# Propagating a wave pulse -- the math

#### Prerequisites

In describing our one-dimensional transverse waves on an elastic string we start conceptually with a simplified model consisting of massive beads connected by taut massless springs, something like the figure shown below.

In order to describe the motion of the string mathematically, we've specified a coordinate system and numbered the beads. The generic unspecified bead is labelled "$i$, with i ranging from $0$ up to $N$".

## Setting up the math

If we want to describe the position of each bead as a function of time we have to write the function as a function of two independent variables: the time, and which bead we are talking about. So in general, we would have 2N functions:

$$x_0(t),\;y_0(t),\;x_1(t),\;y_1(t),\;x_2(t),\;y_2(t),\;...\;x_i(t),\;y_i(t),\;...$$

This is a LOT of functions to deal with — 2N where N might be very large.  But in this model we are ignoring the back and forth ($x$) motions of the beads and only considering the up and down ($y$) motions. So we can drop half of the functions (the $x$'s) and only consider the functions that describe the up and down motion of the beads:

$$y_0(t),\;y_1(t),\;y_2(t),\;...\;y_i(t),\;...$$

or more concisely, $\{y_i(t), i = 0,...N\}$.

This is still too many. To make this "easier" we are going to do something that is necessary but can be confusing: we are going to label the bead by where it is along the x axis, instead of by which bead it is in the sequence. This becomes much easier when we go back to a continuous string model. Then the up-down ($y$) displacement of each piece of the oscillating string is a function of which bit of the string we are choosing to look at (labeled by $x$ ) and when we are looking at it (labelled by $t$). So instead of having $x(t)$ = some function of $t$, we have $y(x,t)$ = some function of $x$ and $t$.

The nasty part of this is that by doing this, we are making $x$ into an independent variable. When we were doing the mass on a spring, we were treating $x$ as a dependent variable. So for the mass on a spring, the $x$ was the answer that we calculated after we picked a time. For the elastic string, we calculate $y$ after picking $x$ and $t$ to be whatever we choose. The equations looks somewhat similar but have very different interpretations. This is one place where just reaching for an equation without thinking what it means physically can produce serious nonsense.

## A couple of pulse shapes

Two functions that produce a nice symmetric pulse shape and are easy to calculate are the Gaussian and the Lorentzian functions (created by mathematicians named Gauss and Lorentz respectively). These are general mathematical functions, but we put in parameters that will allow the units to come out correctly as a displacement.

The parameter $A$ represents an amplitude (the height) of the pulse, and $d$ controls the width of the pulse. If we look at a pulse centered around $x=0$ at a time $t=0$, here is what the equations would look like:

$$y_G(x,0) = Ae^{-(x/d)^2}\quad\mathrm{Gaussian}$$

$$y_L(x,0) = \frac{A}{1+(x/d)^2}\quad\mathrm{Lorentzian}$$

And here is what their graphs look like (scaled with $A = 1$ and $d = 1$).

Notice that they both have the same height and widths — at least in the equation — but the Gaussian cuts off faster than the Lorentzian. Both of these shapes are convenient for describing pulses that have a limited extension.

## Making the pulse move

Now we want to write an equation that makes the pulse move down the elastic spring with a constant velocity. To do that, we need to consider what it is that makes the graph of a general function, $f(x)$ shift by some amount. We already encountered this question when thinking about the oscillation of the mass on a spring — shifting the cosine curve so it started at different times. (See the discussion "Interpreting the phase shift" in the page Reading the content in the harmonic oscillator solution.)

Suppose we have the graph of a symmetric pulse with a peak at $x = 0$ as shown in the figure at the left below. If we want to shift the peak, so that it has its peak at $x = 2$ instead of at $x = 0$ as shown in the figure at the left, how can we do it?

One way to think about it is to create a new coordinate system, $x'$ with the origin ($x' = 0$) at the position $x = 2$. Since the coordinates $x$ and $x'$ are only shifted, they must be related by $x' = x + b$ where $b$ is some constant. But what is $b$?

Since we know $x = 2$ corresponds to $x' = 0$, putting these values in gives $0 = 2 + b$ or $b = -2$. The the function $f(x') = f(x-2)$ peaks at $x = 2$. This makes sense since when we put in $x = 2$, we get $f(x-2) = f(2-2) = f(0)$ which is the peak value of the function.

So in general, if we want to shift a function to the right by an amount $a$, we substitute $x-a$ into the function everywhere there was previously an $x$. So the Gaussian function

$$y_G(x,0) = Ae^{-\big(\frac{x-b}{d}\big)^2}$$

has amplitude $A$, width $d$, and its peak is centered at $b$.

If we now make $b$ increase with time linearly, by setting $b = v_0t$, the function will move to the right with speed $v_0$:

$$y_G(x,t) = Ae^{-\big(\frac{x-v_0t}{d}\big)^2}$$

In general, if we have an arbitrary shaped string at a time $t=0, y(x,0) = f(x)$, then writing

$$y(x,t) = f(x-v_0t)$$

will move the shape $f(x)$ to the right (towards more positive $x$) with a speed $v_0$. Similarly, writing

$$y(x,t) = f(x+v_0t)$$

will move the shape to the left (towards more negative $x$) with a speed $v_0$.

Joe Redish 3/28/12

Article 685