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 second is:

Options

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

Answer

C

Explanation

Velocity = accelerations
` " " omega sqrt(a^(2) - y^(2) ) = omega^(2) y `
` sqrt( (2)^(2) -(1)^(2)) = omega (1)`
` rArr " " omega = sqrt3`
` " " T= (2pi )/(omega)`
` rArr " " T = ( 2pi )/(sqrt3)`