A particle executes linear simple harmonic motion with an amplitude of 2 cm . When the particle is at 1 cm from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

Options

A. `(1)/(2pisqrt3)`
B. `2pisqrt3`
C. `(2pi)/(sqrt(3))`
D. `(sqrt3)/(2pi)`

Answer

C

Explanation

`v=omegasqrt(A^(2)-y^(2)) and a=omega^(2)y`
`omegasqrt(A^(2)-y^(2))=omega^(2)yimplies omegasqrt((2)^(2)-1^(2))=omega^(2).1`
`implies sqrt(3)=(2pi)/(T)implies T=(2pi)/(sqrt3)`.