STANDARD Physics Simple Harmonic Motion MCQ Question
The x-t graph of a particle undergoing simple harmonic motion is as shown in the figure. The acceleration of the particle at t = 4/3 s is
√3/32 π² cm s⁻²
-π²/32 cm s⁻²
π²/32 cm s⁻²
√3/32 π² cm s⁻²
Correct Answer
Detailed Explanation
From the graph, the amplitude A = 1 cm and the period T = 8 s. At t = 4/3 s, the displacement x = 1 sin(2π/8 * 4/3) = √3/2 cm. In SHM, acceleration a = -ω²x, where ω = 2π/T. Calculating gives a = √3/32 π² cm s⁻².
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