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In the paragraph

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..

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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 $N\to\infty$.

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I still have confusion.. why would they use u(x) to indicate the distance from the x point? Couldn't they just use x+something ? One-dimensional case is just a line, isn't it? – Kevin Feb 12 '12 at 13:22
Its just a notation. They could use $x+\Delta x$, but this creates problems as it isnt x thats changing, x is the equilibrium position. This creates more problems when you try writung the expression for the next ball ($x+h+\Delta(x+h)$). The issue here is that usually, when we use $\Delta$, then $\Delta(x+constant)=\Delta x$, as change is a constant is zero. But, the displacement of the second ball is not necessarily equal to the displacement of the first. More to follow. – Manishearth Feb 12 '12 at 13:28
See, in the end, if we wanted to use x+something, we would have to define that something for every different ball (using indices or something). For example, if we used $x+\delta_1$, $x+h+\delta_2$, then it boils down to assigning a $\delta_r$ to each ball. Then you would have to get x in terms of r. This is pretty much the same amount of work as having $u(X)=\delta_{at\space X}$. The added bonus of u(x) is that the x is useful in the wave equation. $r$ is useless for a wave, as we don't consider particles at all. – Manishearth Feb 12 '12 at 13:33
Basically, they've done all this to cut down on the equation size. – Manishearth Feb 12 '12 at 13:35
I understood, thank you! – Kevin Feb 12 '12 at 14:03

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