If the maximum acceleration of a particle performing S.H.M. is numerically equal to twice the maximum velocity then the period will be

Options

A. 1.57 s
B. 3.142 s
C. 6.28 s
D. 2 s

Answer

B

Explanation

`a_(m)=2v_(m)`
`Aomega^(2)=2Aomega`
`omega=2`
`T=(2pi)/(omega)=(2pi)/(2)=pi=3.142 s`