A pendulum is hung the roof of a sufficiently high huilding and is moving freely to and fro like a simple harmonic oscillator .The acceleration of the bob of the pendulum is `20m//s^(2)` at a distance of `5m` from the meanposition .The time period of oscillation is

Options

A. `1s`
B. `2pi s`
C. `2s`
D. `pi s`

Answer

D

Explanation

in case of `SHM vec a= - omega^(2) vecy`
Given `a = 20m//s^(2)` at `y = 5 cm`
Hence `vec|a| = omega^(2)vec|y|`
`20 = omega^(2) xx 5`
or `omega^(2)= 4 rArr omega = 2 rad//sec`
But `omega = (2pi)/(T)`
`2 = (2pi)/(T) otr T = pi` sec