# Hanging Slinky Analysis 1: Sums to Integrals Last time, we (re-)introduced the Hanging Slinky problem, designed a few years back as a physics lab but suitable for a math class, say Algebra II or beyond. We looked at the length of the hanging slinky as a function of the number of slinks that hang down, and it looked seriously quadratic.

I claim that knowing that the real-world data is quadratic will help you explain  why the data has that shape. That is, “answer analysis” will guide your calculations.

I beg you to work this out for yourself as much as you can before reading this. I made many many many wrong turns in what is supposed to be an easy analysis, and do not want to deprive you of that—and the learning that comes with it.

## Qualitative Analysis

Still here? Okay, here we go, waving our hands qualitatively: When we look at the slinky, we can see it’s more spread out at the top and more scrunched together at the bottom.

Put another way, that the last 10 slinks take up less vertical space than the next-to-last 10 slinks—and that pattern continues. That is, with every slink we go up from the end, the space gets bigger.

If that’s the case, the graph will have to curve upwards—and sure enough, it does.

But why should it do that? Here we do need to think about the physics. To wit: each slink is supporting all the slinks below it, and none of those above. (This means, by the way, that the bottom n slinks of a long hanging slinky are the same as a short hanging slinky of n slinks.) So the nth slink is supporting times as much as the first slink. So it will be stretched to times the length of the first slink, and the total length will be (1 + 2 + 3 + … + n) times that length.

We could look up a formula for that, and find that $L(n) = L_1 \frac{n(n+1)}{2}$

If we plot that, it curves upwards too, and is in fact quadratic.

## Calculus Preview

The astute, experienced reader will recognize that we’ve made an approximation. Introductory physics class usually works with massless springs. And that’s completely wrong here; the point is that the slinky is stretching under its own mass. So our approximation—in this case, our conceptual model—is to treat our massy spring as a slew (n) of small massless springs, in series, supporting tiny masses of (M/n) each.

There is some anxiety about the ends. Is it right to put a whole mass at the end? Won’t that stretch the bottom slink more than it would stretch in reality? Sure. But then we say, just increase n. Slice the slinky into smaller and smaller pieces. Our approximation will get closer and closer to the smooth, continuous model, just as $\lim_{n\to\infty}\frac{n(n+1)}{2} = \frac{n^2}{2}$.

Which is the same thing that happens when you use a Riemann sum to figure out an integral. But before we do calculus on this, let’s put the physics back in.

## Once More, with Real Physics

Suppose our hanging slinky has mass M and that there are n slinks. Hooke’s law states that $F=k \Delta x$ where F is the force, $\Delta x$ is the amount the spring stretches, and k is the “spring constant.” We have a golden opportunity to make a crucial mistake right here. I will save you from it: the spring constant is not as constant as you think. Consider this: suppose we have a spring and if you apply a force of 10 Newtons, it stretches 2 cm. The spring constant k is F/x, or 5 (N/cm). Now cut it into 2 springs. If we weld these springs together in series so they look just like the original spring, and stretch it 2 cm, each spring will experience 10 N but stretch only 1 cm. So each of their spring constants is not 5 but 10. If you cut a spring in half, the spring constant doubles. This means that k is a property of the particular spring, not a property of the material.

If this reminds you of series and parallel circuits, bravo!

Now we can figure out the total length of the hanging spring. Let’s add up the n displacements. We’ll use $\Delta x = F/k$ and see something like this: $(\Delta x)_{total} = \frac{F_1}{k_1} + \frac{F_2}{k_2} + \frac{F_3}{k_3} + ... + \frac{F_n}{k_n}$

From what we said before, $k_m = m k$. The weight (force) of one slink is $F_1 = Mg/n$. And since every slink holds all its predecessors, $F_m = m F_1$. So: $(\Delta x)_{total} = \frac{F_1}{nk} + \frac{2 F_1}{nk} + \frac{3 F_1}{nk} + ... + \frac{n F_1}{nk}$

We do the sum and wave our limit hands to get $\displaystyle (\Delta x)_{total} = \frac{Mg}{n} \frac{1}{nk} \frac{n(n+1)}{2} \approx \frac{Mg}{2k}$.

I think this is correct, but it does not actually answer the question we want to answer. It’s still useful, however: Since the mass is spread evenly over the whole spring, the massy spring stretches half the distance it would if the whole mass were at the bottom.

But that doesn’t tell us how the length depends on n. I mean, the n‘s cancelled—does that mean that the length of the hanging slinky is independent of n? Of course not, and that is the glory of answer analysis. I know in advance—because of the data—that I’m looking for a quadratic.

The problem is that M and k are functions of n. If we let N be the number of slinks in the entire slinky, including the part that’s not hanging down, and define M as the entire mass, and k as the spring constant of the entire spring, then $M(n) = M (n/N)$ and $k(n) = k (N/n)$ (this latter requiring a little thought).

Substituting, we get $\displaystyle L(n) = \frac {M (n/N) g}{ 2 k (N/n)} = \frac {Mg}{2k} \frac{n^2}{N^2}$.

## Doing it with Calculus—and a sweet trick

There are many ways to screw this up if you try to convert our sum-and-limits technique to an integral. The problem arises because as you make infinitesimal bits of spring, the spring constant blows up. Of course you can get it right, but I found a way that makes more sense to me. This must have been done countless times before, but I can’t find them, so I can’t refer you to them or give what must be a proper term. You can tell me and I’ll edit this.

The problem is how we define the spring constant in $F = k \Delta x$. The constant k is larger when the spring is stiffer, when it takes more force to stretch the spring the same amount. But we could just as easily have defined it in terms of stretchiness. I’ll use s for stretchiness, and say that $\Delta x = s F$.

Now, when s is larger, the spring is stretchier, and the same force stretches the spring more. Obviously, s = 1/k, but it’s taking the perspective of finding the stretch instead of finding the force. And its units are centimeters per Newton, which totally makes sense.

This alone will make our sum easier. Now when we cut our springs into parts, the stretchiness s of each sub-spring is proportional to its length. But let’s take a step further and get a property of the material.

If the mass of the whole spring is M, we can define a material stretchiness $\sigma = s/M$, which is now stretchiness per (kilo)gram of spring (cm/N kg). Now we can find the stretch of a spring using our modified, inverted, universalized Hooke’s Law: $\Delta x = \sigma m F$, where m is the mass of that spring.

Now the integral is a (relative) piece of cake. We can find the stretch of an infinitesimal spring of mass dm: $\mathrm{d}x = \sigma F \mathrm{d}m$.

The new constant $\sigma$ is now a property of the material, not the spring. So it’s an actual constant we can pull out of the integral. The force F is still a function of mass; in fact, it’s the mass up to that point in the spring. That is, if we’re integrating over the spring’s total mass M, from the tip to where it’s supported, $F(m) = mg$. So $\Delta x = \int \mathrm{d}x = \int_0^M \sigma mg \, \mathrm{d}m = \frac {\sigma M^2 g}{2}$.

This is the same as the result we got before, but is prettier. If we know the mass M of the slinky, we can use the data to calculate sigma.

## The Wrinkle

If you’re lucky, you stop here. But if you are foolish enough to look more deeply into your data, you’ll notice something disturbing. Here is our data again, but with a residual plot: Oh, no. The residuals show a clear pattern. If you’re a parabola-residual-plot veteran, you know that since the residuals look kinda like a parabola, the coefficient (P) must be too small. But if you bump P up so the residuals are straight, they are no longer flat. The model will rise above the data, and the residuals will trail down to the right.

One response is to shift the curve to the right, as in this graph which shows a new model: The residual plot is the same scale as the one above: adding a new parameter to the model, the horizontal shift dC, seems to be warranted by the data—but is it warranted by the physics?

I love this situation, by the way, and it comes up all the time: we have interesting data, and can model them with a parabola. We could stop there.

But if we can, we can dig more deeply and figure out a good reason for the data to be quadratic of the form $y = Px^2$. We can find the meaning of our parameter P. And we could stop there.

But if we dig even more deeply, and notice that the model, which looks great at first, doesn’t quite match the data, we have the chance to learn something new. You can read my explanation in the next post. 