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MHT CET 2020 (16 Oct, Shift 1) · Energy in SHM

Energy in SHM question from MHT CET 2020 (16 Oct, Shift 1)

Q65MHT CET 2020 · 16 OCT · SHIFT 1Easy

Energy in SHM

For a particle performing SHM when displacement is xx, the potential energy and restoring force acting on it is denoted by EE and FF, respectively. The relation between x,Ex, E and FF is

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Answer: (b)

  1. E=12kx2E = \dfrac{1}{2}kx^2 and restoring force F=−kxF = -kx.
  2. 2EF=kx2−kx=−x⇒2EF+x=0\dfrac{2E}{F} = \dfrac{kx^2}{-kx} = -x \Rightarrow \dfrac{2E}{F} + x = 0 (option c).
  3. If only the magnitude F=kxF = kx is used, 2EF−x=0\dfrac{2E}{F} - x = 0 (option b).
  4. The key takes magnitudes: (b).
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