1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple mass spring lagrangian

  1. Dec 12, 2006 #1
    1. The problem statement, all variables and given/known data
    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.


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


    3. 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!
     
  2. jcsd
  3. Dec 13, 2006 #2

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  4. Dec 14, 2006 #3
    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
     
  5. Dec 14, 2006 #4

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  6. Dec 15, 2006 #5
    So, the potential would then be ( correct me if I am wrong ) U= 1/2*k*(Δx^2+Δy^2) ?
     
  7. Dec 15, 2006 #6

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?