Recent content by bomerman218

Its much easier if you use work energy to solve this problem. Also the tensions are not equal- only in the static case but the relative position, velocity, and acceleration are.

\frac{1}{8}u''+\frac{1}{2}u'+\frac{25}{2}=0 u(0)=\frac{1}{7} u'(0)=\frac{6}{7}

Yeah I can. Unfortunately when I was solving the governing differential equation I was not given the solution so I'm wondering how I would approach this problem not knowing the solution.

\frac{1}{7} e^{-2t} \cos(4 \sqrt 6 t)+\frac{\sqrt 6}{21} e^{-2t} \sin(4 \sqrt 6 t) =\frac{\sqrt 15}{21} e^{-2t} \cos(4 \sqrt 6 t+\arctan \sqrt 6/2) I am having trouble figuring out how to prove this relation. Any help would be greatly appreciated. My initial thought was to use this...