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MHT CET 2026 (16 Apr, Shift 2) · Energy in SHM

Energy in SHM question from MHT CET 2026 (16 Apr, Shift 2)

Q6MHT CET 2026 · 16 APR · SHIFT 2Easy

Energy in SHM

The maximum potential energy of a block executing S.H.M. is E and amplitude of oscillation is 'A'. The kinetic energy of the block at amplitude A/2 is

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Answer: (a)

  1. Maximum PE = total energy E=12kA2E = \dfrac{1}{2}kA^2.
  2. At x=A/2x = A/2: PE=12kA24=E4PE = \dfrac{1}{2}k\dfrac{A^2}{4} = \dfrac{E}{4}
  3. KE=E−E4=3E4KE = E - \dfrac{E}{4} = \dfrac{3E}{4}
  4. So the answer is (a) 3E4\dfrac{3E}{4}.
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