Two particles execute SHM about the same equilibrium position with equal amplitude A and time period T. At t = 0, one particle is at y₁ = +A and the other at y₂ = -A/2, and they are approaching each other. After what time do they cross each other?
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Question

Two particles execute SHM about the same equilibrium position with equal amplitude A and time period T. At t = 0, one particle is at y₁ = +A and the other at y₂ = -A/2, and they are approaching each other. After what time do they cross each other?

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G1 Physics · Accepted Answer

Take particle 1 as x₁=A cosωt. For particle 2, x₂(0)=−A/2 and it is approaching particle 1, which fixes its initial phase. Solving x₁=x₂ for the first positive time gives ωt=π/3. Hence t=T/6. Correct answer: D.

Two particles execute SHM about the same equilibrium position with equal amplitude A and time period T. At t = 0, one particle is at y₁ = +A and the other at y₂ = −A/2, and they are approaching each other. After what time do they cross each other?