Re: [Phys-l] sound waves and beam flexures

[John Denker <jsd@av8n.com>]

On 02/22/2010 11:22 PM, Bernard Cleyet wrote:
> Before I embarrass myself:    Beam flexure normal modes are not due
> to traveling sound waves.  By definition sound waves are described by
> a second order partial differential eq.,  while beam flexures are by
> a fourth order one, yes????

I think you understand the physics.  So don't get too
tangled up in definitions.

Sound should be defined by the physics.  It is not
"defined" by this-or-that equation.

And even the names of the equations are shifty.  In
the context of QM, the Schrödinger equation is called
"the wave equation" even though it is very different
from "the wave equation" that is used to describe
acoustic plane waves in introductory classes.

And speaking of plane waves:  for non-planar sound
waves, such as the spherical waves spreading out from
a point source, the physics is different.  Specifically,
it is dispersive (whereas plane waves in the same medium
would be non-dispersive).  "The wave equation" must be
reformulated to take this into account.

  Note that in the presence of dispersion or dissipation,
  the definition of "wave" that you learned in high school 
  crashes and burns.  The wave does not simply move from
  place to place while keeping its shape.

Turning (finally :-)) to the question that was asked:
There *are* transverse sound waves in a beam.  The physics
of such sound is nontrivial.  Among other things, it 
is quite dispersive.  It is not described by "the wave
equation" you learned as a child ... but that's OK,
practically nothing else is either.

> e.g.  banging a beam on it's end (axially) produces a sound
> compression wave, while banging it perpendicularly to its axis,
> ...

This has direct technological application.  The gong
in a cuckoo clock is implemented using transverse
waves i.e. flexural waves on a long piece of steel
wire.  This exploits the high dispersion to create
a much richer sound than you would get otherwise.