Question

A spring has natural length 0.75 m and a 5 kg mass. A force of 8 N is needed to keep the spring stretched to a length of 0.85 m. If the spring is stretched to a length of 1 m and then released with velocity 0, find the position of the mass after t seconds.

Answer 1

Use Hooke's law to find the spring constant. If it takes 8N to stretch the spring by 0.1m, then

$8\text{ N}=k(0.1\text{ m})\implies k=80\dfrac{\text N}{\text m}$

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

$y''(t)+80y(t)=0$

where $y(t)$ 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

$y(t)=C_1\cos4\sqrt5t+C_2\sin4\sqrt5t$

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 $y(0)=\dfrac14$ and $y'(0)=0$.

$y(0)=\dfrac14\implies\dfrac14=C_1\cos0+C_2\sin0\implies C_1=\dfrac14$
$y'(0)=0\implies 0=-4\sqrt5C_1\sin0+4\sqrt5C_2\cos0\implies C_2=0$

So the spring's motion is dictated by the function

$y(t)=\dfrac14\cos4\sqrt5t$