Re: SLINKY

[John Denker <jsd@MONMOUTH.COM>]

At 04:07 AM 1/23/00 -0500, Ludwik Kowalski wrote:
> Thanks again, John.
>
> As, always I am happy to learn new thinks which
> can help me in a classroom. Your derivation (see
> below) was clear.

:-)

> I did not know that the speed of
> longitudinal waves is described by a formula which
> differs from the transverse formula in "s instead
> of T".

Be careful, there are three pieces of physics here, and you mentioned only
two;  we have not only
  a) transverse waves driven by tension, and
  b) longitudinal waves driven by compression/extension, but also
  c) transverse waves driven by stiffness

> To predict the transverse speed we must know the
> tension T (in addition to linear density rho, which I call mu);

Yes, if you insist on ignoring flexional stiffness.

> the magnitude of the spring constant k has nothing to do with it.

OK;  evidently k is the compressive spring constant and you are still
ignoring flexional stiffness.

> But, as you "proved", to
> predict the longitudinal speed we simply replace T
> by s, the spring stiffness (which appears in Hooks
> law for flexion).

Ooops, now you mention flexion and stiffness, which is inconsistent with
the foregoing statements and assumptions.  My (s) is absolutely not the
flexional stiffness, for at least two reasons:
  *) My (s) is a *compliance* per unit length, so it measures how
unresisting a spring is, not how resisting it is, and
  *) My (s) has to do with pure compression, not flexion.



> After all in both cases, longitudinal and transverse,
> the restoring force is directly proportional to the
> corresponding displacement.

I agree.

> If k is not relevant in
> one case then I would expect s to irrelevant in
> another. I would expect the longitudinal wave in a
> slinky to depend on T rather than on s.

I disagree.  Here's why:

The tension from the left termination and the right termination apply
forces to every element of the spring, but as long as we consider only
longitudinal motions these two forces are equal and opposite at all places
and all times.

Applying tension to a spring might change its length, but this does not (to
first order) change its spring constant.  (This is direct consequence of
the linearity of Hooke's law.)  It is the compressive spring constant that
drives the compressive longitudinal waves.

> It turns out that s can be measured as easily as k.

Huh?  My (s) is just (L/k) where (L) is the length.  Measuring (s) is
slightly more work than measuring (k), but it's not rocket science.

> Thus your mathematically derived formula can
> be subjected to an acceptable experimental validation.

Sure.

> A good student project or lab. Does anybody do this?

I dunno.

> By the way, my first classroom waves demo is based on
> a long stretched spring. I firmly attach one end to
> a hook, increase the length of the spring by stretching
> it (about 30%) and wiggle the other end up and down.
> A standing wave (over a distance of about 10 m) can
> easily be produced. The speed was never measured but
> I assume that the standard formula, v=sqr(T/mu), is
> applicable. Would you agree?

As I said in previous messages, it depends on what medium you're
using.  Any slinky I've ever owned would be destroyed if extended to 10m
length.  Depending on the medium, I wouldn't be surprised to find that the
flexional stiffness makes a nontrivial contribution to the speed of the
transverse waves.

> P.S. Actually, the dependence of v on k does exist
> but it is hidden. Two springs whose relaxed lengths
> are identical, but whose k are different, would result
> in different mu for the same T (or different T for
> the same mu).

Not necessarily.  One spring could be wound of hollow tubing while the
other was wound of solid rod with the same mass per unit length; the former
would be much greater flexional stiffness and longitudinal spring constant.

Note that a slinky is quite sophisticated;  it is wound of strips that are
much wider than they are thick;  this has the effect of increasing the
ratio of transverse flexional stiffness versus longitudinal compressive
spring constant -- by well over an order of magnitude, as best as I recall
the dimensions;  stiffness goes like the *cube*.

> Each of them, however, should behave like
> a cable, or a rope, whose T and mu are identical.

Not unless you can arrange to:
 *) Ignore the compressional spring constant, and
 *) Ignore flexional stiffness.

A piano uses long, thin wires under tremendous tension in order to minimize
the contributions of these elastic effects -- but even then elasticity
contributes enough to require the stretching of octaves as discussed back
in November.