Why is the energy at a different frequency than the angular frequency in a simple harmonic oscillator?

I have a displacement function a simple harmonic oscillator: $$x(t) = A\cos(2\pi t),$$ The potential energy is then $$E_p(t) = \frac{1}{2} sA^2\cos^2(2\pi t),$$ where $s$ is the spring constant. So I have an angular frequency of $2\pi$, and I get the following result for the Fourier transform of the displacement in Wolfram Alpha: $$\hat{x}(\omega) =\sqrt{π/2} A δ(ω - 2 π) + \sqrt{π/2} A δ(ω + 2 π)$$ So I have a peak at a frequency of $\omega=2\pi$ which is to be expected given by displacement function.

Now when I take the Fourier Transform of the potential energy in Wolfram Alpha I get: $$\hat{E}_p(\omega) = \frac{1}{2} sA^2(\sqrt{π/2} δ(ω) + 1/2 \sqrt{π/2} δ(ω - 4 π) + 1/2 \sqrt{π/2} δ(ω + 4 π))$$

So as I have delta functions at $4\pi$ it seems all the energy is contained at $4\pi$, even thought the frequency of the oscillator is $2\pi$! So it works out this way mathematically, but what is happening physically, I would have thought that because the oscillations happen with angular frequency $2\pi$, that naturally the potential would energy also show a peak at $2\pi$? And why do we also have a peak at $\omega = 0$?

• It's worth being clear that the kinetic energy varies as $\frac{1}{2}sA^2\sin^2 (2\pi t)$ so that the total energy remains constant. And of course, you can address the question just as clearly from the kinetic energy side where the resolution is wrapped up in the interchangeability of the two directions of motion rather than that of the two direction of spring distortion. – dmckee Sep 6 '17 at 19:40

Mathematically, you could get to you result quicker if you consider the identity

$$\cos^2\alpha = \frac{\cos 2\alpha + 1}{2}$$

Physically, let's take a specific example, like a swinging pendulum.

The pendulum swings up at one end of its swing, where it has maximum g.p.e. Then it swings down through the bottom where it has maximum k.e., then back up to a point of maximum g.p.e. Then swings back through the bottom to its starting point.

The main point is that there's two points of maximum potential energy for each swing of the pendulum (and two points of maximum kinetic energy as it swings left-to-right and right-to-left). So you should expect the potential energy vs. time function to have twice as many peaks and troughs as the position vs. time function or velocity vs. time function.

• BTW, I think his example was a spring, which is a little different because one potential energy is in compression, the other is in tension. That may be confusing him a bit. This answer is spot on though. – JMac Sep 6 '17 at 19:09
• @JMac Yes my example was for a spring but this answer is great, I get it now! Why is there also potential energy a peak at a frequency of zero though, that seems strange? – Riggs Sep 7 '17 at 5:11
• There's a k.e. maximum at position 0, not a p.e. maximum, assuming you've chosen to put the zero of your coordinate system at the same place I did. – The Photon Sep 7 '17 at 5:29
• @ThePhoton The command I entered in WolframAlpha to get the potential energy was "Fourier transform 1/2 b A^2 cos^2(2*pi * x) with respect to x"...this leads to the results in my post where we have a peak at $0$. So I understand why we have twice as many peaks and troughs when looking at the potential energy but I don't see why we have potential energy peaking at zero frequency? – Riggs Sep 7 '17 at 17:53
• Because in the identity I gave above for $\cos^2\alpha$ there is a term $+1/2$. That means there is a peak in the Fourier transform at DC. – The Photon Sep 7 '17 at 17:58