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The kinetic energy of a particle executing simple harmonic motion oscillates with a frequency of 50 Hz . The time period of the particle's displacement is:

Options

  1. A0.02 s
  2. B0.01 s
  3. C0.04 s
  4. D0.08 s

Correct answer

C. 0.04 s

Step-by-step solution

Given, the frequency of kinetic energy, f_ KE = 50 Hz . In simple harmonic motion, the kinetic energy becomes maximum twice in one complete cycle of displacement (once when passing through the mean position in the positive direction, and once in the negative direction). Therefore, the frequency of kinetic energy is twice the frequency of the simple harmonic motion ( f_ SHM ). f_ KE = 2f_ SHM f_ SHM = f_ KE 2 = 50 2 = 25 Hz The time period ( T ) of the particle's displacement is the reciprocal of its frequency: T =

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