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Finding eigenfrequencies of a coupled pendulum

  1. Jan 9, 2012 #1
    1. The problem statement, all variables and given/known data
    This problem is from the 1992 GRE. A tube is free to slide on a frictionless wire. On each end of the tube is attached a pendulum. The mass of the tube is [itex]M[/itex]. The length and mass of the pendula are [itex]l[/itex] and [itex]m[/itex], respectively.




    2. Relevant equations
    It is given that one of the eigenfrequencies is [itex]\sqrt{\frac{g(M+2m)}{lM}}[/itex]. I want to verify this.


    3. The attempt at a solution
    I found the Lagrangians and made the suitable small angle approximation. Here is my work:
    The position of either pendulum bob, using the initial position of its pivot as an origin is\\
    [itex]
    \begin{eqnarray*}
    x = s + l\sin\phi\\
    y = -l\cos\phi\\
    \end{eqnarray*}
    [/itex]

    The derivatives are
    [itex]
    \begin{eqnarray*}
    \dot x = \dot s + l\dot\phi\cos\phi\\
    \dot y = l\dot\phi\sin\phi
    \end{eqnarray*}
    [/itex]

    The total kinetic energy is
    [itex]
    \begin{eqnarray*}
    T = \frac{1}{2}M\dot s^2 + m(\dot x^2 + \dot y^2)\\
    = \frac{1}{2}M\dot s^2 + m((\dot s + l\dot\phi\cos\phi)^2 + (l\dot\phi\sin\phi)^2)\\
    = \frac{1}{2}M\dot s^2 + m(\dot s^2 + 2l\dot s\dot\phi\cos\phi + l^2\dot\phi^2\cos^2\phi + l^2\dot\phi^2\sin^2\phi)\\
    = (\frac{1}{2}M+m)\dot s^2 + m( 2l\dot s\dot\phi\cos\phi + l^2\dot\phi^2)\\
    \end{eqnarray*}
    [/itex]

    The potential energy from the height of the bobs
    [itex]
    \begin{eqnarray*}
    U = 2mgy\\
    =-2mg\cos\phi
    \end{eqnarray*}
    [/itex]

    The Largrangian is
    [itex]
    \begin{eqnarray*}
    \mathcal{L} = T - U\\
    = (\frac{1}{2}M+m)\dot s^2 + m( 2l\dot s\dot\phi\cos\phi + l^2\dot\phi^2) + 2mg\cos\phi\\
    \end{eqnarray*}
    [/itex]

    Find the Lagrangian equation relative to [itex]s[/itex]
    [itex]
    \begin{eqnarray*}
    \frac{\partial \mathcal{L}}{\partial \dot s} = (M+2m)\dot s + 2lm\dot\phi\cos\phi\\
    \frac{d}{dt}\left ( \frac{\partial \mathcal{L}}{\partial \dot s} \right )
    = (M+2m)\ddot s + 2lm(\ddot\phi\cos\phi - \dot\phi^2\sin\phi)\\
    \frac{\partial \mathcal{L}}{\partial s}=0\\
    \boxed{(M+2m)\ddot s + 2lm(\ddot\phi\cos\phi - \dot\phi^2\sin\phi) = 0}
    \end{eqnarray*}
    [/itex]

    Find the Lagrangian equation relative to [itex]\phi[/itex]
    [itex]
    \begin{eqnarray*}
    \frac{\partial \mathcal{L}}{\partial \dot \phi} = m(2l\dot s\cos\phi + 2l^2\dot\phi)\\
    \frac{d}{dt}\left ( \frac{\partial \mathcal{L}}{\partial \dot \phi} \right )
    = 2lm(\ddot s\cos\phi - \dot s\dot\phi\sin\phi + l\ddot\phi)\\
    \frac{\partial \mathcal{L}}{\partial \phi} = -2m\sin\phi (l\dot s\dot\phi + g)\\
    \boxed{2lm(\ddot s\cos\phi - \dot s\dot\phi\sin\phi + l\ddot\phi) + 2m\sin\phi (l\dot s\dot\phi + g) = 0}
    \end{eqnarray*}
    [/itex]

    Now, recall the first equation
    [itex]
    \begin{eqnarray*}
    (M+2m)\ddot s + 2lm(\ddot\phi\cos\phi - \dot\phi^2\sin\phi) = 0\\
    2lm(\ddot s\cos\phi - \dot s\dot\phi\sin\phi + l\ddot\phi) + 2m\sin\phi (l\dot s\dot\phi + g) = 0
    \end{eqnarray*}
    [/itex]

    Take the small angle approximation
    [itex]
    \begin{eqnarray*}
    \frac{(M+2m)}{2lm}\ddot s + \ddot\phi - \dot\phi^2\phi = 0\\
    \ddot s + l\ddot\phi + \phi \frac{g}{l} = 0
    \end{eqnarray*}
    [/itex]

    It is clear that the goal is in sight, but my diffeqs skills are somewhat rusty. I can't see exactly how to show that the system has the stated eigenfrequency. Any help would be appreciated :)
     
    Last edited: Jan 9, 2012
  2. jcsd
  3. Jan 9, 2012 #2
    I wanted to add that solving algebraically (if you neglect the [itex]\dot\phi^2\phi[/itex] term), one reaches the following equation:

    [itex]
    \ddot\phi - \frac{g(M+2m)}{l^2M}\phi = 0
    [/itex]

    Which appears to have too many factors of [itex]l[/itex] in the denominator. I suppose the GRE could have been in error, but I want to know for sure.
     
    Last edited: Jan 9, 2012
  4. Jan 9, 2012 #3

    vela

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    The units don't work out with your result.

    In your derivation of the Lagrangian, how come you're assuming the angular displacement of the two pendulums are identical? Don't they oscillate independently of each other?
     
  5. Jan 9, 2012 #4
    Hmm, you're right. I'll have to look through it again.

    The question (I may not have stated this) is looking for normal mode frequencies. According to my understanding, that means that all parts oscillate such that the phase difference between them doesn't change. You're correct that this does not mean they need to have the same angular displacement. I think I must have assumed that the mode mentioned was in fact the mode where the pendula had equal angular displacement.
     
  6. Jan 9, 2012 #5
    It turns out that I forgot to bring along the [itex]l[/itex] into the potential energy; that makes the units work out.
     
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