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A particle of mass 0.2 kg performs S.H.M. of amplitude 0.05 m with period π s. Its maximum kinetic energy (in mJ) is approximately
A5
B25
C1
D10
Answer & Solution
Correct answer: D. 10
1. ω = 2π/T = 2 rad/s.
2. Maximum speed v_max = ωA = 2 × 0.05 = 0.1 m/s.
3. KE_max = ½ m v_max² = ½ × 0.2 × 0.01 = 0.001 J.
4. Converted to mJ this is 1 mJ; the closest option is 1.
_Source: Maharashtra Balbharati Std XII Physics, Ch 5 "Oscillations" §5.6_
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