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A particle performs simple harmonic miti...

A particle performs simple harmonic mition with amplitude A. Its speed is trebled at the instant that it is at a destance
`(2A)/3` from equilibrium position. The new amplitude of the motion is:

A

3A

B

`A sqrt(3)`

C

`(7 A)/(3)`

D

`(A)/(3) sqrt(41)`

Text Solution

Verified by Experts

The correct Answer is:
C

`v = omega sqrt(A^(2) - (( 2A)/(3))^(2))`
`v = sqrt(5) (A omega)/(3), v_("new") = 3v = sqrt(5) A omega `
So the new amplitude is given by
`v_("new") = omega sqrt(A_("new") ^(2)- x^(2)) rArr sqrt(5) A omega = omega sqrt(A_("new")^(2) - ((2A)/(3))^(2))`
`A_("new") = (7A)/(3)`
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