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

Options

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

Answer

A

Explanation

Let new amplitude is A'
Initial velocity
`v^(2) = omega^(2)(A^(2) - ((2A)/(3))^(2))"………..'(1)`
Where A is initial amplitude & `omega` is angular frequnecy.
Final velocity
`(3v)^(2) = omega^(2) (A^('2)- ((2A)/(3))^(2)) "…….."(2)`
From equation & equation `(2)`
`1/9 = (A^(2) - (4A^(2))/(9))/(A^('2)- (4A^(2))/(9))`
`A' = (7A)/(3)`