The problem of interest is worded as follows:
A horizontal square wire frame with vertices $ABCD$ and side length $2a$ rotates with constant angular frequency $\omega$ about a vertical axis through $A$. A bead of mass $m$ is threaded on $BC$ and moves without friction. The bead is connected to $B$ and $C$ by two identical light springs of force constant $k$ and equilibrium length $a$.
By considering a small displacement of the particle away from the midpoint of $BC$, find the Lagrangian.
I'm struggling to find a good coordinate system to work in. I considered taking A to be the origin, and using a polar angle measured from the line produced by the initial position of $AM$ where $M$ is the midpoint of $BC$, but the expression for the angle that the particle makes with the initial line seems far too complicated to be the best way.
Any hints on how to proceed?
You know the explicit polar coordinates of $B$ and $C$ and you know the bead is constrained to move between them, i.e. its position is always of the form $${\bf b}=\lambda {\bf B} + (1-\lambda){\bf C} \text{ for } \lambda \in \left[0,1\right].$$ I would try using $\lambda$ (or $\mu = \lambda-\frac{1}{2}$, the displacement from the centre of $BC$) as the coordinate of the bead. You can calculate the kinetic and potential energy in terms of this.