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JEE Main 2024 (29 Jan, Shift 1) · Energy in SHM

Energy in SHM question from JEE Main 2024 (29 Jan, Shift 1)

Q28JEE Main 2024 · 29 JAN · SHIFT 1Easy

Energy in SHM

When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is x8\frac{x}{8}, where x=x = _________.

Numerical answer type. Work it out, then open the answer.

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Answer: 9

  1. EKE=A2A2−x2\frac{E}{KE} = \frac{A^2}{A^2 - x^2}
  2. At x=A/3x = A/3: EKE=A2A2−A2/9=98\frac{E}{KE} = \frac{A^2}{A^2 - A^2/9} = \frac{9}{8}
  3. x8=98⇒x=9\frac{x}{8} = \frac{9}{8} \Rightarrow x = 9
  4. So the answer is 9.
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