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

Options

A. `(sqrt5)/(2)A`
B. `(sqrt3)/(2)A`
C. `sqrt2A`
D. `sqrt5A`

Answer

C

Explanation

Due to impulse force, 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. (apply energy conservation law)
`(1)/(2)m_(omega)^2(A')=m_(omega)^2A^2`
`A'=sqrt2A`.