A particle is performing SHM and at t = 3T/4 is at x = (√3 A)/2 and moving towards the origin. The equilibrium position is x = 0. What were its position and direction at t = 0?
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A particle is performing SHM and at t = 3T/4 is at x = (√3 A)/2 and moving towards the origin. The equilibrium position is x = 0. What were its position and direction at t = 0?

G1 Physics · Accepted Answer

At t=3T/4, x=(√3/2)A and v<0, so the phase is 2π/3. Going backward by 3T/4 subtracts 3π/2, giving phase 7π/6 at t=0. Thus x(0)=−A/2 and v(0)<0, i.e. the particle is moving away from the mean position. Correct answer: A.