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 $d$.

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 $$(\Delta x)^2 = (\sqrt{(d+l sin \phi_1-l sin \phi_2)^2+(l cos \phi_1-lcos \phi_2)^2}-d)^2?$$

What staggers me is the $l^2$. Isn't the potential energy for a spring just given by $1/2*k*\Delta x$?

  • $\begingroup$ I thought it was a typo carried over from the K.E. term, but on looking at 6 sets of notes on the Web, 3 have l$^2$ and 3 have the P. E term as you think it should be. $\endgroup$ – user108787 Jul 27 '16 at 9:26
  • $\begingroup$ i think as the coupled pendulum oscillates the change in length of the spring will be a function of l and the angular displacements ...like l.times the d(phi).. thereby the expression for delta.x is involving those phi terms... $\endgroup$ – drvrm Jul 27 '16 at 9:43
  • $\begingroup$ I think this is really just a typo. Maybe someone confused the approximated \delta x with the exact one. Using an expansion for sin and cos up to the first term, I get an l^2 which then cancels out with the l^2 form the kinetic term, when using the Euler-Lagrange equations. $\endgroup$ – Quasar Jul 28 '16 at 8:15
  • $\begingroup$ Maybe I should write the answer to this question so it can be closed... $\endgroup$ – Quasar Jul 29 '16 at 21:12

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