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JEE Main 2021 (27 Aug, Shift 1) · Energy in SHM

Energy in SHM question from JEE Main 2021 (27 Aug, Shift 1)

Q59JEE Main 2021 · 27 AUG · SHIFT 1Easy

Energy in SHM

The variation of displacement with time of a particle executing free simple harmonic motion is shown in the figure.

Figure

The potential energy U(x) versus time (t) plot of the particle is correctly shown in figure:

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Answer: (d)

  1. From the graph x=Asin⁡ωtx = A\sin\omega t, zero at O, A, B, C.
  2. U=12kx2=12kA2sin⁡2ωtU = \frac{1}{2}kx^2 = \frac{1}{2}kA^2\sin^2\omega t, never negative.
  3. U is zero at O, A, B, C and maximum midway between them (humps of period T/2T/2).
  4. So the answer is (d).
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