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The phase difference between two waves, represented by y₁=10⁻⁶ sin [100t+(x/(50))+0.5] m and y₂=10⁻⁶ cos [100t+(x/(50))] m, where x is expressed in metres and t is expressed in seconds, is approximately

Asked in NEET 2004 · Wave parameters from the equation

Answer: (1) 1.07 radians

Step-by-step solution

Given: y₁ a sine with an extra phase 0.5, y₂ a cosine with the same argument otherwise.

Idea: turn the cosine into a sine so that the two phases can be compared directly.

cos θ=sin (θ+π/2), so y₂=10⁻⁶ sin [100t+x/(50)+1.57].

φ₁=100t+x/(50)+0.5 and φ₂=100t+x/(50)+1.57.

Δφ=1.57-0.5.

Δφ=1.07 radians.

Why the other options are wrong

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