A particle executes `SHM` along a straight line `4cm` long. When the displacement is `1cm` its velocity and acceleration are numerically equal. The time period of `SHM` is

Options

A. `2 pi s`
B. `(2pi)/(sqrt(3))s`
C. `(2pi)/(sqrt(5))s`
D. `(2pi)/(sqrt(7))s`

Answer

B

Explanation

2A = 4 cm `rArr` A = 2cm
v = a
`omega sqrt(A^(2) - x^(2)) = omega^(2)x`
`2^(2) - 1^(2) = omega^(2) (1)^(2)`
`T = (2pi)/(sqrt(3))`