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. `(sqrt(5))/(2)A`
B. `(sqrt(3))/(2)A`
C. `sqrt(2)A`
D. `sqrt(5)A`

Answer

C

Explanation

Due to impulse, the total energy of the particle becomes `:`
`(1)/(2)momega^(2)A^(2)+(1)/(2)m omega^(2)A^(2)=m omega^(2)A^(2)`
Let `,A'` be the new amplitude.
`:. (1)/(2) m omega^(2)(A')^(2)=m omega^(2)A^(2)`
`rArrA'=sqrt(2)A. Ans. `