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

Explanation

Here `m=1kg, k= 50 Nm^(-1)`
`A=10 cm =0.10 m`
`omega= sqrt((k)/(mk))= sqrt((50)/(1))=707. "rad" s^(-1)`
As the motion starts from the mean position, So the displacement equation can be written as
`x(t)=A sin omegat`
(or) `x(t)=0.10 sin 7.07t`
and `v(t)=(dx)/(dt)=0.10xx7.07 cos 7.07 t`
(or) `v(t)=0.707 cos 7.07 t ms^(-1)`