Two particles P₁ and P₂ execute SHM about the same mean position with x₁ = A sinωt and x₂ = A sin(ωt+π/3). What is the first positive time at which the two particles occupy the same position?
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Question

Two particles P₁ and P₂ execute SHM about the same mean position with x₁ = A sinωt and x₂ = A sin(ωt+π/3). What is the first positive time at which the two particles occupy the same position?

Options

G1 Physics · Accepted Answer

For equal positions, sinωt = sin(ωt+π/3). Using the non-trivial sine equality, ωt = π−(ωt+π/3). Therefore, 2ωt = 2π/3, so ωt = π/3. Hence, t = (π/3)/(2π/T) = T/6. Correct option: B.