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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 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

A

`(sqrt(5))/(2)A`

B

`(sqrt(3))/(2)A`

C

`sqrt(2)A`

D

`sqrt(5)A`

Text Solution

Verified by Experts

The correct Answer is:
C
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. `
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