A particle performs S.H.M of amplitude A along a straight line. When it is at a distance √3A2 from mean position, its kinetic energy gets increased by an amount 12mω2A2 due to an impulsive force. Then its new amplitude becomes nA. Find n:
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Solution
Due to impulse, the total energy of the particle becomes : 12mω2A2+12mω2A2=mω2A2
Let A′ be the new amplitude. ∴12mω2(A′)2=mω2A2⇒A′=√2A.
Why this question?
Tip- Mechanical energy in S.H.M. is conserved only if no external force except the one responsible for S.H.M. acts along the line of motion, however the expression for it is 12mω2A2 only, where A adjusts as per the energy changes.