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Re: hanging spring



At 09:26 PM 12/18/02, you wrote:
Sorry to come back to this so late, but I only recently
got around to working on it. I think the discussion left
the topic incomplete.

To restate the problem: A spring is held vertically by
the upper end, and released. What is the acceleration
of the highest infinitesimal part ?

Consider a spring of mass M, that stretches by dL,
modelled as N masses separated by N-1 springs
of spring constant k.

The uppermost mass, of size M/N supports the load
(N-1)Mg/N when held in equilibrium.

When released a = [(N-1)Mg/N] / (M/N) = (N-1)g,
so that a gets very large as N is increased.

However, if we consider the time response, we
get a finite impulse.

Calculate k (the i'the spring supports the load iM/N),
and the total stretch is dL, so k = (N-1) N M g / (2 dL).

Taking the period of SHM for the topmost mass as a
measure of the time response T = 2 pi sqrt( M/(Nk))
so T =2 pi sqrt( 2 dL/ ( N (N-1) g ) )

Taking delta V = a T ~ sqrt( (N-1)/N ),
which clearly goes to 1 as N is increased.

Moreover, the momentum transfer ~ (M/N) delta V ~ 1/N
which goes to 0 as N increases.

Two more comments:
1. The same model applies to a hanging bar of material.
2. Real materials or springs are not infinitely divisible
into uniform entities.

Al Bachman

Providing a clean-cut answer may not be quite PC, but it makes me feel good.
But let me muddy the water anyway:
The extended spring provides a mechanical wave transmission medium
which is mismatched at the ends, encouraging reflections.
So
3. An impulse traverses the spring, and is reflected at the ends, repeatedly.

(This point was touched on in the thread. Always nice to introduce easily
demonstrable transmission line effects, which often go unremarked.)

Brian Whatcott
Altus OK Eureka!