Hooke's Law, Sine, and Differential Equations

When I was an undergrad, I had to take an intro physics course. I liked it quite a bit, but there were a few concepts that I felt were brushed over a bit quickly. Outside of my math and computer science classes, I didn't tend to ask many questions, so I just sat on these questions for a long time. One of them was about how the kinematic equations relate to differentiation/integration (I eventually found my answer here). Another was on the equation that governs the motion of springs, Hooke's Law.

Hooke's Law can be formulated a few different ways, but I like to think of it as the force a spring exerts on an object it's attached to. Formally, it looks like this $$F=-kx$$ where $F$ is the force exerted by the spring, $k$ is a constant that indicates the stiffness of the spring, and $x$ is the spring's offset from its rest length.

An object attached to a spring.

In the above illustration, we can see $-x$, which pulls the attached object back towards the spring's rest length. A stiffer spring with a larger $k$ will pull harder. Note that if the red block was closer to the gray block so that the spring was forced to compress, the red block would be pushed away from the gray block instead of towards it. Springs apply force to objects they're attached to in a manner that brings them closer to their rest length.

When I first learned about this, I found it pleasing that spring forces result in such a simple equation. What I found odd, though, is that, when a spring pulls on an object, the object's position over time forms a sine wave:

Green: $kx$ (for some small k). Red: $\sin x$.

You don't typically see a linear function and a trigonometric function sitting side-by-side! I found this pretty mystifying. The way I thought of it was something like this:

1. $Force = -kx$
2. Force acts on acceleration
3. Acceleration is the second derivative of position
4. Integration is the inverse operation of differentiation
5. Therefore, the second integral of $-kx$ is $\sin x$ -- that is, $\iint kx = \sin x$

That's pretty dubious! No wonder I was flummoxed. ¹ The problem I encountered without realizing it is, even though $F=kx$ looks linear, $x$ is doing a lot of work. It doesn't represent an independent variable, as seen in the above graph. Rather, it represents a distance from the spring's offset, something like $x = r - p(t)$, which is rest length, $r$, minus position at time $t$, $p(t)$ (though even that expanded formula only works when the spring in question is at the origin -- otherwise we'd need yet another term).

The reason that's weird is that F is dependent on the position of the object. Position depends on acceleration, which depends on the force acting on the object, which depends on the object's position! We have a strange little feedback loop, here. Intuitively, though, a feedback loop is exactly the kind of thing that might produce something like a sine wave. ²

Here's how I reframed this to get the right answer and to have it make a bit more sense.

Consider our simplified case: a spring at the origin, who's rest length is 0 and for whom $k=1$. Assume an object with unit mass ($m=1$) is attached to this spring, that the spring is the only force acting on it. By definition, acceleration at some time $t$ is equal to the second derivative of position at that same time:

$$a(t) = \frac{d^2}{dt^2}p(t)$$

Also, by Newton's second law, force is mass times acceleration:

$$\Sigma F=ma$$

and since $m = 1$,

$$a = \Sigma F$$

and since our spring force is all that's acting on the object,

$$a = F_{spring}$$

Since our spring is at 0 (the origin), and has a rest length of 0, that means when the object is at 5, the spring will pull it back towards the origin, exactly in proportion to the object's distance from the origin. So, $x = 5$, and $F=-kx=-(1)*5=-5$. When our object is at -5, the spring will again pull it back towards the origin, this time moving it in the positive direction, so $x = -5$ and $F=-(-5)=5$. In other words, $x = p(t)$ and $F = -x = -p(t)$.

This is our key insight! Let's see what happens when we plug in our value for F:

$$a = F = -p(t)$$

$$a = -p(t)$$

Replacing $a$ for it's derivative form, we have:

$$\frac{d^2}{dt^2}p(t) = -p(t)$$

So, the formula for our position function, $p(t)$, is a function whose second derivative is $-p(t)$. We need a function whose second derivative is itself, negated! We have several options that fit the bill, but the simplest I know is $\sin x$ (and all the other ones I know are trig functions, anyway).

So, $$\frac{d^2}{dt^2}p(t) = -p(t) \therefore p(t) = \sin x$$

To summarize: The force equation $F=-kx$ appears to be linear with respect to the independent variable $x$, but $x$ is not an independent variable. $x$ is dependent on an object's position, $p(t)$. Force depends on position, position depends on acceleration, and acceleration depends on force. That feedback gives us our "springing," oscillating motion. By examining position and acceleration through the lens of derivatives, we can see why $p(t) = \sin x$ makes sense as a consequence of Hooke's Law.

I've never taken a differential equations course before (humblebrag), so I don't know how to formalize this, but I found things "clicked" much better once I put these pieces together to get this result. I wish I could go back and explain this to my younger self!

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 ¹ I would have figured this out sooner if I'd noticed that I wasn't integrating with respect to anything. I was probably thinking it was with respect to $dx$, but I didn't have a clear idea in my head right away.

 ² My first pass at solving this problem was to use the kinematic equation for position: $p(t) = p_0 + v_0t + \frac{1}{2}at^2$, plugging in $a = -p(t)$. I was hoping I'd end up with an infinite series $p(t) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ...$, which is the Taylor series definition of sine. It didn't even come close to working out, in no small part because acceleration must be constant for the kinematic equations to hold. Still, if anyone knows a way to get a Taylor series definition of position using Hooke's Law, I'd love to see it!