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:

Options

A. `(A)/(3)sqrt(41)`
B. `3A`
C. `Asqrt(3)`
D. `(7A)/(3)`

Answer

D

Explanation

Applying conservation of energy
`(1)/(2) m omega^(2) ((2A)/(3))^(2) + 9 xx (1)/(2) m omega^(2) [A^(2) -((2A)/(3))^(2)]`
`= (1)/(2) m omega^(2) A_(1)^(2)`
`rArr A_(1) = (7A)/(3)`