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ExamsJEE MainPhysics

A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance 2A/3 from equilibrium position. The new amplitude of the motion is:

  1. A/3 √41
  2. 3A
  3. A√3
  4. 7A/3

Correct answer: 7A/3

Solution

When the speed of the particle is increased while it is at a distance of 2A/3 from the equilibrium, the total mechanical energy of the system increases. Since the total energy in simple harmonic motion is proportional to the square of the amplitude, the new amplitude can be calculated using the relationship between the initial and final speeds and positions, leading to the result of 7A/3.

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