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A block of mass m is attached to a massless spring of spring constant k and executes simple harmonic motion on a smooth horizontal surface. The time period of oscillation of the kinetic energy of the block is:

Options

  1. Am k
  2. B2 m k
  3. C4 m k
  4. D2 m k

Correct answer

A. m k

Step-by-step solution

The time period of simple harmonic motion for a spring-mass system is given by T = 2 m k . In one complete oscillation of the block, its velocity becomes zero twice (at the extreme positions) and maximum twice (at the mean position). Therefore, the kinetic energy completes two full cycles in one time period of the SHM. The time period of oscillation of the kinetic energy is half the time period of the SHM: T_ KE = T 2 = 1 2 ( 2 m k ) = m k . Answer: m k

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