The maximum speed of a particle performing a linear S.H.M. is 0.16 m/s and the maximum acceleration is 0.64 `m//s^(2)`. The period of S.H.M. is

Explanation

Given : `v_(max)=0.16m//s`, `a_(max)=0.64m//s^(2)`, `T=?`
We know that,
Maximum velocity, `v_(max)=Aomega`
Maximum acceleration `=Aomega^(2)`
`:.omegaA=0.16` .....`(i)`
`omega^(2)A=0.64`
`omegaAxxomega=0.64`
`0.16xxomega=0.64` [Using the value from equation `(i)`]
`omega=4`
Now, `omega=(2pi)/(T)`
`(2pi)/(T)=4`
`T=2pi//4`
`=pi//2`
`=1.57sec`.