Two waves are y₁ = 4 sin(kx-ωt) and y₂ = 2 sin(kx-ωt + 2π/3). The amplitude and phase of the resultant wave, relative to y₁, are:
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Question

Two waves are y₁ = 4 sin(kx-ωt) and y₂ = 2 sin(kx-ωt + 2π/3). The amplitude and phase of the resultant wave, relative to y₁, are:

G1 Physics · Accepted Answer

Let the phase of y₁ be zero. The two phasor amplitudes are 4 and 2 with phase difference 120°. Resultant amplitude: A = √[4² + 2² + 2(4)(2)cos120°] = √[16+4-8] = √12 = 2√3. For the resultant phase φ: tanφ = [2sin120°]/[4 + 2cos120°] = [2(√3/2)]/[4-1] = √3/3. Thus, φ = π/6. Therefore the resultant is characterized by A = 2√3, φ = π/6. Correct option: D.