# Why is the potential energy of a spring the same when it is compressed and stretched?

I'm giving a high school lecture and I want to introduce the potential energy of a spring. My students have not learned the Hooke's Law and the notion of integral is too advanced. I'm really trying to justify with a hand waving argument that the energy is given by

$$U = \frac{1}{2}kd^2.$$

To do so I let them realize that stretching/compressing the spring will change its energy. So this lets me justify why it only depends on the properties of the spring captured by $k$ and the deformation $d$.

Then by looking at the units of energy they should realize that the deformation $d$ has to be squared and that the constant $k$ takes care of the remaining units.

But if a student argues that $k$ could be defined with other units so that the dependence in $d$ is linear, I could answer that the energy should be identical whether the spring is stretched/compressed so that only $|d|$ or $d^{2n}$ are possible solutions. I see how to justify that $|d|$ is not a physical solution because it would create a cusp in the energy in $d=0$ and that nature does not like that (at their level at least). Additionally, having $n=1$ is just the simplest case.

My missing argument is therefore how to justify that the energy is the same when a spring is stretched/compressed by $d$.

• What do you mean by a discontinuity in $d = 0$? The function $|d|$ is continuous there, no problem. It is not differentiable... But continuous it is, so you might have to explain it a little further. Hand-waving suggestion: nature does not like "cusps". Sep 7, 2017 at 0:44
• "But if a student argues that kk could be defined with other units so that the dependence in dd is linear" - if integrals are too advanced and they haven't yet covered Hooke's Law how likely is a question like this? I agree with the others, Hooke's Law F = kx, is worth the few seconds.
– Mick
Sep 7, 2017 at 1:19
• "But if a student argues that k could be defined with other units...", sorry, but I can't see any mathematics at all, from that point on. I can see a lot of pseudo-mathematical nonsense, though. Sep 7, 2017 at 23:36
• This is the XY problem. You have started justifying that formula using several bad ideas instead of the one correct idea, Hooke's law. Now you want to help to demonstrate one of these steps, which isn't necessarily true. This is not going to eventually lead to an adequate answer. Your real problem is to understand and explain Hooke's law.
– jwg
Sep 8, 2017 at 9:15
• Agreed with @jwg. I think you are doing your students a disservice with your approach. Spring energy is quadratic because increasing $d$ increases both the average force and the distance. Better to teach them that, than to instill the notion that physics is just a bunch of random, unrelated concepts where you just make stuff up as you go along. I'm not sure why you would even want to discuss a spring's energy if the students don't even know the most basic thing about a spring's force. Sep 10, 2017 at 8:55

One way is to explain how a spring actually works.

A coil spring is a large wire that is wound into a helix. When you compress or extend a spring, from the wire’s perspective, you aren’t really pushing or bending. Instead, you are twisting the wire one way or another.

Twisting a bar clockwise or counterclockwise should be the same thing.

• Your post is apropos to coil springs. There are other forms of springs that do bend, not twist. In fact, a bar anchored on one end is a very common form of spring. Of course, for OP's purposes, looking only at coil springs is acceptable (and maybe preferable). Sep 6, 2017 at 16:30
• @PaulSinclair Good point. You could do this with any appropriately designed system as long as it behaves like a "spring". I just find coil springs are the easiest example, where everyone has seen them and understands approximately what they do.
– JMac
Sep 6, 2017 at 16:40
• It might help to also remember that all of the formulae we use in physics are really models that approximate the actual world closely enough that we can use the results the models give us to do things, like build working machines and accomplish tasks with those machines. The models/formulae shouldn't be taken to indicate the exact, actual behavior of anything. Sep 6, 2017 at 21:37

Unfortunately, there isn't a physical reason that the energy must be the same at +d and -d, because in general it is not. All springs will be nonlinear and non-symmetric if you stretch them enough. The reason we can write $F=-kx$ is that you can always linearize the force for sufficiently small displacements, and then you just assume you're working in the linear regime. You might just have to assert to the students that for small enough displacements, the energies are the same in each direction. Alternatively, you could use another system for which there would be no reason to believe that the energies aren't symmetric, such as a pendulum.

One aspect you might mention is that the potential energy of a particle is related to the force the particle feels (since mathematically the force is the negative gradient of the potential energy). This point is powerful for intuitive arguments. For example, you can argue that the energy shouldn't be |d| because then the restoring force would be the same whether the spring is stretched a little or a lot, which seems unreasonable.

If possible, you could also have students test this with a spring in the classroom, feeling the force by holding the spring displaced by different amounts in either direction to get them on board with the idea that the force is symmetric on either side of the equilibrium point and that it increases as you pull (or push farther).

• There is a physical reason why it is linear, by the way. Generally with a spring you are elastically deforming the material. Depending on the material used, there is actually a fairly large portion of the stress strain relationship that is very linear. Assuming you're willing to ignore very small errors due to secondary stress effects and material imperfections, there are established physical reasons why they are linear in the operating range.
– JMac
Sep 6, 2017 at 18:31
• @JMac Good point! I knew there was a reason springs are generally quite linear, but I couldn't put my finger on it. Perhaps it's a bit subtle for the students, though? Sep 7, 2017 at 4:28
• The energetic symmetry (for sufficiently small stretching) is quite physical: A spring at rest is in equilibrium, thus the leading term in a Taylor expansion cannot be anything but linear (though that might vanish) Sep 7, 2017 at 13:14
• @TobiasKienzler Yes, I agree. Unfortunately, there's no physical reason why energy must be symmetric for all +d and -d, because I could choose d to be in the nonlinear regime in general. Sep 7, 2017 at 18:40
• @Gilbert Agreed - actually the contrary is the case, you can only compress a spring a small amount compared to the potential straight wire you might end up with by pulling (unless it breaks before). But it seems we're both aware of Hooke's Law's disclaimer "Warning: may only be valid for small displacements" Sep 8, 2017 at 5:58

This is really a question of practical vs. theoretical. Any spring actually will have a non-linear effect. But in theory we ignore it for a basic understanding. So we assume it is linear. Here are some ways to approach this: If the spring body is hidden from your eyes and all you have access to is a movable handle connected to the free end of the (hidden) spring, then you can't tell which way the spring distorts (stretch or compress) when you move that handle--it takes the same amount of energy to distort it the same distance. Besides that, consider that a short section of the spring (for example, a short section of the spring wire, assuming a coil spring), bends one way when the spring is stretched and the other way when the spring is compressed. Bending a piece of metal compresses one side and stretches the other, or stretches the one side and compresses the other. If the spring is homogeneous, then this is the same sort of distortion.

• I like the last bit of your argument with the short section of a spring... Sep 6, 2017 at 13:53
• "you can't tell which way the spring distorts (stretch or compress) when you move that handle" - For any plausible real-world (coil) spring I tend to disagree. Moving the handle in the direction that compresses the spring will eventually have a hard-stop where the spring reaches maximum compression. Moving the handle in the other direction will encounter no such thing, at least not while the spring remains a spring. It would be impossible to tell if the handle's range of motion is restricted to something that works equally well in both directions, but that design aspect wasn't stipulated. Sep 8, 2017 at 15:04

Your students are correct. Without Hooke's law, springs would store energy linearly.

Ie, if the force to extend or compress a string was constant regardless of how far away from "rest" the spring is, you'd end up with k|d| energy stored by moving a spring d distance.

And a spring-like device that behaves that way is physically possible and reasonable (to a pretty good approximation). If your argument does not respect this fact, your argument is wrong, and any convincing of your students you pull off is in error.

This, however, is an opportunity. Put down two different equations for the energy stored in a spring, one with |d| and one with d^2.

Work out how you'd determine which is more correct. What does each equation predict?

I assume they know that energy = force times distance. With |d| you should be able to show that a small distance change near "rest" and a small distance change "far from rest" should involve an equal amount of force being applied.

With d^2 this isn't the case; a small distance movement near rest is going to be less force than one far away from rest.

So now we have a prediction:

• If E ~ k|d|, then the force a spring applies near rest is the same as the force a spring applies far from rest.

• If E ~ k d^2, then the force a spring applies near rest is much smaller than a force a spring applies far from rest.

You could go further and work out the in the d^2 case, the force is roughly proportional to d, but you don't have to.1

Now, take two springs. Place one near rest. Place one far from rest. Attach them so the one far from rest tries to pull the near rest away from rest.

If |d| hypothesis is correct, this system should be in equilibrium, as both springs apply the same force.

If d^2 hypothesis is correct, the system isn't, and the one far from rest should pull the one near rest.

And ... done. We just experimentally proved that |d| isn't true, and that d^2 is at least consistent with observations. (We have not proven that d^2 is correct, that takes more work).

1 Suppose the energy of a spring = 10 J * (d / 1 m)^2.

Energy at 1 cm and 2 cm is 0.0001 J and 0.0004 J, giving us roughly 0.0003 J delta.

Energy at 1.01 m and 1.02 m is 1.0201 J and 1.0404 J, giving us roughly 0.0203 J delta.

Energy at 2.01 and 2.02 m is 4.0401 and 4.0804 J, giving us a 0.0403 J delta.

If we divide the J delta by the average distance of these test locations, we get F ~ 0.02 * d N/m assuming force doesn't vary over small distances.

As noted, this isn't a requirement to the above argument.

• Thanks for your answer but I think you missed the point of my question... I know what the theory is and how spring works. I was just willing to find a simple explanation (which does not involve equation) on why the potential energy of a spring is the same when stretched/compressed. I was not trying to argue against my students. I still have not given that lecture this year. Sep 6, 2017 at 19:38
• I like how this answer turns the problem around and considers how something with E ~ k |d| would behave, and how it's different from what we know about how springs behave qualitatively. Sep 7, 2017 at 9:28
• I like how this answer engages in actual physics rather than received wisdom. Sep 9, 2017 at 16:15

I don't know if I'm correct, but what comes to my mind is the following:

You can tell them to take (imagine) two identical springs, and get them in contact with each other on one side and attached to walls on other sides. Now apply force at the point of contact of springs horizontally, such that one gets stretched and other compressed.

Now, the distance by which the springs have been compressed is same. Moreover, the work done on both of them is same (adding all the forces, including spring force on each other), which implies that the energy stored or potential energy is same.

Please correct me if I went wrong somewhere, or ignored something.

Thanks.

• You went wrong by assuming that the work done on both of them is the same. You need to prove this assumption. Sep 7, 2017 at 2:27
• I din't want to use mathematics in my answer or a formal justification, but I believe it can be easily proved by considering that each spring is acted upon by a spring force due to another spring and force applied by us. Also the displacement is exactly same, which implies that work done on both the springs is equal. Sep 7, 2017 at 12:30
• Imagine one is a much stiffer spring, now the displacement is the same but the work is not equal. How do you know that is not also true of identical springs? Probably because you were thinking about Hooke's law, but you can't assume Hooke's law in order to justify Hooke's law. Sep 7, 2017 at 20:03

You are going way too far with the Occam's razor! For this very basic physics the best is to make small live experiments and leave it to Nature's authority.

Take a spring hang it somewhere and stretch it by some distance $d$ with some weight $W$. The work done by the load is: $d\times W = U$. By measuring $d$ for a few $W$ and then computing $U$ you can easily verify the quadratic dependency when elongating the spring.

What about negative $d$? For that we need to repeat the experiment compressing the spring. By fixing the spring to a surface and loading weights on the top of it you can easily find out the same quadratic dependency.

Finally the key point is that whether the spring is compressed or elongated, the force and the displacement have always the same (positive or negative) sign, therefore we always have positive work on the spring and positive energy stored in it.

If you want to avoid calculus, give them a geometric understanding: plot the $F vs. x$ line and state that the energy is the area inbetween the x-axis and the curve. They all know the area of a triangle (hence $1/2kx^2$)

However...

The expression depends on the spring. Some springs are constant force (constant-force springs), so the expression for potential energy would not be a quadratic.

So the best way to avoid mathematics is to just assert that Hook's law is valid for most springs, for small amounts of stretch. If they ask why the energy is the same in compression and tension, just say that there are springs for which this is not true.

Just draw a graph of "spring length" vs "force exerted". Hopefully they'll agree that if the spring has been stretched a distance of zero, it will exert no force. Hopefully they'll also agree that if you stretch the spring in the positive direction, the force is negative, and vice versa. And hopefully they'll agree that if you increase the distance stretched, you'll increase the force. So you should be able to get them to agree the graph looks something like this: The precise form of the force doesn't matter, just the overall shape. Then say, "we'll be focusing on what happens when the spring is only streched a small amount." And then draw the tangent line at zero: Hopefully they'll all agree, just from the picture, that this is a good approximation when the displacement is small. Then you can say that, in this approximation, the force exerted by the spring is clearly equal and opposite when the spring is compressed or when it's stretched. Thus, the spring should store the same amount of energy. This is kind of a gentle way to introduce the idea of a first order Taylor expansion, which is what Hooke's law is based off of, but in a way that is easy to understand graphically. It's much easier to understand than a second order Taylor approximation to the energy, since it's not obvious why you would want to approximate things by parabolas if you've never seen a Taylor series. But it's pretty obvious why you might want to approximate things by straight lines.

• Your argument is that if the force of the spring is symmetrical and smooth, it can be linearized. However, the question is how to justify that the force is symmetrical.
– Pere
Sep 8, 2017 at 22:54
• @Pere You don't need to assume the force is symmetrical, just that it is smooth. Non-symmetrical functions still have Taylor expansions! Sep 9, 2017 at 2:02
• @Pere It's true that it might be better to draw something obviously non-symmetric on the board, though. The graphs I showed looked a little too symmetric. Sep 9, 2017 at 2:03
• @Pere Edited to include asymmetric graphs. Sep 9, 2017 at 2:07