Hello! So far we have considered unbounded continuum

media. We have seen that these media could bear waves, propagating waves.

But at some point these waves in real systems are going to reach a boundary. What happens then?

And what happens if waves are confined between two boundaries? That is the topic

of the week we start now. Let us see first what happens when a wave

reflects on a boundary.

To make things simple, I will stay with waves on dimension one continuum media, as we did

before. We did that with cables, beams, sound and surface waves.

Remember that we could make a very important distinction between non-dispersive waves and

dispersive waves. Non-dispersive waves existed on the tensioned

cable but also for sound in pipes, or on the surface of shallow water.

The model system for all this was the tensioned cable.

Dispersive waves could be found in many different cases: bending beams, deep water

and the model system was the tensioned cable resting on elastic foundations.

Let us work with the tensioned cable, with or without elastic foundation

What kind of boundary condition can we have on a cable?

The simplest would be that the cable is fixed and cannot move. That would mean that at this

point, we have y=0 for all times. Here is a semi-infinite cable which is fixed at x=0,

so that y(0,t)=0. You can imagine other boundary conditions.

For instance, a free end (with tension) would be dy/dx=0.

How does this condition interact with waves on the cable? What happens when a wave hits

the boundary? We can take first the simple non-dispersive

case. You remember the d'Alembert solution that

states that waves are always the sum of two functions, one of x-ct and one of x+ct,

y=F(x-ct)+G(x+ct). Can we have any form of F and G?

Well, this satisfies the equation for the tension cable in the medium.

But now we have a condition that the displacement should be equal to zero at x=0.

This means that for any time t F(-ct)+G(ct)=0.

These two functions F and G are now related. If I take an argument z, then F(-z) = -G(z).

They cannot be chosen independantly. The displacement that satisfies both the cable

equation and the boundary equation must therefore be

y(x,t) = -G(-x+ct)+G(x+ct).

What does this look like? Well, we have a left going part and the right

going part is just the opposite. Why is that? Because the sum of the two will exactly

cancel at x=0. Let us take a simple function G like here.

A pulse, G is zero everywhere except on a small zone.

At early times, on the cable, I will see the pulse G moving left: The second part of the

solution does not materialize on the cable When the pulse reaches the boundary, the second

part of the solution becomes non-zero and exactly compensates the first part, at the

point x=0. This is how the condition is satisfied. And then, for later time I will only see the

second part. What happened? the wave reflected, and at

reflection it changed sign. This is a very common phenomenon. And now

you know why the sign changes. It is because we need a wave opposite in sign and direction

to exactly balance the wave so that the boundary condition is satisfied.

Of course, you can do the same for other boundary conditions.

If the cable is free, dy/dx=0 you will get y(x,t) = G(-x+ct) + G(x+ct). The wave will reflect

but not change sign.

Of course, this is true for any form of waves, any G function. So, let us have a harmonic

function G(x+ct) = sin[k(x+ct)].