A particle executes simple harmonic motion along the x-axis with amplitude A and angular frequency ω. At an instant its displacement from the mean position is +A/2 and it is moving towards the positive extreme. Which option correctly gives the magnitude of its velocity and its acceleration at that instant?
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Question

A particle executes simple harmonic motion along the x-axis with amplitude A and angular frequency ω. At an instant its displacement from the mean position is +A/2 and it is moving towards the positive extreme. Which option correctly gives the magnitude of its velocity and its acceleration at that instant?

Options

G1 Physics · Accepted Answer

For SHM, v² = ω²(A²−x²). At x = A/2, v = ω√(A²−A²/4) = √3Aω/2. Acceleration is a = −ω²x = −Aω²/2. The negative sign shows that acceleration is directed towards the mean position. Correct option: A.