# Kinetic energy and potential energy variation over distance in SHM

When you compute the average potential energy of a horizontal spring mass system from the mean position to the positive amplitude A, the value comes out to be $\frac{1}{6}kA^2$. For the average kinetic energy over the same range and direction, it is $\frac{1}{3}kA^2$, which is double the average potential energy. What the physical explanation of the different average values of PE and KE?

P.S. No mathematical explanations please, i.e. area under the graph, etc, only explanations in terms physics of the event are appreciated. The question does not involve time averages of PE and KE. Snapshots of derivations can be uploaded if requested.

• It's not clear to me what you mean by average value, especially when you specifically say it's not a time average. Can you extend your question to explain exactly what quantities you are calculating and comparing? Jul 30, 2014 at 6:22
• Just like in the plot of KE and PE vs X, where X is the distance from the mean position. Here taking the average of either KE or PE with distace we can get the average of the Y axis quantities wrt the X axis quantity, i.e. distance. Jul 30, 2014 at 7:22

One can't discuss the difference between factors of $1/3$ and $1/6$ without mathematics. The difference between these two numbers – and generally, any fact about any numbers – is all about mathematics.

If all values of $x$ between $0$ and $A$ were equally likely, the average value of $kx^2/2$ would be $kA^2/6$ as you say because the average value of $X^2$ for $X$ uniformly distributed between zero and one is $$\int_0^1 dX\,X^2 = \left.\frac{X^3}{3} \right|^1_0=1/3$$ However, when the harmonic oscillator (you talk about a spring which is a harmonic oscillator) oscillates, it oscillates harmonically, via sines and cosines, so it spends much more time near the $|x|=A$ extreme points where the speed is low than it spends in the vicinity of $x=0$ where the speed is high.

If you compute the average value (over time) of $kx^2/2$ in this oscillating motion, the result will be proportional to the average value of $\cos^2 t$ over time which is equal to $1/2$. So the average kinetic energy of the oscillating motion will be $kA^2/4$ – a number that doesn't appear in your list of results at all.

Similarly, the maximum value of the kinetic energy is $mv_{\rm max}^2/2 = kA^2/2$ at the maximum achieved exactly when the potential energy $kx^2/2$ has the minimum value (zero). Similarly, the minimum value of the kinetic energy is $0$ exactly when the potential energy $kx^2/2$ is maximized i.e. at $|x|=A$.

The average contribution of the kinetic and potential energy to the total energy is the same for the harmonic oscillator – both averages are $kA^2/4$ – a fact that is guaranteed by the "virial theorem".

No factor of $1/3$ ever appears in the correct average values for the harmonic oscillator (just like it doesn't appear in the right solution to the sleeping beauty problem).

• Excellent answer! I completely agree with your reason for why the averages should be different because the oscillator spends more time near $x=A$. With this logic, since potential energy is maximized at $x=A$, it would be reasonable to conclude that the potential average should be greater. However, as we see, it is the other way around. Why is that so? Dec 14, 2022 at 4:05