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What kind of system does this lagrangian describe?

  1. Feb 17, 2014 #1
    1. The problem statement, all variables and given/known data
    Consider the following Lagrangian:
    \begin{equation} L = \frac{m}{2}(a\dot{x}^2 + 2b\dot{x}\dot{y} + c\dot{y}^2)- \frac{k}{2}(ax^2 + 2bxy + cy^2)\end{equation}
    Assume that \begin{equation} b^2 - 4ac \ne 0 \end{equation}Find the equations of motion and examine the cases a=b=0 and b=c, c=-a. Which kind of physical system is described by this Lagrangian? What meaning does \begin{equation} b^2 - 4ac \ne 0 \end{equation} have?


    2. Relevant equations
    \begin{equation} \frac{\partial L}{\partial q_i} -\frac{d}{dt}\frac{\partial L}{\partial \dot{q_i} } = 0 \end{equation}

    3. The attempt at a solution
    I'm able to solve the maths for this problem (assuming no errors in my calculations). In general for a, c not equal to 0 i get
    \begin{equation} x(t) = \frac{1}{ac-b^2}(A_1 c \cos(\omega t + \phi_1) - A_2 b \cos(\omega t + \phi_2)) \end{equation}
    \begin{equation} y(t) = \frac{1}{ac-b^2}(A_2 a \cos(\omega t + \phi_2) - A_1 b \cos(\omega t + \phi_1)) \end{equation}

    While for the cases a=c=0 and b=0, c=-a i get
    \begin{equation} x(t) = A_1 \cos(\omega t + \phi_1) \end{equation}
    \begin{equation} y(t) = A_2 \cos(\omega t + \phi_2) \end{equation}

    The problem is I can't exactly figure out what kind of system this could be. My first thoughts are some kind of 2D harmonic oscillator or some system of 2 masses coupled with springs. I'm having problem answering what \begin{equation} b^2 - 4ac \ne 0 \end{equation} "means" when I can't figure out what kind of system this is.

    Thanks in advance,
    Brotof
     
  2. jcsd
  3. Feb 18, 2014 #2
    Hi.
    I didn't check your solution but it looks ok... now:
    "2D harmonic oscillator or some system of 2 masses coupled with springs" are examples of two coupled oscillators, so you're basically right.
    For your next question, how can you express the Lagrangian when b^2= 4ac? What does it say about the relative phases of x and y?
     
  4. Feb 19, 2014 #3
    I see that the Lagrangian then can be written:

    \begin{equation}
    L = \frac{m}{2c}(b\dot{x} + c\dot{y})^2 -\frac{k}{2c}(bx + cy)^2
    \end{equation}

    Also, bot my equations of motion reduces to a single one:
    \begin{equation}
    \frac{b^2}{c}\dot{x} + b\dot{y} + \frac{k}{m}(\frac{b^2}{c} x + by) = 0
    \end{equation}

    Hmm. I'm not exactly sure what this says about the relative phases. We get the solution

    \begin{equation}
    (\frac{b^2}{c} x + by) = A \cos(\omega t + \phi)
    \end{equation}.
    Does this necessarily mean they have the same phase? I'm still not able to visualize this system.
     
    Last edited: Feb 19, 2014
  5. Feb 19, 2014 #4
    The above is wrong. If:
    ax2+2bxy+cy2 and b2= 4ac, then you have equivalently:
    (x√a + y√b)2
    And your Lagrangian can be written in terms of a single variable:
    u = x√a + y√b
    In other terms, a degree of freedom has been lost somewhere. Think about this example:
    you have first a system of two masses and three springs, something like that:
    |^^O^^O^^|
    Now replace the central spring by a rigid rod:
    |^^O––O^^|
    See the connection?
     
  6. Feb 19, 2014 #5
    Thanks a lot, that makes sense!

    Im really sorry, I was a bit hasty and have made an error. It's supposed to be
    \begin{equation}
    b^2 = ac
    \end{equation}
    without the factor 4.

    I don't think I'm wrong when taking that into account that I meant b^2 = ac, even though I wrote it a bit more complicated that necessary.

    \begin{equation}
    \frac{k}{2}(ax^2 + 2bxy + cy^2) = \frac{k}{2}(\frac{b^2}{c}x^2 + 2bxy + cy^2) = \frac{k}{2c}(b^2 x^2 + 2bcxy + c^2 y^2) = \frac{k}{2c}(bx + cy)^2
    \end{equation}
    and of course the same way for the kinetic energy term.

    Then we get

    \begin{equation}
    bx + cy = A\cos(\omega t + \phi)
    \end{equation}

    Anyway, the help is appreciated!
     
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