>> Let's go back to Kenny's question for a second, because Kenny's question was a really good one. It was what is the difference between a horizontal spring with a mass going back and forth versus a hanging spring with a mass going up and down? And what we said was you can in fact ignore gravity in this case and just treat it exactly like that. Now, let's explain this one more time from the point of view of forces. So, when I first put the spring and I hang it there, there's some length associated with it. If I then put a mass on it such that it comes to rest, there is a restoring force given by K X naught. How far did the spring stretch out from where it was? But we know there is also gravity acting down on it. And if I think about the forces, I have K X naught going up. I have M G going down. All of that has to equal 0 if this thing is at rest. And so we get K X naught is equal to M G. But now let's do the following. Let's take that spring and stretch it out. Okay? And let's stretch it out a distance X from where it was. Remember, it already stretched out a distance, X naught, from that first case. And so what are the forces that are acting on it now? Well, we've got some force going up. Whoops, sorry. We've got some force going up but now it's really K X plus X naught. And then we've got M G going down. You see that over there okay? Yeah. K X plus X naught. And so now in this case I can say the sum of the forces is what? It's K X plus X naught minus MG. Which is K X plus K X naught minus M G. But from the earlier case, we knew that K X naught was equal to M G. And so that cancels with that and we just get K X. What's the restoring force on this spring? It is K X if you measure from that new equilibrium position. And therefore it's exactly the same as going from X = 0 out to X in the horizontal case. You can do energy considerations if you'd like as well but I think this is a little simpler to understand. The restoring force here is just K X. All you've done is change where your equilibrium position is located.