Re: test problem

[John Mallinckrodt <ajmallinckro@CSUPOMONA.EDU>]

Brian is (as he noted) nitpicking, but his nitpicks are
nonetheless technically correct.  I think you are simply
misinterpreting his remarks.  He means for us to consider in
particular the case of a spring that is "coil bound" at its
"normal" length, i.e., one that can not be compressed and that
requires a nonzero applied force before it begins to elongate at
all.

Such a coil bound spring will stretch by less than mg/k when the
mass is hung from it and will "run into itself" at the top end of
the oscillation if the oscillation amplitude is greater than the
downward equilibrium displacement caused by the mass.

John Mallinckrodt      mailto:ajm@csupomona.edu
Cal Poly Pomona          http://www.csupomona.edu/~ajm

On Tue, 4 Dec 2001, Bob Sciamanda wrote:

> Brian objected:
> > No, the questions are unreasonable. If the extra downward displacement is
> > greater than the equilibrium displacement, then the possibility of coil
> > binding is not excluded on the upward rebound.
>
> I don't understand this objection - I would think coil "binding" (?) is a
> function of other things having to do with the coil design and structure. As
> g is made to go to zero (go to different planets, or just turn down gravity)
> "the extra downward displacement" becomes infinitely "greater than the
> equilibrium displacement". You imply this worsens the situation!
>
> >     For this same reason, the spring constant is not determined by only
> the
> > force required for a specified displacement.
>
> I understand this statement even less.  Surely the static k is still
> "determined by only the force required for a specified displacement".  Do
> you refer to some "dynamic" k associated with the binding during
> oscillation?