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A particle performs SHM of amplitude A a...

A particle performs SHM of amplitude A along a straight line. When it is at distance `sqrt(3)/2` A from mean position, its kinetic energy gets increased by an amount `1/2momega^(2)A^(2)` due to an impulsive force. Then its new amplitude becomes.

A

`(sqrt5)/(2)A`

B

`(sqrt3)/(2)A`

C

`sqrt2A`

D

`sqrt5A`

Text Solution

Verified by Experts

The correct Answer is:
C

Due to impulse the total energy of the particle becomes
`(1)/(2)momega^2A^2+(1)/(2)momega^2A^2=momega^2A^2`
Let `A'`. Be the new amplitude
`(1)/(2)momega^2(A')^2=momega^2A^2impliesA'=sqrt2A`
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