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Re: Elastic potential energy (repeat)



On Tue, 20 Nov 2001, RAUBER, JOEL wrote:

John M responded:

Not at all, but the correct equation for any choice of
the "reference stretch" xo is

elastic potential energy = (k/2)x(x + 2*xo)

This reduces to the familiar (k/2)x^2 with the convenient (but
not mandatory) choice, xo = 0.

If I'm not mistaken, there is a minus sign error,
I get for (1dim motion along x axis, F=-k(x-xo);

elastic potential energy = (k/2)x(x-2*xo)

I suspect there may be some misunderstanding of terms. My "xo" is
intended to indicate how much the spring is *stretched* at the
arbitrarily chosen "reference position" and my "x" is a coordinate
with origin at the reference position that is positive in the
direction of *increasing* stretch. With these definitions, the
relaxed spring corresponds to x = -xo and the force equation is F
= -k(xo+x). The resulting equation has the desired features
1) that the potential energy is zero at the reference position and
2) that it recovers the usual formula (k/2)x^2 when the stretch
at the reference position is 0.

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