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I've recently been re-learning some physics, and a question came to me when looking over Hooke's law:

In the following I am always assuming that the force required for permanent deformation is sufficiently large and thus irrelevant for my purposes.

Hooke's Law gives that the force applied to a spring is linearly proportional to the displacement, i.e. $F \propto x$. I am wondering how one might accomplish a system where the proportionality is quadratic ($F \propto x^2$). What about cubic? ... What about for any arbitrary degree? ($F \propto x^n$) by using only linear components such as ideal springs that obey Hooke's law.

One thought that came to my mind was that this might be possibly by using some combination of multiple springs, some in series and some in parallel. But I am not certain if this would work.

Alternatively, in my research prior to posting this I discovered other types of springs such as a torsion spring, possibly something like this could be used in combination in some way?

The primary focus of this question is to understand whether a system built out of ideal spring-like components that react with forces linear to their displacement, can as a whole, exhibit a force that reacts nonlinearly to its displacement.

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    $\begingroup$ Welcome! This is a very interesting question, one tip though: you may want to better focus the allowed range of solutions you are interested in here. You mentioned doing it with several springs in parallel/series which is great, but when you also just ask in general "how one might accomplish" this, then there are probably too many ways. Some that come to mind involve either using different sorts of materials with nonlinear properties, or incorporating gearwheels into the design to control the rate of the spring's extension... $\endgroup$
    – Amit
    Jun 1, 2023 at 14:39
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    $\begingroup$ Variable rate spring $\endgroup$ Jun 1, 2023 at 15:01
  • $\begingroup$ It depends. If you are allowing tangential motion and collisions, then even a system made from ideal linear springs will behave in a non-linear fashion. $\endgroup$ Jun 1, 2023 at 20:59

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Here's a set-up that will give a cube law response for small displacements. Join 2 springs (each of natural length $l$ and obeying $T=kx$) end-to-end and anchor the far end of each, so that the separation of the anchorages (A and B, say) is $2l$.

Now displace the join by $y$ ($y\ll l$) at right angles to the line AB.

You can show using Pythagoras's theorem that, for $y\ll l$ and therefore spring extension $x\ll l$, $$y^2=2lx$$ You can also show that, for $y\ll l$ the force, $F$, needed to displace the join is related to the spring tension, $T$, by $$F=2T\frac yl.$$ Even though the springs obey $T=kx$, you'll find that $$F=\tfrac k{l^2}y^3.$$ Two points to note: (a) The non-linearity is geometric in origin. (b) The range of displacements, $y$, over which the cube law holds is limited. I suspect that the limited range will be the case with any configuration of Hookeian springs.

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  • $\begingroup$ I am trying to follow the steps you outlined, and I am not getting the same approximation for $F$, I'm getting a squared relation to the small displacement $y$. I found that the spring displacement $x$: $$x = \sqrt{l^2+y^2}-l \approx \frac{y^2}{2l}$$ Which is the same as you wrote (last step is by Taylor expansion). But then isn't computing the total force of springs reaction is done by just substituting $x$? So: $$F = 2kx = \frac{ky^2}{l}$$ ? $\endgroup$
    – Amit
    Jun 1, 2023 at 15:44
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    $\begingroup$ @Amit No. $F\neq 2kx$. $F$ is applied to the join in a direction at right angles to AB. You need to do some force resolution (or equivalent) to relate it to the tension $T$ in the springs and hence to the extension, $x$. $\endgroup$ Jun 1, 2023 at 16:06
  • $\begingroup$ Got it! It's actually $\frac{2ky^2}{2l} \cdot \sin\varphi$ where $\varphi$ is the angle of displacement of the join. Thanks. $\endgroup$
    – Amit
    Jun 1, 2023 at 16:16
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    $\begingroup$ And $\sin\phi\approx\frac yl$ $\endgroup$ Jun 1, 2023 at 16:24
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If motion isn't constrained to one dimension, Philip Wood's answer nicely illustrates that the force is not necessarily proportional to displacement. If motion is constrained to one dimension only, it can be shown that the net force of a system of ideal Hookean springs is proportional to displacement. Here is a "proof" by explicit construction of a system of linear equations for any network of springs.

Suppose we have a network of $n$ springs connecting a series of $m$ "nodes". The springs are oriented horizontally, and can only push/pull the nodes left or right. We define a parameter $A_{ij}$ as follows. If spring $j$ is not connected to node $i$, $A_{ij} = 0$. If spring $j$ is connected to node $i$ from the right (pulling it to the right when extended), $A_{ij} = 1$. If spring $j$ is connected to node $i$ from the left, $A_{ij} = -1$.

Now consider spring $j$ with spring constant $k_j$. It is connected between nodes $a$ and $b$. Let $x_a$ and $x_b$ the amounts by which nodes $a$ and $b$ are displaced from their equilibrium positions. Node $a$ lies to the spring's left and $b$ to its right. The amount the spring is stretched by is $x_b - x_a$, so its tension (the force it pulls with) is $t_j=-k_j(x_a-x_b)$. Since $A_{aj} = 1$ and $A_{bj} = -1$, we can write $$t_j = \sum_{i=1}^m -k_jA_{ij}x_i.$$ Note: this tension is in addition to any tension the spring may have had in the equilibrium configuration of the system.

Finally, in equilibrium, the total force on node $i$ must be zero. Including any external force $F_i$ that you apply on node $i$, we can write $$\sum_{j=1}^n A_{ij}t_j + F_i=0.$$ $$\sum_{\ell=1}^m\sum_{j=1}^nk_jA_{ij}A_{\ell j}x_\ell = F_i.$$ This is a linear system of $m$ equations (one for each node) in $m$ unknowns (node displacements $\{x_i\}$). These equations make it clear that if $\{x_i\}$ are the displacements for applied forces $\{F_i\}$, $\{\alpha x_i\}$ are the displacements for applied forces $\{\alpha F_i\}$.

It may not be obvious, but a network of ideal springs is entirely analogous to a circuit of resistors, also a linear system. Displacements are analogous to node voltages, spring forces to currents, spring constants to conductances (inverse resistance), and external forces to current sources. The parameters $A_{ij}$ constitute a matrix called the incidence matrix.

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Piano wires exhibit this sort of behaviour- specifically, the larger ones for low notes, that are made of a longitudinal core with a transverse outer coil. The friction between these components increases the wire's resistance to stretching in a non-linear way.

There are many papers on this topic- e.g.

https://www.researchgate.net/publication/241345038_Piano_string_modeling_From_partial_differential_equations_to_digital_wave-guide_model/link/54da45f20cf261ce15cc5e10/download

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