# 10.3 The physics of springs  (Page 3/11)

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Our solution to this problem involves experimental stacking. Due to the fact that our experiment is easily repeatable under different conditions, we can arrive at more observations by simply altering the forces acting on the table. This, however, should not alter any of the spring constants. By doing so, we generate a second set of data which is still dependent on the spring constants. We are left with 16 springs, but we now have 28 degrees of freedom: an overdetermined system. The experiment can be repeated s times to minimize experimental error as well, yielding 14 s degrees of freedom and the same 16 values in K .

With this experimental technique in hand, we can revisit our primary equation:

${A}^{T}KAx=f.$

In order to solve for K , we re-express part of our equation:

$KAx=Ke=\left[\begin{array}{c}{k}_{1}\\ & \ddots \\ & & {k}_{16}\end{array}\right]\left[\begin{array}{c}{e}_{1}\\ ⋮\\ {e}_{16}\end{array}\right]=\left[\begin{array}{c}{e}_{1}\\ & \ddots \\ & & {e}_{16}\end{array}\right]\left[\begin{array}{c}{k}_{1}\\ ⋮\\ {k}_{16}\end{array}\right].$

This simple substitution lets us rewrite our equation as

${A}^{T}diag\left({e}^{\left(i\right)}\right)k={f}^{\left(i\right)}-{f}^{\left(0\right)}$

which allows us to solve for k .

At this point, we can apply our experimental technique of stacking. Using s experiments, we may construct the appropriate matrices such that

$B=\left[\begin{array}{c}{A}^{T}diag\left({e}^{\left(1\right)}\right)\\ {A}^{T}diag\left({e}^{\left(2\right)}\right)\\ ⋮\\ {A}^{T}diag\left({e}^{\left(s\right)}\right)\end{array}\right],\phantom{\rule{1.em}{0ex}}f=\left[\begin{array}{c}{f}^{\left(1\right)}-{f}^{\left(0\right)}\\ {f}^{\left(2\right)}-{f}^{\left(0\right)}\\ ⋮\\ {f}^{\left(s\right)}-{f}^{\left(0\right)}\end{array}\right].$

We now recall Hooke's Law:

$f=Bk.$

With f and B , we are now ready to solve for k . However, due to the fact that our system is overdetermined, there does not have to be a unique solution. To find the solution of best fit, then, we turn to the least squares method so as to find k that satisfies

$\underset{k\in {\mathbb{R}}^{16}}{min}{\parallel Bk-f\parallel }^{2}.$

We go about this using the standard method of normal equations: we multiply both sides of the Hooke equation by B T :

${B}^{T}Bk={B}^{T}f.$

This allows us to use the Moore-Penrose psuedoinverse of B solve for k , our vector of spring constants:

$k={\left({B}^{T}B\right)}^{-1}{B}^{T}f.$

Bar graphs of k are presented from a laboratory implementation of this technique. [link] shows the calculated spring constants using the first two experiments from Data Set A. These spring constants have 246.7% error (see "Notes: Our Data Sets, Measuring Spring Constants, and Error" for notes on data sets and calculating error). [link] shows the measured spring constants.

Due to our technique stacking of s experiments, our solution should minimize experimental error. However, it should be clear that, due to the amount of experimental error involved, we are extremely unlikely to arrive at an exact solution to the problem. It is important to note that, though wildly inaccurate, we can correctly identify the stiff spring in the system.

## Our question

In theory, this works out beautifully. Unfortunately, our experiments are carried out in the real world, so the entire process is rife with error. Measurements of both the masses and the positions of the nodes may be imprecise, and the alignment of the webcam may be off (See [link] ), introducing the keystone effect, all of which dirties the displacement data. The forces may not be perfectly aligned along the horizontal and vertical (See [link] ), and the masses we hang may not be exactly what we believe, introducing error in the force vector. Furthermore, Hooke's Law is an approximation, valid only in a certain range (although we keep our forces within that range). We also assume that the springs lie at angles of 0, $\pi /4$ , or $\pi /2$ to the nodes, a belief that is reflected in the adjacency matrix and not at all accurate (See [link] ). Our model also approximates the elongations of the springs linearly and assumes that the pennies are massless points, which introduces further error.

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