Variation of potential and kinetic energy over distance in SHM.

[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.

 

[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?

 

[sophiecentaur]

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.

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?

 

[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.

 

[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 traveled and the reverse is true for the PE. Probably a √2 in each case so the total ratio is 2 ??

 

[olivermsun]

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.

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

 

[sophiecentaur]

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

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.

 

[olivermsun]

He means the "change" in KE?

 

[sophiecentaur]

He means the "change" in KE?

? 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.

 

[olivermsun]

? 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.

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

 

[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.

 

[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.

 

[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 $$ { k A^2 \over 2 } - {k A^2 \over 6} = {k A^2 \over 3} $$

 

[sophiecentaur]

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

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

 

[olivermsun]

The statement easily follows from #11. Max potential energy is $$ k A^2 \over 2 $$

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.

 

[voko]

It doesn't easily follow

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

 

[olivermsun]

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

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."

 

[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.

 

[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.

 

[thephysicist]

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 $$ { k A^2 \over 2 } - {k A^2 \over 6} = {k A^2 \over 3} $$

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.

 

[thephysicist]

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 traveled and the reverse is true for the PE. Probably a √2 in each case so the total ratio is 2 ??

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

 

[voko]

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.

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.

 

[voko]

How do you type the mathematical symbols and equations in PF posts? Sorry and thanks.

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.

 

[olivermsun]

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.

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!

 

[olivermsun]

How do you type the mathematical symbols and equations in PF posts?

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).

 

[olivermsun]

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.

One intuitive way you can you can think about the discrepancy is the following:

We know that the time averaged PE and KE are the same (1/2 of the total each). The KE is large near the origin, when the spring is unloaded, while the PE is large near the ends of the travel, where the KE has gone into compressing the spring.

The mass spends much more of its time near the ends of travel, where it's moving slowly (small distance per time), than it does near the origin, where it's moving quickly (large distance per time). Therefore, in order for the KE and PE per time to be equal, the KE has to be much larger than the PE per distance.

(If you diagram the motion on the unit circle, you can get the numbers without using any calculus at all!)