Wave equation from hooke law - simple question In the paragraph http://en.wikipedia.org/wiki/Wave_equation#From_Hooke.27s_law
it is said, regarding the u(x) function, that 

Here u(x) measures the distance from the equilibrium of the mass
  situated at x.

I can't understand where this point x is.. is is the left end where the hypothetical array holds? Why is it called "equilibrium of the mass" point? I'm missing something..
 A: They've done a bit of reuse of variables here. I'm using x and X to reduce confusion.
Basically, they have taken an array, with separation h between each point. Each mass is initially at equilibrium. Let us consider a ball at a distance x from the beginning of the grid, and solve the problem in its vicinity. Now the equilibrium positions of the particles in the neighborhood are x, x+h, x+2h, etc. Now, they have defined u(X) as the displacement from equilibrium of the mass whose equilibrium position is X. So, u(x) is the displacement of particle at x, u(x+h) is displacement of particle at (x+h), and so on. Then they've used this notation go derive stuff.
The reason that they've chosen a particle at x and solved for its neighborhood, instead of taking the particle at the leftmost end, is that the final wave equation must be a function of x and t. (more accurately, $x-vt$). So then they get a general expression for displacement of particle at x (after limiting h to zero, we get a continuous row of particles, instead of a discrete one, so the notation of x becomes more appropriate)
Note that actually there is no leftmost end, as $N\to\infty$.
