Dispalcement (x) and velocity (v) of a particle exectuting S.H.M are related as `2v^(2)=9-x^(2)`. The time period of particle is

Options

A. `4pi`
B. `2pi`
C. `4sqrt2pi`
D. `2sqrt2pi`

Answer

D

Explanation

`2v^(2) = 9 - x^(2)`
Differentiating with respect to time t, we have
`4v(dv)/(dt) = 0 - 2x(dx)/(dt)`
`rAr 4va = -2xv`
`rArr 4va = -2xv`
`rArr a = -((1)/(2))x`
For S.H.M., `a = -omega^(2)x`
`therefore omega = (1)/(sqr(2))`
`rArr (2pi)/(T) = (1)/(sqrt(2))`
`rArr T = 2sqrt(2)pi`