# A basic question on the derivation of the wave equation

Today I saw the derivation of the wave equation in class, and I did not understand the following step.

We are modeling a uniform-density string as being made up of tiny masses spaced a small amount $h$ apart connected by springs obeying Hooke's law. Let $y_i(t)$ be the vertical position of the $i$'th such particle. The mass of this particle is $(h/L) M$ where $L$ is the total length, $M$ is the total mass. On the other hand, the displacement of the string between the $i$'th and the $i+1$st particle can be shown to be approximately proportional to $y_{i+1}(t) - y_i(t)$.

It seems to me that this leads to the equation

$$\frac{hM}{L} y_i''(t) = k \left( [y_{i+1}(t) - y_i(t)] + [y_{i-1}(t) - y_i(t)] \right)$$

but that is, in fact, wrong; the derivation by instructor gives sticks a $1/h$ on the right hand side.

Can someone explain to me (i) why the right hand side needs a $1/h$ (ii) intuitively, what is happening?

My first guess is that the spring constant $k$ should be proportional to the length of the spring, keeping everything else constant? But that seems weird to me; it seems to me like if you have a spring with spring constant $k$, then cutting the same spring into two halves should result in two springs each with half the length but the same spring constant. Can someone unconfuse me?

Thanks.

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## Just addressing the question of the spring constant of the cut down spring pieces:

For a typical material spring you will find that cutting it in half result in two springs with twice the spring constant.

To understand why imagine that you tie the two half back together again. At their natural length the system look like:

O~~~x~~~O


where O is an end, x the place you tied them together and ~ a piece of spring. Now you pull them apart by a distance $2s$:

O~~~~x~~~~O


because the two halves are identical they each stretch by $s$. You also know that each spring exerts the same force $F$ on x (because x is not accelerating and neither is either segment of spring).

So for the combined spring

$$F = -k_c (2s)$$

and for each half-spring

$$F = -k_h s$$

Thus $k_h = k_c*2$, and you can generalize this to a fragment of length $h$ taken from a total length of $L$ to find that you need a factor of $L/h$.

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Thanks, that unconfused me. –  alex Feb 5 '11 at 3:48

Probably you got into trouble because of the detailed quantity $\frac{hM}{L}$, which is simply the mass of small weights. However, how many of them are there? Let's suppose you string is made up of $N$ small weights, each weight separated by a small space of length $h$. Therefore, we can see that the total length is $L=N h$. Likewise, the total mass should be $M=Nm$, right?. Having this in mind, we can find that the expression $\frac{hM}{L}$ is actually the mass $m$.

Notice also that your $k$ is actually the stiffness of each spring. However, the total stiffness is given by $K = \frac{k}{N}$. Substituting this value in your equation, we have

$$\frac{hM}{L} y_i''(t) = KN \left( [y_{i+1}(t) - y_i(t)] + [y_{i-1}(t) - y_i(t)] \right)$$

However, we know can put $N$ in terms of $h$, that is, $N = \frac{L}{h}$.

$$y_i''(t) = \frac{K L^{2}}{M} \left( \frac{y_{i+1}(t) - y_i(t) + y_{i-1}(t) - y_i(t)}{h^{2}} \right)$$

There you have the $h$ you were looking for.

By the way, take a look at the difference inside the parentheses. They are very interesting because those terms provide a nice way to understand the Laplacian.

UPDATE: I realized I didn't answer your specific question "Why does the right hand needs an (extra) h?". Well, you're equating two forces, therefore, the right hand needs to have something involving second order derivatives. You need that $h$ to obtain such derivative (that is, the whole thing inside the parentheses) when the separation between weights tends to $0$. However, in my opinion, it's easier to understand the derivation from first principles. Also, you could make this clearer if you write explicitly the dependence of $y$ on $h$.

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By the way, I didn't finish the derivation because I think you already know it. –  Robert Smith Feb 5 '11 at 2:25