Spring potential energy


The idea of potential energy is that, by the changing relative position of two objects, kinetic energy can be changed in a reversible way. We know that if we hang a mass from a spring and pull it down, that displacement creates a stretch of the spring that can accelerate the mass, creating kinetic energy. When we release the mass, the spring pulls it up and the mass starts to move. If the spring has little or no internal friction, the mass will go up, come back down, turn around, and start moving up again. So the process looks reversible. Let's see if we can construct the PE of a stretched spring.

The work done by a spring

In the example we described in the preamble — a mass hanging from a spring — the mass and spring were moving in a way that changed both kinetic energy and gravitational potential energy. This complicates the situation. Let's do our usual physics trick for understanding and analyze the simplest case we can imagine: a cart moving on a horizontal track attached to a spring as shown in the figure at the right, ignoring friction and air resistance. This simplifies the physics and lets us isolate the potential energy of the spring directly. To simplify the math, we'll describe the position of the cart by a 1D coordinate axis, $x$, whose origin is placed so that when $x = 0$, the spring is at its rest length. With this choice of coordinates, the coordinate $x$ becomes the stretch or squeeze of the spring. Since when $x$ is positive the spring is pulling back, Hooke's law in this coordinate system becomes simply

$$F_{spring \rightarrow cart} = -kx$$

This creates a problem for calculating the work, since the force changes as we move. But since it changes linearly, we can get away with a bit of a trick without having to do an integral explicitly.

Consider a small change in the cart's position, $Δx$.  The work done by the spring will be:

$$W = F_{spring \rightarrow cart}\Delta x = -kx\Delta x$$

Here's our problem: What value of $x$ should we use? Our $Δx$ means that the position is changing from initial (i) to final (f):

$$Δx = x_f - x_i$$

Since $x$ changes linearly, we might guess that the appropriate value to use is the average:

$$\langle x \rangle = (x_f + x_i)/2$$

This turns out to be correct. [We can look at the integral explicitly. The average value is the constant that makes the integral exact, so in the case of a linear force, this average approximation is exact.] The result is

$$W = F_{spring \rightarrow cart}\Delta x = -k \langle x \rangle \Delta x = -k \bigg(\frac{x_i + x_f}{2}\bigg)(x_f - x_i)$$

When we multiply it out, we find a work that depends only on the initial and final positions:

$$W = - {1 \over 2} k \big(x_f^2 - x_i^2 \big)  = -{1 \over 2} k x_f^2 + {1 \over 2} k x_i^2 $$

Potential energy of an ideal spring

Since the work only depends on the initial and final positions (and not, for example, on the direction the mass is going),  we can define a potential energy for the stretched spring:

$$W = -{1 \over 2} k x_f^2 + {1 \over 2} k x_i^2  = - \Delta U_{spring}$$

$$\Delta U_{spring} = U_{spring}(x_f) -   U_{spring}(x_i) = {1 \over 2} k x_f^2 - {1 \over 2} k x_i^2$$

(where the parenthesis means "$U$ is a function of $x$ evaluated at the value of $x$ indicated" — NOT a multiplication).

So in general, for an arbitrary stretch or squeeze, $x$ (positive for a stretch, negative for a squeeze) we can conclude

$$ U_{spring}(x) = {1 \over 2} k x^2$$

Don't forget our assumptions! This simple result is only true for stretches and squeezes of a spring for which Hooke's law holds. For larger stretches and squeezes, we have to do the integral of the work explicitly (and maybe even numerically). But it will be true for a spring that can be treated by Hooke's law in any context, even when there are other forces.

Sense-making note: Notice that we derived this result using a coordinate system in which $x$ means the amount that the spring is stretched or squeezed. It's important that you realize that this is a physical not a mathematical equation. In some cases, instead of "$x$" we might have "$y$" or even "$y-L_0$". But it always means the amount that the physical spring is stretched or squeezed!

Why the spring is more important than it looks

Springs may seem to be rather uninteresting objects. After all, how often do you actually use a spring? Especially in biology. But the fact that the spring PE is a simple parabola makes it much more important.

Any physical system where there is a stable position (such that a deviation from that position results in forces pushing back toward the stable position)  has a PE with a minimum and a lowest point. An example of this is a typical Interatomic force. Small deviations in the PE from the stable point can be approximately described by a parabola — so the spring provides a good model for everything from the vibrations of electric fields in light to the binding forces in molecules. At the right is shown a spring PE (parabola in red) fit to a typical inter-molecular potential. For small motions near the minimum, the spring PE works pretty well.

Workout: Spring potential energy


Joe Redish 11/3/11


Article 448
Last Modified: February 20, 2019