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.

Options

A. (a)`(sqrt(5))/(2)A`
B. (b)`(sqrt(3))/(2)A`
C. (c)`sqrt(2)A`
D. (d)`sqrt(5)A`

Answer

C

Explanation

at `x = (sqrt(3))/(2)A`
`KE = 1/2 momega^(2) (A^(2) - (3)/(4)A^(2)) = 1/8 momega^(2)A^(2)`
`KE` is increased by an amount of `1/2momega^(2)A^(2)`. Let now amplitude be `A_(1)` then total `KE`
`KE_(1) = 1/8momega^(2)A^(2) + 1/2 momega^(2)A^(2)`
`= 5/8momega^(2)A^(2) = 1/2momega^(2)(A_(1)^(2) - (3)/(4)A^(2)) rArr A_(1) = sqrt(2)A`