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A particle is executing simple harmonic motion with a frequency f . Match the physical quantities in List-I with their frequencies of variation in List-II. List-I List-II (A) Velocity (I) 0 (B) Kinetic energy (II) f/2 (C) Potential energy (III) f (D) Total energy (IV) 2f Choose the correct answer from the options given below:

Options

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

Correct answer

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

Step-by-step solution

For a particle executing SHM with frequency f , its displacement, velocity, and acceleration all vary sinusoidally with the same frequency f . Thus, the frequency of velocity is f . Kinetic energy and potential energy involve the square of velocity and displacement, respectively. Because of the trigonometric identity ^2( ) = 1 - (2 ) 2 , squaring the sinusoidal function doubles its frequency. Therefore, the frequency of both kinetic and potential energy is 2f . The total energy of a particle in SHM remains constant

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