Variation of potential and kinetic energy over distance in SHM.

1. Jul 30, 2014

thephysicist

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 to be (1/6)kA^2. For the average kinetic energy over the same range and direction, it is (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.

2. Jul 30, 2014

Nathanael

Hello, and welcome to Physics Forums.

I think I am misunderstanding what you mean by "average kinetic energy over the same range and direction," because my result is that the average kinetic energy is also (1/6)kA^2

Would you mind explaining why the average KE is 1/3?

3. Jul 30, 2014

sophiecentaur

You have asked for a non-mathematical explanation about a quantity that is mathematical (Kinetic energy has a mathematical definition.) The "Physics of the event" is hardly describable at any significant level without using Maths. Do you have a problem with Maths? Contrary to some opinions, Maths makes the describing of things easier and not harder.
Where would your snapshots be derived from? A simulation possibly?

4. Jul 30, 2014

thephysicist

It's not about that I do not like the math but my instructor wanted me to come up with an explanation purely based on physics to describe the discrepancy. I have solved it on a whiteboard and kept the snapshots as it is difficult to write the maths online. All I needed was an explanation of the avg of kinetic energy being double that of potential energy when you take the average over distance and not over time, like we explain things in simple English.

5. Jul 30, 2014

sophiecentaur

Oh. That makes more sense now and it's a question worth answering. Sorry I mis interpreted the OP in terms of Mathphobia!!
Energy, averaged over distance doesn't often have a lot of meaning but this concept is often used in the issue of vehicle stopping distance. If the brakes produce a constant force then the energy dissipated per unit distance is constant but, as the KE must be dropping at the same rate per unit distance, the speed must be following a non linear (square-root) law with distance.
I guess the hand waving 'reason' for the averages in SHM will be the fact that the highest KE is when the greatest distance is travelled and the reverse is true for the PE. Probably a √2 in each case so the total ratio is 2 ??

6. Jul 30, 2014

olivermsun

Can you explain what it means to have an average kinetic energy over the same range and direction?

7. Jul 30, 2014

sophiecentaur

I don't think he means direction, actually. the PE and KE changes are both symmetrical over the four quadrants and only depend on displacement from mean position.

8. Jul 30, 2014

olivermsun

He means the "change" in KE?

9. Jul 30, 2014

sophiecentaur

??? I can't find that in the OP He is taking the KE values at each point and integrating it over distance. If you do it, integrating over time, you get the same values for PE and KE but not when you integrate over distance.

10. Jul 30, 2014

olivermsun

Well that's what I thought originally, but then how do you know the KE between 0 and A?

11. Jul 30, 2014

voko

Total energy is constant $(H)$, hence average kinetic energy is total energy less average potential energy (whatever sort of average that is). If we calibrate potential energy so that it is zero at the equilibrium position, then total energy is simply the max potential energy, and average kinetic energy is the difference between max potential energy and average potential energy, again for any kind of average. The rest of the argument requires some rudimentary calculus, I am afraid.

12. Jul 30, 2014

olivermsun

So I guess what we need is for the OP to explain what is meant by the original statement and how (s)he got it.

13. Jul 30, 2014

voko

The statement easily follows from #11. Max potential energy is $$k A^2 \over 2$$ and average potential energy is $${\int\limits_0^A {k x^2 \over 2} dx \over A } = {k A^2 \over 6}$$ so average kinetic energy is $${ k A^2 \over 2 } - {k A^2 \over 6} = {k A^2 \over 3}$$

14. Jul 30, 2014

sophiecentaur

Won't it be Total Energy - PE at any position
and PE is kx2/2

15. Jul 30, 2014

olivermsun

It doesn't easily follow unless you assume that the mass is released from position A, i.e., $v_0 = v(A) = 0$. Then the average PE and KE during the return motion from A to 0 are as the OP stated.

As I'm sure you're aware, for the general harmonic oscillator you can have any total energy you want, hence any KE over the range 0 to A. That's why I was asking the OP to clarify.

16. Jul 30, 2014

voko

It does, if you read it carefully: If we calibrate potential energy so that it is zero at the equilibrium position.

17. Jul 30, 2014

olivermsun

So if I pull the mass out to x = 10A and release it, won't the system have a vastly different total E, and therefore a different KE between x = A and 0 (where it's close to maximum KE), compared to the case where I release the mass from x = A?

What does that have to do with the PE being referenced to zero at the origin? You'd expect that anyway since the OP described the system as a "horizontal spring mass system."

18. Jul 30, 2014

voko

olivermsun, I fail to see your point. If potential energy is calibrated as stated, then total energy = max potential energy = max kinetic energy, and they all depend solely on the amplitude, no matter what initial conditions were. Average energies, as I remarked earlier, depend on the nature of averaging.

19. Jul 30, 2014

olivermsun

The relationship in the original post, which you so kindly derived in #13, is only true for a certain initial condition which wasn't stated explicitly in the original post. That's why I asked what the original poster had in mind.

Because it depends on the initial condition, I don't think the result would "easily follow" from your #11, no matter how carefully I read your post.

That's all, no biggie.

Last edited: Jul 30, 2014
20. Jul 30, 2014

thephysicist

Thanks for the answer. I think that's the right one. It was meant to be a simple question though I made it look bigger than it should have been. How do you type the mathematical symbols and equations in PF posts? Sorry and thanks. I'm new to PF.

21. Jul 30, 2014

thephysicist

Thank you very much sophiecentaur. I got my answer. You got it right.

22. Jul 31, 2014

voko

No, that is not true, and I have already addressed this. I think you misunderstand what "amplitude" means. It is the max deviation from equilibrium, regardless of initial conditions.

23. Jul 31, 2014

voko

Click the "Quote" button on my post, and you will see the code. The convention for that code is known as "LaTeX", which you have or will have to know as a physics student.

24. Jul 31, 2014

olivermsun

Actually, "amplitude" has a few different common usages, but you're right, I should have assumed that's what (s)he meant given the results posted. Anyway, thanks for clearing this up!

Last edited: Jul 31, 2014
25. Jul 31, 2014

olivermsun

The prettier and more flexible way is to type LaTeX code, as voko mentions.

A quick-and-dirty way for simple stuff is to use the symbols on the side and look in the ∑ menu in the editing window (you may have to Go Advanced to see these options).