A partical executes SHM on a straigh line path. The amplitude of oscillation is 2cm. When the displacement of the particle from the mean position is 1cm, the magnitude of its acceleration is equal to that of its velocity. Find the time period of SHM, also the ms. velocity and ms. acceleration of SHM.

Options

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

Answer

C

Explanation

Here,`r=2cm,`As per question
`omegasqrt(r^(2)-y^(2))=omega^(2)y or omegasqrt(2^(2)-1^(2))=omega^(2)xx1`
`omegasqrt(3)=omega^(2) or omega=sqrt(3)`
`(2pi)/(T)=sqrt(3) or T=(2pi)/(sqrt(3))`