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Homework Help: Lagrangian of pendulum

  1. Apr 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider a pendulum of mass m and length b in the gravitational field whose point of attachment moves horizontally [tex]x_0=A(t)[/tex] where [tex]A(t)[/tex] is a function of time.

    a) Find the Lagrangian equation of motion.
    b) Give the equation of motion in the case of small oscillations. What happens in that case when [tex]A(t)=cos\left(\sqrt{\frac {a} {b}}t\right)[/tex]

    2. Relevant equations

    [tex] {\cal L} = T - U [/tex]

    3. The attempt at a solution

    a) The position of the pendulum would be given by:

    [tex] x = A(t) + bsin\left(\theta\right) [/tex] [tex] \dot{x} = \dot{A}(t) + b\dot{\theta}cos\left(\theta\right) [/tex]
    [tex] y = bcos\left(\theta\right) [/tex] [tex] \dot{y} = -b\dot{\theta}sin\left(\theta\right) [/tex]

    The kinetic energy [tex]T[/tex] would be equal to:

    [tex] T = \frac {m} {2} \left({\dot{A}(t)}^2 + 2\dot{A}(t) b \dot{\theta} cos\left(\theta\right) + b^2\dot{\theta}^2\right) [/tex]

    and taking the zero potential to be at [tex] x = 0 [/tex] I get that the potential is equal to :

    [tex] U = -mgy = -mgbcos\left(\theta\right) [/tex]

    And the Lagrangian would be:

    [tex] {\cal L} = T - U = \frac {m} {2} \left({\dot{A}(t)}^2 + 2\dot{A}(t) b \dot{\theta} cos\left(\theta\right) + b^2\dot{\theta}^2\right) + mgbcos\left(\theta\right) [/tex]

    I would like to know if I have represented the position of the pendulum the right way because I get a non-linear differential equation for part b and I doubt that's right. Thanks!
  2. jcsd
  3. Apr 4, 2012 #2
    I'm sure the system will be nonlinear. What you should assume is that the oscillations are small, θ << 1. Then you can, by construction, drop all the nonlinear terms.
  4. May 7, 2012 #3
    May i know what will be the phase plot for the same? I mean how should i proceed to get the phase plot for the same pendulum as above.
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