Simple Mass Spring Lagrangian: Equations of Motion for a Two-Dimensional System

[GreenLRan]

Homework Statement

A particle of mass m is suspended between two springs, each of spring constant k. In equilibrium position each spring is horizontal and has length L. The springs can swivel in the x,y plane as well as stretch or contract. Treating this as a two dimensional problem and ignoring gravity, write the Lagrangian for the system and find the equations of motion.

Homework Equations

L=T-U, (d/dt (pL/px')) - pL/px =0 T= kinetic energy, U= potential energy, p= partial derivative symbol, '= prime symbol

The Attempt at a Solution

I set T= 1/2*m*x'^2, U= -2*(1/2*k*Δx^2),
then deriving the equations of motion I got mx''+2kΔx=0 and my''+2kΔy=0 (wrong answer)
(I'm not sure if there are supposed to be two equations)
Thanks!

 

[OlderDan]

It looks to me like you have left a lot out. This is a two dimensional problem. I only see one dimension represented in your T equation, and it appears you have tried to equate the extensions of the springs. They need not be the same.

 

[GreenLRan]

ok well I understand that the kinetic energy is T= 1/2*m*(x'^2+y'^2), but the "can swivel in x,y plane" part is confusing me. Would the potential energy of the springs only be in the Y direction ? Anything would be helpful. Thanks

 

[OlderDan]

ok well I understand that the kinetic energy is T= 1/2*m*(x'^2+y'^2), but the "can swivel in x,y plane" part is confusing me. Would the potential energy of the springs only be in the Y direction ? Anything would be helpful. Thanks

The potential energy is a function of the elongation of both springs. The "swivel" means the mass can move anywhere in the x-y plane permitted by the spring connections. If the x-axis is the line joining the connection points of the springs and the equilibrium position is at x = 0, then the PE is the sum of the elastic energies of the springs relative to equilibrium. The x-axis and the springs form a triangle that can be any shape, but the one side is the distance between the points of connection of the springs. The vertex opposite that side can be anywhere in the x-y plane, as long as you don't violate the linearity of the spring forces.

 

[GreenLRan]

So, the potential would then be ( correct me if I am wrong ) U= 1/2*k*(Δx^2+Δy^2) ?

 

[OlderDan]

So, the potential would then be ( correct me if I am wrong ) U= 1/2*k*(Δx^2+Δy^2) ?

With a bit of algebra I think you can show that U has that form, except without the 1/2 factor. The spring constant of each spring is k, so the effective spring constant is 2k.