The acceleration of a particle executing SHM is `0.10 ms^(-2)` at a distance of `8xx10^(-2)m` from the mean position. Calculate its time period.

Explanation

`a=0.10 ms^(-2)u=8xx10^(-2)m`
We know that `a=-omega^(2)y` i.e.
`0.1=-omega^(2)(0.08)`
`omega^(2)=|(-0.1)/(0.08)|=1.25" i.e. "((2pi)/(T))^(2)=1.25`
`or (2pi)/(T)=sqrt(1.25)=1.118" or "T=(2xx3.14)/(1.118)`
`T=5.617s.`