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MHT CET 2025 (23 Apr, Shift 2) · Energy in SHM

Energy in SHM question from MHT CET 2025 (23 Apr, Shift 2)

Q19MHT CET 2025 · 23 APR · SHIFT 2Easy

Energy in SHM

A particle is executing S.H.M. of amplitude ' AA '. When the potential energy of the particle is half of its maximum value during the oscillation, its displacement from the equilibrium position is

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Answer: (d)

  1. 12kx2=12⋅12kA2\dfrac{1}{2}kx^2 = \dfrac{1}{2}\cdot\dfrac{1}{2}kA^2
  2. x2=A22⇒x=±A2x^2 = \dfrac{A^2}{2} \Rightarrow x = \pm\dfrac{A}{\sqrt{2}}
  3. So the answer is (d).
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