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 $\mu$. 


*

*Is the membrane isotropic?

*Suppose we use another tiling (like hexagonal) before taking the continuous limit, will this sheet behaves the same way?

*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 $k$?

*What is the equation of motion for the square sheet with spring constant $k$?

*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.

*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.
 A: 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.
A: 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.
A: 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.
