cscott said:
How is this meaningful in terms of the integrals I provided? Why aren't these integrals OK in terms of <E>space = <E>time?
There is no reason the expectation values of the kinetic energy and potential energy averaged over time should be the same as those averaged over space. In your first set,
<br />
<KE>_{time} = \frac{1}{T} \int_0^T \frac{1}{2} k A^2 \cos^2(\omega t) dt = \frac{1}{4}kA^2
<br />
<PE>_{time} = \frac{1}{T} \int_0^T \frac{1}{2} k A^2 \sin^2(\omega t) dt = \frac{1}{4}kA^2 ,
time is a "uniform" variable, so every infinitesimal interval is "weighted" the same. You find that <KE>_{time} = <PE>_{time}, in accord with the virial theorem for a potential of the form U = k·(x^2) .
However, for your second set,
<br />
<KE>_{space} = \frac{1}{A} \int_0^A \left[ \frac{1}{2}kA^2 - \frac{1}{2}kx^2 \right] dx = \frac{1}{3} kA^2
<br />
<PE>_{space} = \frac{1}{A} \int_0^A \frac{1}{2} kx^2 dx = \frac{1}{6} kA^2 ,
the displacement from equilibrium is
not uniform, in that the simple harmonic oscillator (mass on a spring, pendulum bob, or what-have-you) spends disproportionately more time far from equilibrium than close to it. So the spatial average for potential energy receives most of its contribution from the relatively small sections of the oscillation where PE is greatest, while most of the contribution to the spatial average for kinetic energy comes from a fairly broad region around the equilibrium point.
After all, if you use x = A sin(wt) for the displacement from equilibrium, then you have the differential dx = Aw cos(wt) dt , which gives most of the "weight" in the spatial average to the two-thirds of the cycle around the equilibrium point.
The important thing here is that you found the same expected value for
total mechanical energy over the entire cycle, whether temporally-
or spatially-averaged.