: (Take the angular frequency of initial waves same as ω) (a) [√(3), π/6] (b) [2√(3), π/6] (c) [6, 2π/3] (d) [6, π/3]

Answer: (b). A² = 4² + 2² + 2(4)(2)cos 2π/3 = 16 + 4 - 8 = 12 ⇒ A = 2√(3). tan δ = (2sin 120°)/(4 + 2cos 120°) = √(3)/3 = 1/√(3) ⇒ δ = π/6. So the answer is (b) [2√(3), π/6].

The amplitude and phase of a wave that is formed by the superposition… | JEE Main 2025 (8 Apr, Shift 2)

Q6JEE Main 2025 · 8 APR · SHIFT 2Difficult

Reflection, Superposition & Interference

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 $y_2(x, t) = 2\sin\left(kx - \omega t + \frac{2\pi}{3}\right)$ , are:

(Take the angular frequency of initial waves same as $\omega$ )

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Answer: (b)

$A^2 = 4^2 + 2^2 + 2(4)(2)\cos\frac{2\pi}{3} = 16 + 4 - 8 = 12 \Rightarrow A = 2\sqrt{3}$ .
$\tan\delta = \frac{2\sin 120^\circ}{4 + 2\cos 120^\circ} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} \Rightarrow \delta = \frac{\pi}{6}$ .
So the answer is (b) $\left[2\sqrt{3}, \frac{\pi}{6}\right]$ .

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