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A particle executes S.H.M. with amplitude A and angular frequency ω. Its maximum acceleration is
Aω²A
BωA
CωA²
DA/ω²
Answer & Solution
Correct answer: A. ω²A
1. For S.H.M. x = A sin(ωt), acceleration a = −ω² x.
2. Magnitude is maximum when |x| is maximum, i.e. at x = ±A.
3. So a_max = ω²A.
_Source: Maharashtra Balbharati Std XII Physics, Ch 5 "Oscillations" §5.3_
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