A double pendulum consists of light, inextensible strings, AB and BC each of length l. It is fixed at one end A and carries two particles, each of mass m, which hang under gravity. The pendulum is constrained to move in a vertical plane. The angle between the vertical and AB is [itex]\theta[/itex], which the angle between BC and the vertical is [itex]\phi[/itex]. Show for smll angles about the equilibirium position:
[itex]d^2\theta/dt^2 +g/l(2\theta -\phi)=0[/itex]
[itex]d^2\phi/dt^2 +2g/l(\phi - \theta)=0 [/itex]
Newton's second law.
I shouldn't need to use Langrangian mechanics.
The Attempt at a Solution
I've managed to derive the first equation, first by assuming the tension in BC is mg (small angle approximation) then resolving forces about the top mass, and using the small angle approximation for sin. But I just can't get the second result out. I've tried to do the same this (resolving forces) but I must be making a mistake somewhere. Any help would be appreciated. Thanks