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Re: Elastic potential energy (repeat)



See comment below:

Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor
----- Original Message -----
From: "RAUBER, JOEL" <JOEL_RAUBER@SDSTATE.EDU>
To: <PHYS-L@lists.nau.edu>
Sent: Tuesday, November 20, 2001 11:35 AM
Subject: Re: Elastic potential energy (repeat)


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

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);

This is only true if the spring is relaxed at x = xo; ie; xo is not
arbitrary.


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

One can take advantage of the ability to add an arbitrary constant to
put
this into a more "elegant" form

namely add (k/2)xo^2 to get

elastic potential energy = (k/2)(x-xo)^2 which is in a familiar form
compared to the usual

(k/2)x^2 , which is to say in words

"elastic potential energy = half k times (distance spring is stretched
or
compressed) squared" for any choice of coordinate origin.