Re: Missing Energy

[Mark Sylvester <msylvest@SPIN.IT>]

At 15:51 20-05-99 -0400, Michael Edmiston wrote:
> I believe this problem is somewhat analogous to a spring pendulum.  If
> I do work to displace the mass of a spring pendulum, in the long run,
> where does my work go?  Initially it is passing back and forth between
> potential energy and kinetic energy.  But assuming any amount of
> friction, it eventually ends up as thermal energy.  If we want to
> imagine the absence of "nonconservative forces" (i.e. we want to rule
> out any thermal energy changes) then the pendulum oscillates forever.

I think a closer mechanical analogy is to have two springs each with one end
anchored, and the free ends linked together. Move the anchored ends apart to
stretch both of them , and then take the join in your fingers and stretch
one spring beyond the equilibrium length, thereby simultaneously reducing
the stretch of the other. Now you have something like the two capacitors
with different p.d.'s. Let go, and the springs return to the lengths where
the tensions are equal. It seems easy to see that the PE lost by the more
stretch one is less than the PE gained by the less stretched one. Just
imagine the Hooke's law graph for the two springs (let them have the same k
for convenience). The delta x is the same for each, but arriving
respectively from left and from right, so the areas are different.

> Now let's look at the two capacitors.  When we hook them together we
> make a loop.  That means we have inductance.  Therefore we have made an
> RC circuit.

You Mean LC of course. For the springs we just have to decide not to neglect
the ir mass, and we have an oscillating system.

As with the capacitors, the interesting thing - and the subject of Stuart's
question - is why the dissipated energy depends only on the initial and
final conditions, and not on the dissipation factor.


Mark


Mark Sylvester
United World College of the Adriatic
34013 Duino TS
Italy
msylvest@spin.it