L'(unstretched) = root(2)*L
L'(stretched) = root((L+x)^2+(L+y)^2)
So delta L' = root((L+x)^2+(L+y)^2) - root(2)*L
Squaring delta L' is still going to leave me with a handful of linear terms. Here's the potential I got (skipped a few steps since this is difficult to type out, and leaving out...
As far as I can tell, the potential energy I have written is simply wrong. I just don't know why. Throwing out gravity makes it simpler, but the issue is still that I can't use the typical step of simply pulling out matrix K using the definition I gave. A 3x3 matrix wouldn't help either, since I...
Homework Statement
Two equal masses (m) are constrained to move without friction, one on the positive x-axis and one on the positive y axis. They are attached to two identical springs (force constant k) whose other ends are attached to the origin. In addition, the two masses are connected to...