Question:

A particle is moving in a straight line with simple harmonic motion of amplitude ‘A’. At a distance B from the mean position, the particle receives a blow in the direction of motion which instantaneously doubles the velocity. The new amplitude A' will be,

  • \(\sqrt{4A^2-3B^2}\)
  • 2A
  • \(\sqrt{3A^2-4B^2}\)
  • A
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The Correct Option is A

Solution and Explanation

The correct option is (A): \(\sqrt{4A^2-3B^2}\)
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