tahayassen said:
What would the point of that be? :| They would still be two separate terms, right?
Yes, but only one will be time dependent.
If you consider first a horizontal spring with a mass on the end (i.e. ignore gravity). The total energy, at any time, is:
E_{Total}= E_{K} + E_{SP}
where E_{K} is kinetic energy and E_{SP} is spring potential energy. More explicitly we have:
E_{Total} = 1/2 mv^{2} + 1/2 kx^{2}
where v is the velocity of the mass, and x is the extension of the spring. Assuming we're not losing energy through friction etc., then the total energy is constant from the Conservation of Energy Principle. We can confirm that by substituting in expressions for x and v if we want.
So we have quite a simple situation. In terms of problem solving, we know that:
1/2 mv^{2} + 1/2 kx^{2} = constant
at all times, which is a pretty useful identity.
What I want to show is that we can recover this simple situation even when gravity is involved. I'll do this by working relative to the initial, equilibrium extension of the spring due to gravity.
When the spring is in equilibrium with gravity, forces are balanced - the downward force of gravity matches the upwards force of the spring. So:
kl = mg
where l is the equilibrium extension of the spring and g is the acceleration due to gravity. So the initial extension is:
l = mg/k
which we'll need shortly.
Now, we set the spring bouncing. At any time we have the total energy:
E_{Total} = E_{K} + E_{GP} + E_{SP}
where E_{GP} is gravitational potential energy.
To simplify things, I'm going to choose E_{GP} = 0 at the equilibrium position - I'm free to choose this anywhere I like.
Now we can say:
E_{Total} = 1/2 mv^{2} - mgx + 1/2 k(x + l)^{2}
where x is the extension beyond the equilibrium extension at any give time. So the total extension is x + l, which is why that expression appears in the term for the spring potential energy.
Now, expand out the final term:
E_{Total} = 1/2 mv^{2} - mgx + 1/2 kx^{2} + kxl + 1/2 kl^{2}
substitute in the expression we calculated for the equilibrium extension, l earlier:
E_{Total} = 1/2 mv^{2} - mgx + 1/2 kx^{2} + mgx + 1/2 kl^{2}
cancel the mgx terms, and we have:
E_{Total} = 1/2 mv^{2} + 1/2 kx^{2} + 1/2 kl^{2}
So now, as you say, we still have three terms instead of two. The first describes the kinetic energy as a function of velocity, and hence of time. The second describes the potential energy as a function of extension from equilibrium, and so also as a function of time. And the final term describes the energy as a function of the equilibrium extension which is a constant in time.
So, in terms of the time evolution of the system, we have got back to a nice, simple, identity:
1/2 mv^{2} + 1/2 kx^{2}= constant
where x is now the extension from equilibrium.
This makes some problems easier to solve.
