# Wave speed

#### Prerequisites

The video below (created by M. Wittmann) shows a pulse traveling on a long taut spring.

(If it runs too fast to see, try maximizing the video and starting it with your space bar.)

The pulse seems to move with a constant velocity. What could that velocity depend on? In the beads & massless springs model, the velocity of a pulse is about the springs pulling on the beads due to the springs not being perfectly aligned, as in the figure below.

How quickly the bead responds depends on how hard its pulled — the tension, $T$, and the inertia of the beads — the mass of a bead, $m$. We expect a bigger tension would make the beads respond faster and a bigger mass would make the beads respond slower.

Now to actually derive the wave speed is not TOO bad — it just requires applying Newton's second law to a small bit of string. But the problem is that since we are describing the motion of the shape of the string, we are writing a differential equation for a function of two variables — $y(x,t)$. The result is a partial differential equation — one that relates not just the position and its time derivative like when we had one mass and Newton's second law — but one that relates the time and space derivatives. This is definitely "beyond our pay grade" here. So we will not go through this. (See the wikipedia article, String vibration for a clear derivation.) Instead, we will see what we can do with dimensional analysis alone.

The wave speed is going to be determined by the force parameter, $T$ (the tension), and an inertial parameter.  For a long spring what should we choose for inertia? In the beaded model it might depend on the mass of each bead and the space between them. But we don't really have a beaded model. What might we use for a continuous elastic string? In this case, we expect the speed to depend on the local properties of the string. If we doubled the length of the string (and therefore doubled the mass) we wouldn't expect the moving pulse to know — at least until it got to the end of the first string. So it wouldn't make sense for the speed to depend on the total mass. So if it is going to depend on the mass, it probably should depend not on the total mass, but on the mass per unit length.  If $m$ is the mass of the whole string, and it has length $L$, we would expect it to depend not on m itself, but on the mass density, which we define to be $μ$ (mu):

$$μ = m/L.$$

(This choice is a good example of how we blend the simple math of dimensional analysis with thinking about the physics of the situation.)

So the question for dimensional analysis is: can we make a speed out of $μ$ and $T$?

Aargh! Be careful here! We have two different "T"s in this problem: $T$ and T. You should be able to tell which is which by the context. Remember that when we do a dimensional analysis we label what kind of quantity everything is: mass (M), length (L), time (T), or charge (Q). And we use these symbols M, L, T, and Q.  These can be confusing since we also use these for variables and for units. But when you see a quantity surrounded by a square bracket "[ ...]" it means that we are asking its dimensionality so what goes on the right are to be interpreted as the symbols MLTQ and NOT as variables or units.

So what are the dimensionalities of the quantities we are trying to relate? These are $μ$, $T$, and $v_0$ (the wave speed). Since $μ$is a mass per unit length, $T$ is a force, and $v_0$ is a velocity, we have the dimensionalities

$[μ] =$ M/L

$[T] = [ma] =$ ML/T2

$[v_0] =$ L/T

We want to use $T$ to get L/T but it has a mass in it. So that suggests we divide it by $μ$ to get rid of the M. This gives dimensions:

$[T/μ] =$ (ML/T2)/(M/L) = L2/T2.

This sort of looks like a velocity, or at least like the square of one. If we take the square root of $T/μ$ it will at least have the right dimensionality (and therefore the right units). This suggests that we propose the hypothesis that the wave speed is given by

$$v_0 = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{TL}{m}}$$

where $T$ = tension of the string, $\mu$ = mass density of the string which equals $m$ = (mass of the string) / $L$(length of the string).

It turns out that when we do the full math (with the partial differential equations!) this turns out to be exactly correct. There are no extra factors of 2 or $π$ or whatever. Those factors would be invisible to our dimensional analysis argument.

Remember! This is the speed that displacements move along the string; it is NOT the speed with which the bits of the string are moving transversely. Those can be anything and depend on how the original motion moving the first bit of string up and down were moved.

Joe Redish 3/28/12

Article 686