3
$\begingroup$

I recently stumbled upon the above image describing partial transmittance, and was wondering what sort of equation would model such a wave propagating through varying mediums. Is there also an equation for continuous mediums, as opposed to the discontinuous transition in the picture?

$\endgroup$
4
$\begingroup$

Yes.

You simply use the wave equation on either side of the interface between the two media, and then you impose appropriate smoothness conditions at the interface. In one dimension, the wave equation is \begin{align} \frac{\partial^2y}{\partial x^2} = \frac{1}{v^2}\frac{\partial^2y}{\partial t^2} \end{align} If, for example, the wave is a transverse wave along a string, then $y = y(t,x)$ can be thought of as the transverse displacement of the string as a function of position $x$ along the horizontal axis and time $t$. The parameter $v$ in this equation is the wave speed which is determined by the properties of the medium (like its density) and is, in general, different for different media.

When you have an interface between two media, like in the animation, one solves the wave equation on either side of the interface and then imposes appropriate smoothness conditions at the interface such as continuity of $y$ and its first derivative. If the interface is at $x=0$, then these conditions would read \begin{align} y(t, 0^-) = y(t, 0^+), \qquad \frac{\partial y}{\partial x}(t,0^-) = \frac{\partial y}{\partial x}(t,0^+). \end{align} You'll probably find the following physics.SE post illuminating:

Boundary conditions on wave equation

$\endgroup$
  • $\begingroup$ How about when there is no interface, but rather a continuous transition from one medium to the next? $\endgroup$ – user1825464 Jan 25 '14 at 21:42
  • 1
    $\begingroup$ @user1825464 Oh in that case you simply attempt to solve the wave equation with $v$ being a continuously varying function of position. For example, this is how you could treat a rope with continuously varying linear mass density. However, the equation becomes a lot harder to solve in that case. $\endgroup$ – joshphysics Jan 25 '14 at 21:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.