Two particles execute SHM on the same line with x₁ = A sinωt and x₂ = A cosωt. What is the first positive time at which their velocities are equal in both magnitude and direction?
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Question

Two particles execute SHM on the same line with x₁ = A sinωt and x₂ = A cosωt. What is the first positive time at which their velocities are equal in both magnitude and direction?

G1 Physics · Accepted Answer

Velocities are v₁ = Aω cosωt, v₂ = −Aω sinωt. For equal velocities, cosωt = −sinωt. Therefore, tanωt = −1. The first positive phase satisfying this is ωt = 3π/4. Hence, t = (3π/4)/(2π/T) = 3T/8. Correct option: C.