Re: Stiffness waves

[John Denker <jsd@MONMOUTH.COM>]

At 01:09 PM 1/27/00 -0500, Ludwik Kowalski wrote:

> The condition for the validity of your derivation [for
> compression waves] was Hooke's law, the restoring force
> must be proportional to displacement. Is this not
> true for an ideal transverse wave (small amplitude)?

The derivation depends not only on Hooke's law, but also on other bits of
physics.  In the compressional case the key notion is that a force applied
at some point causes a uniform extension of everything to the left, and a
uniform compression of everything to the right.  The resulting displacement
field is graphed in the bottom curve in
  http://www.monmouth.com/~jsd/physics/greenfun.gif

That is, the displacement ramps up linearly and then ramps down
linearly.  The displacement for compressional waves is oriented along the
axis of the spring;  we graph it in a perpendicular direction just because
that's how graphs are made.

Note that the same curve is an equally good description of a string under
tension (with zero stiffness).  In this case, the displacement is
perpendicular to the string, so the graph is practically a picture of the
physical situation.

If you differentiate this function twice, you get a delta function at the
point where the force is applied.  This delta function represents the force
per unit length.  We know the mass per unit length, so in conjunction with
this force per unit length we can immediately write down the wave equation
that describes this case:
        (d/dt)^2 y - (d/dx)^2 y = 0
which is the elementary wave equation.  It is linear and nondispersive.

> What prevents me from applying the same to a transverse
> displacement and from using k in Hooke's law (rather than s)?
> I am assuming that there is no tension in a long string of tiny
> masses connected by tiny relaxed springs.

The foregoing analysis applies to compression waves and tension-driven
waves on a string, but...

It _cannot_ be reused for flexy waves on stiff springs.  There is a simple
reason for this: a force at a point on a flexy rod does _not_ cause a
ramp-like displacement field.  What you actually get is shown in the top
curve in the URL cited above.

Note that the flexy rod is clamped at each end so as to ensure y=0 and
dy/dx=0 at each end.  The right end is free to slide in a tube, so there
won't be any tension.

In the upper curve, let's look at the regions where no force is
applied.  We see that the first, second, and third derivatives are
nonzero.  This is clear from the graph, because there is a slope (first
derivative), a curvature (second derivative), and the curvature is changing
(third derivative) -- even in places where no force is applied!  You need
to differentiate that function at least _four_ times in order to get
anything related to the local force per unit length.  (In fact the fourth
derivative is the right thing, but you can't tell that just by
looking.)  So we get a different wave equation in this case:
        (d/dt)^2 y - (d/dx)^4 y = 0
which is highly dispersive.  (Note that it's still linear -- linear in y.)

Here's another dead giveaway that we're dealing with a fourth-order
system:  For the flexy waves we needed to specify _four_ boundary
conditions (y=0 and dy/dx=0 at both ends).  Contrast this with the
compressional waves (a second-order system) for which we needed to specify
only _two_ boundary conditions (y=0 at both ends).


--------------

BTW....

1) Some nomenclature:  For any linear system, the response of the system to
a delta-function excitation is called a Green function.  The given URL
portrays the Green functions for the two different systems.

2) Hint:  The analysis sketched here is only one of eleventeen handy things
you can do with Green functions.