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MHT CET 2025 (22 Apr, Shift 1) · Energy in SHM

Energy in SHM question from MHT CET 2025 (22 Apr, Shift 1)

Q23MHT CET 2025 · 22 APR · SHIFT 1Medium

Energy in SHM

For a particle performing S.H.M.; the total energy is ' nn ' times the kinetic energy, when the displacement of a particle from mean position is 32A\frac{\sqrt{3}}{2}A, where A is the amplitude of S.H.M. The value of ' nn ' is

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Answer: (c)

  1. KEE=A2−x2A2\frac{KE}{E} = \frac{A^2 - x^2}{A^2}
  2. At x=32Ax = \frac{\sqrt{3}}{2}A: KEE=1−34=14\frac{KE}{E} = 1 - \frac{3}{4} = \frac{1}{4}
  3. E=4 KE⇒n=4E = 4\,KE \Rightarrow n = 4
  4. So the answer is (c) 4.
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