# Equivalent spring-constant for infinite square grid of springs

Consider an infinite square grid, where each side of a square is a spring following Hooke's law, with spring constant $k$.

What is the relation between the force and displacement between two points? If they are proportional, what is the equivalent spring constant between the origin and the point $(x,y)$ (integers) ?

Edit 1: I also want to know this: Suppose you make the springs so small that this can be treated as a continuous sheet, at what speed will a wave propagate? Assuming a wave starting as an initial displacement perpendicular to the sheet.

Given some initial state, is there an equation for the time-evolution of the continuous sheet?

Edit 2: Suppose there is a mass at every node, and its $(x,y)$-coordinates is fixed, it only vibrates out of the plane. Consider that we take the continuous limit, such that we get a 2D membrane of mass density $\mu$.

1. Is the membrane isotropic?
2. Suppose we use another tiling (like hexagonal) before taking the continuous limit, will this sheet behaves the same way?
3. If not, but they are both isotropic, how does one characterize their difference, can they be made to behave the same way by changing the spring constant $k$?
4. What is the equation of motion for the square sheet with spring constant $k$?
5. What is the equation of motion for the square sheet if the springs obey a generalized Force law, $F=kx^n$, where $n$ is a variable.
6. What is the equation of motion for a 3D cubic grid?

I am particularly interested in answers to 1., 2. and 3. I dont expect anyone to answer all these and will also accept an answer which does not explain anything but simply provides a good reference.

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I'm not sure I understand this. If you just pick two points, you're not getting enough information. You need to know the displacements of every node connected to a given point in order to find the force on that point. – Mark Eichenlaub Dec 21 '10 at 13:52

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This is an infinite dimensional problem. Each connection between springs is roughly one degree of freedom. You can analyse the equilibrium case, by assuming all springs must displace equally (thus reducing the nº of degrees to 1). – Bruce Connor Dec 21 '10 at 15:06

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sniping physicists! – Jeremy Dec 21 '10 at 15:28

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This looks like a model for a crystal except it can't be stable like this. You can continuously deform the angle between the sides of the square into a rhombus and eventually (letting the angle go to zero) obtain a one-dimensional model. Reasonable crystal models also add diagonal interaction into the square that makes this angle deformation disadvantageous (because you would make the diagonal spring twice longer if you'd collapse the square). – Marek Dec 21 '10 at 18:23

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@Kalle43 You should simply your question, not complicate it. It is a very complex system you suggested. Reduce it to manageable limits, state assumptions, describe the grid, and you might get your answer. – Bruce Connor Dec 21 '10 at 19:47
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## 3 Answers

I'll answer only the third one (for now at least); the movement with limit to small vertical oscillations will be governed by the drum equation:

$\ddot{s}(x,y)=c^2 \nabla^2 s(x,y)$

where $s(x,y)$ is a vertical displacement in point $(x,y)$ and $c$ is the weave speed; using dimensional analysis I would say that $c\sim\sqrt{\frac{k}{\sigma}}$, where $\sigma$ is the mass density. Of course everything is getting much more complex with larger amplitudes.

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I don't believe this is really correct. The wave equation is derived from the assumption that nodes oscillate around stable equilibrium positions. This is not the case in the spring model which can be also arbitrarily deformed (i.e. it is not rigid). – Marek Dec 21 '10 at 19:38

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@marek please find a copy of Goldstein's mechanics. If it is one of the newer editions (with Poole and Safko) then have a look at Ch. 13 "Introduction to Lagrangian and Hamiltonian Formulations for Continuous Systems and Fields." If you find any inconsistencies in that treatment then please post a question because that would be big news. – user346 Dec 21 '10 at 20:16

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@Marek This is small deformation approximation. – mbq♦ Dec 21 '10 at 20:36

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@mbq: okay, but then it should be noted that it's not valid (at least not obviously) because you can always deform the square into a rhombus without expending energy in this model. While the small approximations are usually done around stable solutions that you can't deform in this way. – Marek Dec 21 '10 at 21:08

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@Marek Small vertical deformation than. However, those horizontal motions in linear approximation will also have a speed proportional to $\sqrt{\frac{k}{\sigma}}$. – mbq♦ Dec 21 '10 at 22:49
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I stick to the first question.

If you only do small displacements, and the two points are along the same line of springs then the effective spring rate is

$$k_{eff} = \frac{k}{N}$$

where $N$ is the number of springs between the points. Why? Well split the problem like this

````(inf)---[k_out]---(A)---[k_in]---(B)---[k_out]---(inf)
````

where `(A)` and `(B)` are the two points, and the springs are replaces with the effective springs `[k_out]` between the points and infinity, and `[k_in]` between the two points. The formula for springs in series is $\frac{1}{k_{eff}} = \frac{1}{k_1}+\frac{1}{k_2}+\ldots$, or $k_{eff}=k/N$ if all the springs have the same rate. So `[k_out]` is zero because $N=\infty$ and whats left to consider is only the springs in-between the points.

Note that the springs out of the line of the points are un-important for small displacements because they only contribute higher order non-linearities.

Completely different equations are needed for the continious sheet. The wave speed has to do with the mass/density of the sheet also, not just the elasticity and the sitffness.

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If on every node of the grid you have a small mass, then you have a model for a two dimensional solid. That would behave like a two dimensional membrane. The equation of motion for every disturbance would be a wave equation. In the case of an one-dimensional grid, the wave velocity for such a wave would be

$$c^2=\frac{kl^2}{m}$$

where l is the distance between two neighbouring masses. In the case of the two dimensional grid, you will probably also have a geometric factor.

Update: Regarding the last edit, if you have a tension that characterizes the membrane, then the velocity is the square root of the tension over the mass density. So the geometry of the thing would play some part both to the tension and the mass density. That is because if you change the shape of the cell, then you assign different surface for every mass and you also assign a different number of brunches with springs to every node thus changing the effective spring constant. These are my qualitative guesses.

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"then you have a model for a two dimensional solid" -> no you don't. See my comment under the question. In short: this model is not stable to perturbation that collapses it to 1D and so it is essentially just a 1D model. – Marek Dec 21 '10 at 19:34

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How does that happen for an infinite grid? Anyway, that doesn't matter for the wave velocity. – Vagelford Dec 21 '10 at 20:09

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""then you have a model for a two dimensional solid" -> no you don't", yes you do @marek. I took a look at your comment and I failed to understand your argument. – user346 Dec 21 '10 at 20:11

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@space_cadet: no you don't :-) @Vagelford: the same way it does for square. These things are best illustrated with a picture. All I can say is: think about the same model but with the elementary face being a rhombus instead of a square. You can continuously deform these models into each other. Therefore these can't be good models for a solid because those models need to be... well, solid :-) – Marek Dec 21 '10 at 21:13

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Yes, but you would have to do it simultaneously for all the infinite grid, so a small perturbation can't do it. Anyway, I didn't say it is a good grid model. It is a good enough for small oscillations. – Vagelford Dec 21 '10 at 21:28
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