# Example: Velocity patterns in a pulse

## Understanding the situation

Consider a transverse pulse moving on an elastic string. If we think of the string as a set of small beads connected by massless springs as in Waves on an elastic string, the beads move up and down, perpendicular to the motion of the pulse.

At any instant of time, a picture of the pulse shows a graph of the displacement of the beads at that instant (a graph for the eye). But at that instant, each of those beads might be moving up and down with different velocities. What does the pattern of those velocities look like? Figuring that graph out helps us understand the motion of the pulse.

## Presenting a sample problem

In the figure below are show 4 graphs of an elastic string as a function of position along the string ($x$ is the horizontal axis). If graph A represents the shape of a right-moving pulse on the string at time $t=0$, which graph could represent the graph of the y-velocity of the bits of the string as a function of $x$ at $t=0$?

## Solving this problem

#### A graphical method

The way to figure this out is to use the fact that the y-velocity of a bead at any position is defined by

$$v_y = \frac{\Delta y}{\Delta t}.$$

So let's consider the initial position of the pulse and then see what happens a small amount of time later. Since it's given that the pulse is moving to the right, if it is given by graph A at $t=0$, that looks like the red curve below. At a small amount of time later the pulse will have moved to the right and look like the blue curve in the figure below.

In order to get insight into the velocity pattern in space, we have to ask: how is each bead moving? We can figure out how they move as the pulse shifts from the red curve to the blue if we imagine that there are  beads on each of the vertical graph lines (every 0.2 along the horizontal axis).

At the red curve's instant of time the beads are on the red curve where it crosses the gray vertical lines of the graph. In the time the pulse moves to the blue curve, each bead will move straight up or straight down until it reaches the blue curve. We've drawn arrows to show that motion in the figure below.

Since the time intervals are the same for all, the $v_y$ velocities for each bead are just proportional to $\Delta y$ - the length of the arrows.

So as we move from left to right, we see that the y-velocity starts at 0, gets increasingly negative, then less negative until it's 0, just about at $x=0$. (The reason it is not exactly at $x=0$ is because we biased the time to look at a later time. If we had looked at the shift from an earlier time to a later time, we would have gotten the velocity to be 0 at $x=0$.)

As $x$ becomes increasingly positive, the velocity grows until it's fairly large and then it gets smaller until it goes to 0.

This pattern looks like graph D and this tells us how to interpret graph D as a velocity graph in terms of how each bead is moving.

#### An algebraic solution

We can also solve the problem algebraically. A common and convenient pulse shape is a Gaussian:

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

We know to make the pulse move to the right, we simply replace $x$ everywhere by $x-v_0t$. For the Gaussian this gives

$$y_G(x,t) = Ae^{-(b(x-v_0t))^2}.$$

The instantaneous y-velocity is just

$$v_y(x,t) = \frac{dy}{dt}.$$

Note that the velocity is a function of both $x$ (which bead we are considering) and time. We can take the time derivate by using the chain rule with

$f(x,t) = Ae^{g(x,t)}\quad$ with $\quad g(x,t) = -b((x-v_0t))^2$

The chain rule tells us that

$$\frac{df(g(t))}{dt} = \frac{df}{dg} \frac{dg}{dt}.$$

Since the exponential is its own derivative

$$\frac{df}{dg} = f$$

and the derivative of $g$ is straightforward to calculate by expanding it out:

$$g(x,t) = -bx^2 + 2bv_0tx -bv_0^2t^2$$

so $$\frac{dg}{dt} = 2bv_0x - 2bv_0^2t.$$

So setting $t=0$ we get

$$v_y(x,0) = (2bv_0)A xe^{-(bx)^2}$$

The result is a Gaussian times $x$ times a bunch of constants. This makes the result negative for negative $x$, 0 at $x=0$, and positive for positive $x$. You can easily go to the Desmos graphing calculator and plot $y = xe^{-x^2}$ to confirm that this result looks like graph D.

Joe Redish 6/11/19

Article 692