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!

Normal modes of oscillator

  1. Mar 21, 2009 #1
    1. The problem statement, all variables and given/known data

    I'm reading Landau's Mechanics, in section 23, he discusses the oscillations with more than one degree of freedom, the Lagrangian is

    [tex]L = \frac{1}{2}\left(m_{ik}\dot{x}_i\dot{x}_k - k_{ik}x_ix_k\right)[/tex]

    where [tex]m_{ik},k_{ik}[/tex] are symmetric constants, and the summation over [tex]i,k[/tex] in the above equation is understood.
    By substituting the form of solutions

    [tex] x_k = A_k\exp(i\omega t)[/tex]

    we get the system of linear equations,

    [tex]\sum_i\left(-\omega^2m_{ik} + k_{ik}\right)A_k = 0\quad\cdots(*)[/tex]

    In order to have non-trivial solution, the determinant of the following matrix should be zero,

    [tex]\left| k_{ik} - \omega^2m_{ik} \right| = 0\quad\cdots(**)[/tex]

    The roots of [tex]\omega[/tex] are denoted as [tex]\omega_\alpha[/tex].
    Then comes my question. He said that "The frequencies [tex]\omega_\alpha[/tex] having been found, we substitute each of them in eq(*) and find the corresponding coefficients [tex]A_k[/tex]. If all the roots [tex]\omega_\alpha[/tex] of the characteristic equation are different, the coefficients [tex]A_k[/tex] are proportional to the minors of the determinant (eq(**)) with [tex]\omega=\omega_\alpha[/tex]."

    My question is the sentence with the underline, I think this is a problem of linear algebra actually, but I can't come up with any idea and can't find the material in Wiki.

    Thanks for your solution or reference!
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?



Similar Discussions: Normal modes of oscillator
  1. Coupled Oscillator (Replies: 0)

Loading...