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

y1 = 2sin(3x – 2t) & y2 = 3sin(3x – 2t + π/4). Find the expressio

1.	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» 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).

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