## February 22, 2014

### Lying

Sometimes, for the sake of pedagogy, it is best to suppress some of the ugly details, in order to give a clear exposition of the idea behind a particular concept one is trying to teach. But clarity isn’t achieved by outright lies. And I always find myself frustrated when our introductory courses descend to the latter.

My colleague, Sonia, is teaching the introductory “Waves” course (Phy 315) which, as you might imagine, is all about solving the equation

(1)$0 = \left(\frac{\partial^2}{{\partial t}^2} - c^2 \frac{\partial^2}{{\partial x}^2}\right) u(x,t)$

This has travelling wave solutions, with dispersion relation

(2)${\omega(k)}^2 = c^2 k^2$

If you study solutions to (1), on the interval $[0,L]$, with “free” boundary conditions at the endpoints,

(3)$\left.\frac{\partial u}{\partial x}\right\vert_{x=0,L} = 0$

you find standing wave solutions $u(x,t) = A \cos(k x)\cos( c k t)$ where the boundary condition at $x=L$ imposes

(4)$\sin(k L) = 0\qquad \text{or}\qquad k L = n\pi,\, n=1,2,\dots$

The first couple of these “normal modes” look like

(5)$x=0$

To “illustrate” this, in their compulsory lab accompanying the course, the students were given the task of measuring the normal modes of a thin metal bar, with free boundary conditions at each end, sinusoidally driven by an electromagnet (of adjustable frequency).

Unfortunately, this “illustration” is a complete lie. The transverse oscillations of the metal bar are governed by an equation which is not even approximately like (1); the dispersion relation looks nothing like (2); “free boundary conditions” look nothing like (3) and therefore it should not surprise you that the normal modes look nothing like (4).

Unfortunately, so inured are they to this sort of thing, that only one (out of 120!) students noticed that something was amiss in their experiment. “Hey,” he emailed Sonia, “Why is the $n=1$ mode absent?”

Rather than the 2nd-order wave equation, the transverse vibrations of the thin bar are governed by a 4th-order equation
(6) $0 = \left(\frac{\partial^2}{{\partial t}^2} + b^2 \frac{\partial^4}{{\partial x}^4}\right) u(x,t)$

The dispersion relation, ${\omega(k)}^2 = b^2 k^4$ admits both real and pure-imaginary wavenumbers. So the general standing-wave solution has the form (for real $k$) $u(x,t) = [A_1 \cosh(k x)+A_2 \sinh(k x)+A_3 \cos(k x)+A_4 \sin(k x)]\cos(b k^2 t)$ “Free” boundary conditions for (6) are a pair of conditions at each boundary, $\left.\frac{\partial^2 u}{{\partial x}^2}\right\vert_{x=0,L} = 0,\qquad \left.\frac{\partial^3 u}{{\partial x}^3}\right\vert_{x=0,L} = 0$ Imposing the boundary conditions at $x=0$ yields $A_3=A_1,\quad A_4 =A_2$ To satisfy the boundary conditions at $x=L$ then requires $\det\begin{pmatrix}\cosh(k L) - \cos(k L)&\sinh(k L) - \sin(k L)\\ \sinh(k L) + \sin(k L)& \cosh(k L) - \cos(k L)\end{pmatrix}=0$ or

(7) $\cosh(k L)\cos(k L) = 1$

which is nothing like (4). The first few solutions to (7) are $k L = 1.50562\pi,\, 2.49975\pi,\, 3.50001\pi$, and the lowest mode has a vague (and somewhat accidental) resemblance to the $n=2$ mode of (5).

Analyzing the solutions to (6) is very interesting, but arguably way more complicated than we ought to be doing for students still struggling to understand (1). But assigning them the task of studying the vibrating bar experimentally, and telling them that it’s governed by (1), is just a complete disservice.

What were the folks who designed the lab thinking?

Posted by distler at February 22, 2014 3:55 PM

TrackBack URL for this Entry:   https://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/2701