Re: SLINKY

[John Denker <jsd@MONMOUTH.COM>]

At 11:49 AM 1/23/00 -0500, Ludwik Kowalski wrote:
> The textbooks I know
> totally ignore contribution of stiffens to the speed of transverse
> waves.

Possibly that's because the physics of
    a) transverse waves driven by tension, and
    b) longitudinal waves driven by compression/extension,
is basically the same physics, and is easy and elegant, while the physics of
    c) transverse waves driven by stiffness
is different physics, requires a more sophisticated analysis, and leads to
more complicated results.

Here's the deal:  for case (a) or (b), we get a wave equation of the form
       (d/dt) (dy/dt) - (d/dx) (dy/dx) c^2 = 0
which admits running-wave solutions of the form
        f(x,t) = F(x+-ct)
for any F(), and which has the simplest possible dispersion relation
        omega = +- c k

IN CONTRAST... the physics of flexure is more complicated.  See _The
Feynman Lectures on Physics_ volume II section 38-4 for an introduction.

There are, alas, some people on this list who rely on intuition and seem
allergic to doing calculations or experiments.  Some of them may consider
it intuitively obvious that flexy waves must be analogous to compressional
waves.  Well, too bad, it doesn't happen to be true.

Take a long coil spring and paint the first foot red.  Load it (in
extension) with a suitable weight.  Measure the extension of the red
section.  Take a spring that is identical except for being twice as long,
and again paint the first foot red.  Load it with the identical
weight.  Observe that the extension of the first foot is the same as before.

Now do the same thing except use a flexional spring, clamped horizontal at
the near end and loaded with a weight at the far end.  Paint the first foot
red.  Repeat with a double-length spring, and observe that the same weight
produces *more* deflection of the first foot.  The force is the same, but
it's got more leverage and produces more flexion.

For flexy waves we have an equation of the form
       (d/dt)^2 y - const * (d/dx)^4 y = 0
and yes, that's a fourth-order spatial derivative.  This is a linear
differential equation (linear in y) so we retain the superposition
principle (which means this is much simpler than, say, fluid dynamics,
where the equations are nonlinear).  But it is highly anharmonic, i.e.
highly dispersive.  Because of the dispersion, there cannot be running-wave
solutions that retain their shape.  The dispersion relation is
        omega = const * k^2

*) A familiar obvious example of flexy waves is the musical saw
  http://homepages.enterprise.net/scruss/musicalsaw.html
Obviously the waves are transverse and dominated by flexion not tension.

*) More familiar but less obvious examples include violins and practically
all  other stringed instruments, which have bodies and/or sounding boards
that have carefully sculpted "tap tones" i.e. flexional modes (in addition
to the more-obvious resonances of the strings and air-spaces).

> And their "derivation" of the formula v=sqr(T/mu) is
> nearly artificial (based on the centripetal force). I would not
> accept the derivation without knowing that it happens to lead
> to correct answer.

Huh?  You gotta get some better books.  There is an easy and perfectly
legit derivation for transverse waves on a string under tension;  in fact
it closely parallels the derivation I posted for the compressional waves on
a spring.  Just redefine y() to be the *transverse* displacement from the
rest position and turn the crank.