Re: Elastic potential energy

[Bob Sciamanda <trebor@VELOCITY.NET>]

If I choose the relaxed position for my zero of PE and measure x from
there I would define the PE function:
PE1(x) = INT{0,x}k*x*dx

If instead I choose the position (x=a) for my zero of PE and measure u
from there (so that u=x-a) I would define the PE function:
PE2(u) = INT{0,u}k*(a+u)*du

Performing the integrals =>

PE1 = .5*k*x^2

PE2 = k*a*u + .5*k*u^2

using x = u + a (by definition)

It follows that PE1 = PE2 + .5*k*a^2

IE; the difference is just a constant: the work integral evaluated over a
path joining the two chosen origins where the PE functions are forced to
be zero. This is true for any PE function associated with any conservative
force - linear, quadratic or whatever.

> The PE is really arbitrary to within an additive constant - but no
> additive constant can cancel out the *linear* term k*x1*x that arises
> because of the translation, and that you need to fix your problem.
>
> It's a pity that arbitrariness of PE is always discussed only with
> reference to the gravitational PE near the Earth. There the PE is
> linear, so when you translate the origin only a constant term arises,
> which can be cancelled out by an additive constant.

Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor
----- Original Message -----
From: "Paolo Cavallo" <ton0621@IPERBOLE.BOLOGNA.IT>
To: <PHYS-L@lists.nau.edu>
Sent: Friday, November 16, 2001 10:33 AM
Subject: Re: Elastic potential energy