A particle executes S.H.M. with an amplitude of 2 cm. When the particle is at 1 cm from the mean position the magnitude of tis velocity is equal to that of its acceleration. Then its time period in second is :

Options

A. `2pi sqrt(3)` s
B. `(2pi)/(3)sqrt(3)s`
C. `(sqrt(3))/(2pi)s`
D. `(1)/(2pi sqrt(3))s`.

Answer

B

Explanation

`omega sqrt(r^(2)-y^(2))=omega^(2)y`
`sqrt(r^(2)-y^(2))=omega y`
`r^(2)-y^(2)=omega^(2)y^(2)`
`implies" "4-1=omega^(2)" "implies" "omega=sqrt(3)` rad/s.
`:." "(2pi)/(T)=sqrt(3)implies" "T=(2pi)/(sqrt(3))=(2pi)/(3)sqrt(3)` s.
So correct choice is (b).