Question

A mass of 2 kg is attached to the spring constant `50Nm^(-1)`. The block is pulled to a distance of 5cm from its equilibrium position at `x=0` on a horizontal frictionless surface from rest at `t=0`. Write the expression for its displacement at anytime t.

Answer 1

Here, `m=2kg, k=50N//m, r=5cm=5xx10^(-2)m`
`omega=sqrt((k)/(m))=sqrt((50)/(2))=5rad//s`
Using , `x=rsinomegat, ` we have `x=[5xx10^(-2)sin5t]m`