A particle executies linear simple harmonic motion with an amplitude `3cm` .When the particle is at `2cm` from the mean position , the magnitude of its velocity is equal to that of acceleration .The its time period in seconds is

Options

A. `(sqrt5)/(2 pi)`
B. `(4 pi)/(sqrt(5)`
C. `(2 pi)/(sqrt3)`
D. `(sqrt5)/(pi)`

Explanation

`v = omega sqrt((A^(2) - x^(2)))`
`omega^(2) x = omega sqrt(A^(2) - x^(2))`
`omega = (sqrt(A^(2) - x^(2)))/(x) = (sqrt((3)^(2) - (2)^(2)))/(2) = (sqrt(5))/(2)`
`(2 pi)/(T) = (sqrt5)/(2)`
`T = (4 pi)/(sqrt5)`