Question

y1 = 2sin(3x – 2t) & y2 = 3sin(3x – 2t + π/4). Find the expression for resultant displacement caused by superposition of the two waves.

(a) 3.6sin(2x-3t+3π/7)

(b) 4.63sin(3x-2t+3π/5)

(c) 3.6cos(3x-2t+3π/7)

(d) 4.63cos(3x-2t+3π/5)

I have been asked this question during an interview for a job.

I would like to ask this question from Principle of Superposition of Waves in portion Waves of Physics – Class 11

Answer 1

The correct answer is (b) 4.63sin(3x-2t+3π/5)

Explanation: y = y1 + y2= 2sin(3x-2t) + 3sin(3x-2t+π/4)

= Asin(3x-2t+φ).

A^2 = A1^2 + A2^2 + 2A1A2cosθ,                                                                                                                                           where θ = π/4, A1 = 2 & A2 = 3.

∴ A = sqrt(4+9+12/√2) = 4.63m.

Tanφ = A2sinθ / (A1 + A2cosθ)

= (3/√2)/(2+3/√2) = 0.51.                                                                                                                                                             ∴ φ = 27rad.

∴ y = 4.63sin(3x-2t+8.6π)                                                                                                                                                        = 4.63sin(3x-2t+3π/5).