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MHT CET 2026 (20 Apr, Shift 2) · Energy in SHM

Energy in SHM question from MHT CET 2026 (20 Apr, Shift 2)

Q3MHT CET 2026 · 20 APR · SHIFT 2Easy

Energy in SHM

For a particle executing S.H.M., the potential energy is nn times the kinetic energy when its displacement from mean position is (223)A\left(\frac{2\sqrt{2}}{3}\right)A, where AA is the amplitude of S.H.M. The value of nn is

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Answer: (d)

  1. PEKE=x2A2−x2\dfrac{PE}{KE} = \dfrac{x^2}{A^2 - x^2}
  2. x2=89A2⇒A2−x2=19A2x^2 = \dfrac{8}{9}A^2 \Rightarrow A^2 - x^2 = \dfrac{1}{9}A^2
  3. n=8/91/9=8n = \dfrac{8/9}{1/9} = 8
  4. So the answer is (d) 8.
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