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Following the derivation on the relevant Wikipedia page, I am having a bit of trouble moving from the following line, with the case of 3 particles in a row: $$ \frac{\partial^{2}}{\partial t^{2}} u(x+h,t) = \frac{k}{m} \left[u(x+2h,t) - u(x+h,t) - u(x+h,t) + u(x,t) \right] $$

Which I understand. The force on the middle particle is equivalent to the vector sum of the Hooke's law forces on the two around it, but when you generalize it to: $$ \frac{\partial^{2}}{\partial t^{2}} u(x+h,t) = \frac{K L^{2}}{M} \left[ \frac{u(x+2h,t) - u(x+h,t) - u(x+h,t) + u(x,t)}{h^{2}} \right] $$ (where the notation is directly borrowed/explained at the Wikipedia page.)

I understand how the new variables $K, L$, and $M$ are defined. My question is, why the $u(x+2h,t) - u(x+h,t) - u(x+h,t) + u(x,t)$ doesn't change when you generalize to multiple particles.

My intuition is that, regardless of how many particles exist around the central particle, the only forces that are relevant to it are the ones adjacent to the particle itself, the same ones in the $N = 3$ case. Is that a right way of thinking?

I would like, if possible a rigorous explanation of how we move from the first line to the second, which is both conceptually and mathematically correct.

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  • $\begingroup$ It's called the nearest neighbor approximation. $\endgroup$ – Peter Diehr Apr 21 '16 at 2:01
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You are not changing anything other than writing the spring constant for one spring in another way. You are still considering the same three particles you started with.

The sequence is as follows.
One spring of spring constant $k$ is in series with $N-1$ other springs.
So the spring constant of the $N$ springs $K= \frac k N$.
This comes about because if you apply a force $F$ to one spring and get an extension of $x$ then the same force applied to $N$ springs in series will give an extension of $Nx$ with each of the $N$ springs extending $x$.
So the spring constant of the whole array of springs $K = \frac{F}{Nx} = \frac k N \Rightarrow k = NK$

The mass of the whole array is $M = Nm \Rightarrow m = \frac M N$ and the length $L = Nh \Rightarrow N= \frac L h$

so from you first equation $\dfrac k m = \dfrac{NK}{\left (\frac M N \right)} = \dfrac {N^2 K}{M} = \dfrac {L^2 K}{Mh^2}$

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  • $\begingroup$ Right, so even if there are more than 3 along the line complete line, we are just concerned about the original 3? $\endgroup$ – Αδριανός Apr 21 '16 at 19:01
  • $\begingroup$ Nothing has changed except that $\frac k m $ has been replaced by some new constants $\endgroup$ – Farcher Apr 21 '16 at 19:05

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