Question

A block of mass one kg is fastened to a spring with a spring constant `50Nm^(-1)`. The block is pulled to a distance `x=10cm` from its equilibrium position at `x=0` on a frictionless surface from rest at `t=0. ` Write the expression for its x(t) and v(t).

Answer 1

Here, `m=1kg,k-50Nm^(-1)`
`a=10cm=0.10m`
`omega=sqrt((k)/(m))=sqrt((50)/(1))=7.07rad//s`
Since the motion starts from th emean position, so the displacement equation can be given as
`x(t)=asinomegat=-.10sin7.07tm`
and velocity, `v(t)=dx//dt=aomega cosomegat`
`=0.10xx7.07 cos 7.07t `
`=0.707cos7.07tms^(-1)`