Re: Elastic potential energy

[John Mallinckrodt <ajmallinckro@CSUPOMONA.EDU>]

On Fri, 16 Nov 2001, Justin Parke wrote:

> Consider an ideal spring attached to a wall and in a relaxed
> state.  Define x=0 to be the free end of the spring in this
> relaxed state.  The spring is then stretched so that the end
> of the spring is at x1.  Finally, it is stretched even further
> so that the end of the spring is at x2.  The difference in
> elastic potential energy in the spring between those two
> points is 1/2 k(x2^2 - x1^2).
>
> Now consider the same spring stretched to the original x1.
> We now define a new coordinate system whose origin coincides
> with x1.  The spring is stretched to x2 in the old coordinate
> system which we now simply call x, so that x = x2-x1.  The
> difference in elastic potential energy between these two
> points is now 1/k (x2 - x1)^2.

No, it is still (k/2)(x2^2 - x1^2) or, if you prefer (see below),
(k/2)x(x + 2x1). Do the required work integral and see.

> ... Does the relaxed position of the spring define a unique
> origin for a spring or am I missing something here?

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.

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