A particle executing SHM has time period T. Time taken to go from extreme to mid-amplitude position:
AT/8
BT/6
CT/12
DT/4
Answer & Solution
Correct answer: C. T/12
From x = A cos(ωt) at extreme: at x = A/2, A cos(ωt) = A/2, so cos(ωt) = 1/2, ωt = π/3. Time t = T/6 from extreme to A/2... wait. Let me redo: extreme to mid-amplitude (A to A/2). x = A cos(ωt). x = A/2 → cos(ωt) = 1/2 → ωt = π/3 → t = T/(2π) × π/3 = T/6. So T/6. Let me re-check options. The answer should be T/6, option B. Setting B.
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