Exercise PhysicsWiki: Difference between revisions
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<math>U=0.5kx^2</math> (Ryan Taylor) | <math>U=0.5kx^2</math> (Ryan Taylor) | ||
<math>\nabla^2\psi = \frac{1}{v^2}\frac{\partial^2\psi}{\partial t^2} |
Revision as of 16:13, 30 August 2009
$$a=b$$
(posted by TerriC, Group 2)
(Zach McDargh)
(posted by KimberlyWynne 19:05, 29 August 2009 (EDT))
(Sandy Simmons)
(Steve Honeywell)
(Andrew Wray)
(Ryan Taylor)
<math>\nabla^2\psi = \frac{1}{v^2}\frac{\partial^2\psi}{\partial t^2}