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)/(2pisqrt(3))`
B. `2pisqrt(3)`
C. `(2pi)/sqrt(3)`
D. `sqrt(3)/(2pi)`

Answer

C

Explanation

Given: `omegasqrt(A^(2)-x^(2))=omega^(2)x`
`therefore sqrt(A^(2)-x^(2))=omegax=(2pi)/(T)x`
or `T=(2pix)/(sqrt(A^(2)-x^(2)))=(2pixx1)/(sqrt(2^(2)-1^(2)))=(2pi)/(sqrt(3))s`