A particle performing S.H.M. has an acceleration of `0.2 m//s^2` at a distance of 0.05 m from its mean position then what is it's time period ?

Explanation

`Ac c^n=omega^2x`
` 0.2=((2pi)/T)^2xx0.05`
T^2=(4pi^2xx0.05)/0.2`
`T^2=pi^2`
`T=pi sec.`