First off I wasn't sure if I am just stating that I am posting this question here because it appears in my differential equations course. (Although it seems rather like a physics question).
Question
After a mass m is attached to a spring is stretches s units and then hangs at rest in the equilibrium position. The spring is then set in motion. Let x(t) denote the directed distance of the mass beyond equilibrium position. If x(t) > 0 then the spring is stretched, if x(t) < 0 then the string is compressed. Assume that the only forces acting on the system are the weight of the mass and the restoring force of the stretched spring. Determine a differential equation for the displacement x(t) at time t.
Attempt
I drew a free body diagram and got $F_{net} = ma = F_{weight} + F_{spring}$
From the question we have $F_{weight} = +mg, F_{spring} = -kx(t), k>0$
So obviously we should have $ma = -kx(t) + mg \equiv \frac{d^2}{dt^2}x(t) = -\frac{k}{m}x(t)+g$
But I am confused, because the answer is $\frac{d^2}{dt^2}x(t) = -\frac{k}{m}x(t)$
What happened to gravity? It is almost as if they only considered the spring force, and not weight at all!
Best Answer
At equilibrium, the spring’s restorative force cancels that of gravity, so you have $mg=-ks$. Then $\ddot x=-(k/m)s-(k/m)x=-(k/m)(x+s)$. However, we want displacement from the (new) equilibrium position, so we move the origin to $s$, which doesn’t change the $\ddot x$ term. The spring behaves the same way as it would without the constant gravitational force (as long as the displacement stays in the region in which Hooke’s Law holds, of course), but with a new equilibrium point.