The maximum velocity for a particle in S.H.M. is 0.16 m/s and the maximum acceleration is `0.64 m//s^(2)`. The amplitude of S.H.M. is

Options

A. `4xx10^(-2) m`
B. `4xx10^(-1) m`
C. `4xx10^(2) m`
D. `4xx10^(-3)` m

Answer

A

Explanation

`("Max. acceleration")/("Max. velocity")=(omega^(2)A)/(omegaA)=omega`
`therefore omega=(0.64)/(0.16)=4`
`therefore " Max. velocity " = omega A`
`therefore A= (4)/(100)=4xx10^(-2) m`