If a spring with a spring constant of k is hung vertically, and a mass is attached to it the spring will rest in equilibrium at some distance h from the springs original equilibrium length because the spring force must equal the gravitational force. Now lets say I pull the mass down some distance, and I let go, obviously, it will begin to oscillate. But, I am confused whether the center of oscillation will be at distance h, or about the spring's original equilibrium length.
It will be about the lower point.
Imagine the spring is 10 meters long hanging from a tree branch 16 meters off the ground (so the end of the spring is 6 meters up). You climb up a ladder and add a large mass to the spring, stretching it by 5 meters, to 15 meters in total length. It's resting there quietly, one meter off the ground. Then you pull the mass down 1 centimeter, and let it go. What happens? Does it jump back up to 6 meters in the air?