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Question 31133
JEE Physics · Hard

Simple Harmonic Motion: Energy MCQ

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

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EnergyHardQuestion 31133

Question

Spring of constant k is stretched by x₀ and released. When stretched by x₀/2 during oscillation, KE is:
  1. A
    ¾ of initial PE
    Correct
  2. B
    ¼ of initial PE
  3. C
    ½ of initial PE
  4. D
    Equal to initial PE

Correct answer

¾ of initial PE

Explanation

Total E = ½kx₀² (initial PE). At x = x₀/2: PE = ½k(x₀/2)² = ⅛kx₀² = E/4. KE = E − E/4 = 3E/4.

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