Vertical spring

In our previous example, we had to rely on a frictionless surface to keep the mass from from slowing down. But that's hard to do in reality. It's easier just to hang the spring:

The problem is that this slightly complicates the situation, because even when the mass is hanging at rest, it is stretched due to gravity. We can calculate that stretch. In equilibrium the acceleration is zero so the net force is zero. So the two forces, the spring force, and force of gravity must balance each other

$$-ky_e - mg ~=~ 0$$ (1.40)

or in equilibrium, $y_e ~=~ -mg/k$.

So with this in mind, let's write down $F_{net} ~=~ ma$

$$-ky-mg ~=~ ma$$ (1.41)

But let's change variables so that we measure distance from the equilibrium position $y_e$

$$-ky-mg ~=~ -k(y+mg/k) ~=~ -k(y-y_e)$$ (1.42)

Calling $x \equiv y-y_e$, we have

$$-kx ~=~ ma$$ (1.43)

This is identical to the equation for a horizontal spring, eqn. 1.1. So the angular frequency $\omega$ is $\sqrt{k/m}$. So gravity has no effect on the oscillation frequency.