# Two beads moving down fixed wires, connected by a spring

I was going to post this in the homework section, but there wasn't any forum for graduate-level problems. If someone wants to move this thread to a more appropriate forum (assuming there is one), however, please feel free to do so.

Anyway, this is a problem for my PhD qualifying exam that I'm studying for, and hopefully someone can help. I've attached a diagram which should illustrate the problem well. I'm given two wires at a fixed angle with respect to one another. Two beads of identical mass m are allowed to freely slide down each wire under gravity, and without any friction. The positions of the beads from the intersection of the wires are r1 and r2 respectively. The beads are connected by a spring of spring constant k, and the expansion of the spring (=the distance between the beads) is l.

I'm asked to find the Lagrangian of the system. I'm also asked to find the normal modes of oscillation, but I know that I can do this pretty easily once I know the Lagrangian, so that's the issue I'd like to address.

I know that the kinetic energy of each bead is simply (1/2)mv². The potential, however, is a bit more tricky. The gravitational potential energy is simply mgh (for each bead), and I can find h in terms of each bead's displacement from the top and the angle. However, there's also the spring potential energy, (1/2)kl². I know that l depends on r1 and r2, however, and I don't know how to do this without some complex triangle formula that I probably can't remember.

Anyway, I'd appreciate suggestions on this. I'd also like to request that no one give me the answer directly (if I wanted it I could just look at the solutions manual, but I don't want to see the solution to the whole problem). Thanks everyone!

#### Attachments

• Exam.GIF
1.1 KB · Views: 426

I think Sourabh N is right. With the cosine rule you can write the Lagrangian right down. The "complex triangle formula" is a simple one worth to remember (almost Pythagoras): $$c^2=a^2+b^2-2ab\cos\theta$$.