Average energy of an SHM Why do we usually calculate the average potential or kinetic energy of a simple harmonic motion with respect to time, why not with respect to position?
Why even calculate average energy for an SHM? Does it have any physical significance on practical use?
 A: *

*Why often calculate the average of energies in time, not in position?


I guess that it is rather natural and easier to do following the pathway of mechanics.
Recall that mostly this is the second example in mechanics right after the free-fall problem. Starting with the differential equation:
$$
   m\frac{d^2 x}{dt^2} + k x = 0.
$$
It naturely leads to a solution with $\omega = \sqrt{k/m}$:
$$
  x = A \cos(\omega t) + B \sin(\omega t).
$$
Thence, the velocity $v(t)$, and potential energy $\frac{1}{2} k x^2(t)$, and kinetic energy as functions of time. Therefore, it is natural, logical, and easier to calculate the average of these quantities in time.
Unless, when you have function form like
$$
    \Psi(x, t) = A \cos (k x - \omega t)
$$
in wave mechanics, then you will treat time and position equally.


*

*Does it have any physical significance on practical use?

Yes. The simple harmonic oscillator (SHO) is probably the only system so transparent to most physic students. We can study every detail of this system in both classical and quantum mechanics.
For application purposes, SHO has been applied to thermodynamics, statistical mechanics, solid-state physics, quantum information science... When a potential has a minimum, there is always a region near the minimum that can be treated as approximate SHO.
