A particle performs SHM of amplitude A along a straight line. When it is at a distance 3/2 A from mean position, its kinetic energy gets increased by an amount 1/2mω2A2due to an impulsive force. Then its new amplitude becomes
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a
52A
b
32A
c
2A
d
5A
answer is C.
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Detailed Solution
Due to impulse, the total energy of the particle becomes12mω2A2+12mω2A2=mω2A2Let A' be the new amplitude,12mω2A′2=mω2A2⇒A′=2A