Re: test problem

[Bob Sciamanda <trebor@VELOCITY.NET>]

Of course you are correct if we have such a "coil bound" spring, which has
no "room" to compress from its relaxed condition.  But in the typical
problem a mass on a spring (no gravity - or horizontal) is presumed to be
able to oscillate about its relaxed condition - why should adding gravity
change the presumption in favor of a "coil bound" spring?
"What is gratuituosly assumed can be gratuitously denied."  :)

Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor
----- Original Message -----
From: "John Mallinckrodt" <ajmallinckro@CSUPOMONA.EDU>
To: <PHYS-L@lists.nau.edu>
Sent: Wednesday, December 05, 2001 7:29 AM
Subject: Re: test problem


> 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?