Re: oscillations test question

[Robert Cohen <Robert.Cohen@PO-BOX.ESU.EDU>]

A related question...

If the 200 grams is then stretched a distance x from the new equilibrium point, the potential energy relative to the new 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?

____________________________________________
Robert Cohen; rcohen@po-box.esu.edu; 570-422-3428; http://www.esu.edu/~bbq
Physics, East Stroudsburg Univ., E. Stroudsburg, PA 18301

> -----Original Message-----
> From: Carl E. Mungan [mailto:mungan@USNA.EDU]
> Sent: Wednesday, November 20, 2002 2:16 PM
>
> I'm interested in hearing what folks think of the following question
> in the test bank for Giancoli:
>
> Chap 14 #11. Two hundred grams hung on a spring stretches it 8.4 cm.
> How much energy is stored when stretched 8.4 cm?
>
> Before reading the rest, decide how you would answer this and why.
>
> While a number of answers are possible, the most reasonable answer in
> my mind is zero. By definition, a system at rest in stable
> equilibrium is at a potential minimum.
>
> The purported answer of 82 mJ only considers the increase in the
> elastic potential energy and neglects the decrease in the
> gravitational potential energy.
>
> It is true that no zero points are specified. This is why I can only
> say my answer is reasonable, and other interpretations are possible.
> Comments?

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