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Match List-I with List-II for a particle executing simple harmonic motion with amplitude A . List-I (Energy condition) List-II (Displacement from mean position) (A) Kinetic energy is equal to potential energy (I) A 2 (B) Kinetic energy is three times the potential energy (II) A 2 (C) Potential energy is three times the kinetic energy (III) 3 A 2 (D) Kinetic energy is zero (IV) A Choose the correct answer from the opt

Options

  1. A(A) - (II), (B) - (I), (C) - (III), (D) - (IV)
  2. B(A) - (II), (B) - (III), (C) - (I), (D) - (IV)
  3. C(A) - (I), (B) - (II), (C) - (III), (D) - (IV)
  4. D(A) - (III), (B) - (I), (C) - (II), (D) - (IV)

Correct answer

A. (A) - (II), (B) - (I), (C) - (III), (D) - (IV)

Step-by-step solution

Let x be the displacement from the mean position. Potential energy, U = 1 2 kx^2 and Kinetic energy, K = 1 2 k(A^2 - x^2) (A) K = U A^2 - x^2 = x^2 2x^2 = A^2 x = A 2 (B) K = 3U A^2 - x^2 = 3x^2 4x^2 = A^2 x = A 2 (C) U = 3K x^2 = 3(A^2 - x^2) 4x^2 = 3A^2 x = 3 A 2 (D) K = 0 A^2 - x^2 = 0 x = A Therefore, the correct match is (A) - (II), (B) - (I), (C) - (III), (D) - (IV). Answer: (A) - (II), (B) - (I), (C) - (III), (D) - (IV)

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