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Re: SLINKY

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".

To predict the transverse speed we must know the
tension T (in addition to linear density rho, which
I call mu); the magnitude of the spring constant k
has nothing to do with it. 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).

On the purely intuitive basis I would not expect this.
After all in both cases, longitudinal and transverse,
the restoring force is directly proportional to the
corresponding displacement. 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.

It turns out that s can be measured as easily as k.
Right? Thus your mathematically derived formula can
be subjected to an acceptable experimental validation.
A good student project or lab. Does anybody do this?

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?
Ludwik Kowalski

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). Each of them, however, should behave like
a cable, or a rope, whose T and mu are identical. Right?
----------------------------------

John Denker wrote:

At 05:21 PM 1/22/00 -0500, Ludwik Kowalski wrote:
Thanks for responding; I plan to address this in a class on
Tuesday.

You're welcome.

A cable under tension is a spring with very large k.

Maybe. I'm not sure what you mean or why you bring it up. Let's ignore
that sentence and move on.

I assume that the word "stiffness" refers to the spring constant
k in [N/m].

It scares me when you say "the" spring constant in this context where both
P-waves and S-waves are being discussed. Consider a large coil spring with
*open* windings at rest (not closed like a garage-door spring at rest). It
will have one spring constant for extension/compression, and another for
flexion. The latter is called stiffness. Both will have units of
[nt/m]. (Nitpickers will note there are torsional constants as well, but
we need not discuss those.)

In my version the spring (slinky) is always stretched
to behave like a rope or cable.

In context, I assume that means you intend to treat stiffness as negligible
compared to tension. I doubt that is correct for slinkies; that is, I
suspect that for any tension that does not do irreversible damage to the
slinky, stiffness is non-negligible.

What is "compliance per unit length [1 / nt]"?

According to my dictionary, compliance is strain per unit stress. It has
dimensions of [N/m] and for springs it is the reciprocal of the spring
constant. This is a convenient concept because given a number of springs
in series, the compliance is additive; using thermodynamic language the
compliance per unit length is an *intensive* quantity.

How was your formula derived?

Hmmm. Let me find that envelope. OK, got it.

1) Model the macroscopic spring as a 1-dimensional lattice of tiny lumps
connected by tiny springs. Let the lattice constant be delta_x.

2) Each lump has a rest position. Let the displacement field y(x)
represent the displacement of each lump from its rest position. In this
longitudinal case the displacement-vector "y" is in the same direction as
the location-vector "x" but please ignore that; it will just confuse
you. Instead, please treat them as independent quantities just as you
would in the transverse case (where y would be perpendicular to x).

3) Let the force field F(x) represent the force on the lump at (x) due to
the spring ON THE LEFT. Therefore -F(x + delta_x) is the force due to the
spring on the right.

4) Observe that a uniform translation of the whole lattice (represented by
y(x) = const) produces no force on the lumps.

More generally, we have that the extension of the tiny spring is
extension = (dy/dx) delta_x
and the force produced thereby is (using Hooke's law)
F(x) = -(dy/dx) delta_x / (s delta_x)
where
s = compliance per unit length.

5) Observe that a uniform compression of the whole lattice produces no
*net* force on the lumps. For each lump, the force from the spring on the
left is just cancelled by the force from the spring on the right.

So in fact the *net* force on a lump depends on the sum of the left and
right contributions:
net force = (d/dx) (dy/dx) delta_x delta_x / (s delta_x)

6) The mass of each lump is just
m = rho delta_x.

7) Therefore the acceleration of each lump is
a(x) = net force / m = (d/dx) (dy/dx) 1 / (rho s)
where we note that delta_x has dropped out of the result, as it
should. This leads immediately to a wave equation
(d/dt) (dy/dt) - (d/dx) (dy/dx) / (rho s) = 0
in which we recognize the wavespeed
c = 1 / sqrt(rho s)
QED.