Two particles execute SHM with x₁ = A sinωt and x₂ = A sin(ωt+π/3). What is the first positive time at which their velocities are exactly equal?
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Question

Two particles execute SHM with x₁ = A sinωt and x₂ = A sin(ωt+π/3). What is the first positive time at which their velocities are exactly equal?

G1 Physics · Accepted Answer

Velocities are v₁ = Aω cosωt, v₂ = Aω cos(ωt+π/3). For equal velocities, cosωt = cos(ωt+π/3). The non-trivial condition gives ωt = −(ωt+π/3)+2π. Hence, 2ωt = 5π/3, so ωt = 5π/6. Therefore, t = (5π/6)/(2π/T) = 5T/12. Correct option: D.