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Question 31181
JEE Physics · Hard

Simple Harmonic Motion: SHM Advanced MCQ

Solve this quality-checked JEE multiple-choice question, then review the correct answer and explanation.

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SHM AdvancedHardQuestion 31181

Question

A particle under SHM has displacement x = 4sin(2t + π/6) m. At t = 0, KE and PE are (m = 1 kg):
  1. A
    KE = 24 J, PE = 8 J
    Correct
  2. B
    KE = 12 J, PE = 4 J
  3. C
    KE = 16 J, PE = 0
  4. D
    KE = 8 J, PE = 8 J

Correct answer

KE = 24 J, PE = 8 J

Explanation

At t=0t=0, x=2x=2 m and v=4√3v=4√3 m/s. Hence PE=12mω2x2=8PE=\frac12mω^2x^2=8 J and KE=12mv2=24KE=\frac12mv^2=24 J.

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