Lagrangian of a coupled pendulum - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-08-17T21:39:09Z https://physics.stackexchange.com/feeds/question/270165 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/270165 1 Lagrangian of a coupled pendulum Quasar https://physics.stackexchange.com/users/124665 2016-07-27T07:59:31Z 2016-07-27T08:22:27Z <blockquote> <p>I am trying to find the Lagrangian for a coupled pendulum: the two pendulums have the same characteristics (length $l$ and mass $m$) and are attached to the same roof at a distance $d$. In addition, the two weights are coupled by an ideal spring of characteristic constant $k$ and rest length $d$.</p> </blockquote> <p>Taking the two angles $\phi_1$ and $\phi_2$ as generalized coordinates, why is the potential energy part for the spring given as $$U=\frac{l^2k}{2}(\Delta x)^2,$$ where $$(\Delta x)^2 = (\sqrt{(d+l sin \phi_1-l sin \phi_2)^2+(l cos \phi_1-lcos \phi_2)^2}-d)^2?$$</p> <p>What staggers me is the $l^2$. Isn't the potential energy for a spring just given by $1/2*k*\Delta x$?</p>