Phaden wrote:which is???

remember from up top im only y12 so a few things may need to be explained a little

y does equilibrium effect energy?

Hrm.... I think the simplest way to say it is that, yeah, equilibrium doesn't affect the total energy in the system (just how things are moving from potential to kinetic and back, which I assume you understand just fine). If we've got that oscillating body with no friction, all the energy is always in the system, no matter what form it's in, no matter whether or not the system in equilibrium. It is important not to lose that point in the discussion of how to deal with and watch the energy. Conservation of energy does *not* depend on equilibrium, be careful not to misread anyone's posts as saying that. But being in non-equilibrium states does mean we have to watch which substitutions and equalities we make.

Phaden wrote:sorry to rebring this up but i was just thinking on this today and still dont fully get it.

at the bottom of the oscillations the kinetic energy is 0 for a moment in time and this formula becomes true.....

doesn't it?

to illustrate what i mean, if you are walking forward and then change and walk backwards for a brief moment of time your velocity is 0 and your kinetic energy is 0

I'll take my shot at answering this, since I'm not sure if it's clicking for you yet. I may not help any more than the other explanations you've been getting, but heck, I like hearing myself talk.

At this point, the bottom of the oscillation, your line of equations has a *different* problem. The first line is actually true now. If there's no kinetic energy, and no loss to friction, yeah, everything that was gravitational is now potential energy in the spring, so mgx = 1/2kx

^{2}However,

Look at what you do with F in that proof. First in step two you say F = kx, and run the substitution. Later in step 4 you say F = mg , and run *that* substitution. As everyone's mentioned, at the bottom of the oscillation the forces *don't* balance, since yeah, no equilibrium, there is still a net force. The force of the spring, kx, does not equal mg. That's now an illegal step.

As a matter of fact, you can work out what the forces are at this point and go through your proof as it works at that bottom point of the oscilation and see what happens. Let's say the bottom of the oscillation is some distance x down (sidenote: remember to be careful about your variables. This is a different x than the 'distance x to the balance point' I was talking about in previous posts)

So, the spring compresses a distance and comes to a stop at the bottom of the oscillation (soon to spring back up, but heck, we don't need to worry about what it will do later, we're just worried about that instant when it stops, now)

let's get a few statements down first:

E

_{gained} = (1/2)*kx

^{2} F

_{spring} = kx

E

_{lost} = mgx

F

_{gravity} = -mg (minus sign, since gravity's force is downwards)

How are F

_{s} and F

_{g} related, though? They're not equal, the situation isn't in equilibrium, so the forces don't match. Instead, there's some net difference. If you think about the symmetry of the situation (heh, there's a physics term you'll hear more and more as time passes. symmetry.) We can realize that the net force at the top of the spring is equal and opposite to the net force at the bottom of the spring. The net force at the top of the spring comes from gravity only (no spring compression at the top of the oscillation in this problem) *So*, we know that

F

_{net} = mg (positive, since at the bottom it's equal and *opposite* to what it is at the top)

F

_{s} + F

_{g} = F

_{net}kx - mg = mg

(1/2)kx = mg

now that we've worked out that relationship at the bottom of the oscillation, we can run through your proof. We can now define F = (1/2)*kx = mg at the bottom of the oscillation.

E

_{lost} = E

_{gained} mgx = (1/2)*kx

^{2}mgx = (1/2) * (2*F/x) * x

^{2}mgx = Fx

mgx = (mg) *x

mgx = mgx

Constants equal themselves, physics works, and all is right with the world.