A particle performs Simple harmonic motion of amplitude A along a straight line. When it is at a distance 31/2/2 A from mean position its kinetic energy gets increased by an amount 1/2 mw2A​2 due to an impulsive force. Then it's new amplitude becomes what?

Dear Student,

Please find below the solution to the asked query:

If the particle performs SHM with amplitude A along the straight line, than at any point the total energy should be,

E=K+P=12KA2=12mω2A2.
If the additional kinetic energy of magnitude 12mω2A2 by means of an impulsive force, then the total energy of the particle changes to,

E'=E+12mω2A2=12mω2A2+12mω2A2 12mω2A'2=12mω22A2 A'2=2A2  A'=2A

 

 

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