Re: oscillations test question

[Bob LaMontagne <rlamont@POSTOFFICE.PROVIDENCE.EDU>]

Robert Cohen wrote:

> I asked...
>
> > If the 200 grams is then stretched a distance x from the new
> > equilibrium point, the potential energy relative to the new=20
> > equilibrium point is equal to 1/2 kx^2.  What do you call the
> > quantity 1/2 kx^2?  The difference in the elastic-gravitational
> > potential energy?

Please forgive me if I'm missing the point of the question. Given the way the question
is phrased, it doesn't seem that 1/2 k x^2 has any definite meaning. The spring
elastic energy is measured from the UNSTRETCHED position of the spring - and is a non
linear quantity.

When the mass is hung on the spring, the spring stretches by an amount d = mg/k.
(Non-conservative work was done by your hand taking the mass to the equilibrium point
- otherwise the mass would oscillate!)

If we call this position x=0 and now pull the mass downward by an additional amount,
x, the potential energy term is 1/2 k (d + x )^2 - m g x.   Because of the cross term
in the square, 1/2kx^2 does not seem to take on any easily identifiable meaning (at
least not for an introductory physics class).

Perhaps someone out there has thought about this enough to have come up with an
insightful meaning to the 1/2 k x^2 - please post it if you have.

Bob at PC

I

This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.