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JEE Main 2019 (11 Jan, Shift 2) · Simple Pendulum

Simple Pendulum question from JEE Main 2019 (11 Jan, Shift 2)

Q74JEE Main 2019 · 11 JAN · SHIFT 2Medium

Simple Pendulum

A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K2. Then:

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Answer: (b)

  1. Take amplitude as the angular amplitude θ0\theta_0 (same in both cases).
  2. Max KE = loss of PE from extreme to mean: K=mgL(1−cos⁡θ0)∝LK = mgL(1-\cos\theta_0) \propto L.
  3. Length doubled ⇒K2=2K1\Rightarrow K_2 = 2K_1.
  4. So the answer is (b) K2=2K1K_2 = 2K_1.
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