If a mass of `2.0 kg` is attached to one end of a spring and the spring has a spring constant `1500Nm^(-1)`, then calculate the maximum velocity and acceleration of the mass body for an amplitude `0.05 m`.

Explanation

Given, `m=2kg, k = 1500 Nm^(-1), A=0.05m`
`v_("max")=omegaA=A sqrt((k)/(m))" where k"=momega^(2)=0.05sqrt((1500)/(2))`
`v_("max")=0.05xx10sqrt(7.5)="1.369 ms"^(-1)`.
`a_("max")=omega^(2)A =v omega =1.369xxsqrt((k)/(m))=1.369 xxsqrt((1500)/(2))="37.49 ms"^(-2)`