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 = 12 J, PE = 4 J
    Correct
  2. B
    KE = 12 J, PE = 16 J
  3. C
    KE = 16 J, PE = 0
  4. D
    KE = 8 J, PE = 8 J

Correct answer

KE = 12 J, PE = 4 J

Explanation

At t=0: x = 4sin(π/6) = 2 m. v = 4×2×cos(π/6) = 8×(√3/2) = 4√3 m/s. KE = ½×1×(4√3)² = 24 J. PE = ½×mω²x² = ½×4×4 = 8 J. Hmm: E = ½mω²A² = ½×1×4×16 = 32 J. PE = ½mω²x² = ½×1×4×4 = 8 J. KE = 32−8 = 24 J. Total 32 J = 24+8.