[Phys-L] Re: Spring question

[rtarara at SAINTMARYS.EDU (Rick Tarara)]

The key here, I believe, is that the spring PE is defined in terms of a
displacement.  In your first form, x and x0 are displacements relative to
what?  To the unstretched length?  That is NOT the center of the harmonic
motion, x0 is.  So x0 really is your zero of PE.  If taken as zero, then
both forms agree.  If not, then only the second is applicable to the SHM
that will occur.  I agree however, that the discerning student might get
confused here (mine rarely do).   Once we hang enough weight to slightly
stretch the springs, it is emphasized that THIS is our ZERO point.
Displacements will be measured relative to that position.

Rick

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Richard W. Tarara
Professor of Physics
Saint Mary's College
Notre Dame, Indiana
rtarara@saintmarys.edu
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----- Original Message -----
From: "rlamont" <rlamont@POSTOFFICE.PROVIDENCE.EDU>
To: <PHYS-L@LISTS.NAU.EDU>
Sent: Tuesday, December 14, 2004 10:09 AM
Subject: Re: Spring question


> I thought I'd bring up a slightly different aspect of using springs in lab.
>
> Textbooks emphasize Hooke's Law in the form F = -kx. This is fine for
> springs that are manufactured such that their unloaded configuration is
> such
> that the coils are not touching. However, most springs - and specifically
> the ones we have been using in our simple harmonic motion lab - have a
> relaxed configuration where the coils are touching. This requires hanging
> a
> small but significant weight from the spring before any displacement
> occurs.
>
> We have found that the differential form of Hooke's Law, dF = -k dx, is a
> better starting place. In the lab, one has to use a weight hanger (50 g)
> plus some small additional mass to extend the spring to a comfortable
> position to start reasonably sized oscillations. Students get suspicious
> if
> you apply the traditional form of Hooke's Law to this situation. They have
> been conditioned by the textbooks to view x as the displacement from the
> "relaxed" position - when in fact it can be from any initial position
> (loaded or unloaded). Their suspicions are somewhat justified by the
> appearance of x^2 in the energy term - which is non-linear.
>
> If the loaded equilibrium position is xo, and the spring is then pulled
> down
> to a new position x before being released to oscillate, is the increase in
> the spring potential energy (1/2 kx^2 - 1/2 k xo^2), or is it 1/2
> k(x-xo)^2?
> This leads to two different values of kinetic energy at the center of
> oscillation - which obviously cannot be.
>
> Bob at PC