Use Hooke's law to find the spring constant. If it takes 8N to stretch the spring by 0.1m, then
[tex]8\text{ N}=k(0.1\text{ m})\implies k=80\dfrac{\text N}{\text m}[/tex]
I'm going to assume the spring is fixed to a ceiling, and that any stretching in the downward direction counts as movement in the positive direction. The spring's motion is then modeled by
[tex]y''(t)+80y(t)=0[/tex]
where [tex]y(t)[/tex] is the position of the spring's free end as it moves up and down. Solving this is easy enough: the characteristic solution will be
[tex]y(t)=C_1\cos4\sqrt5t+C_2\sin4\sqrt5t[/tex]
Given that the spring is stretched to a length of 1m (a difference of 0.25m from its natural length), and is released with no external pushing or pulling, we have the two initial conditions [tex]y(0)=\dfrac14[/tex] and [tex]y'(0)=0[/tex].
[tex]y(0)=\dfrac14\implies\dfrac14=C_1\cos0+C_2\sin0\implies C_1=\dfrac14[/tex]
[tex]y'(0)=0\implies 0=-4\sqrt5C_1\sin0+4\sqrt5C_2\cos0\implies C_2=0[/tex]
So the spring's motion is dictated by the function
[tex]y(t)=\dfrac14\cos4\sqrt5t[/tex]
