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Frequency of a simple pendulum

  1. Sep 27, 2011 #1
    A simple pendulum of length L and mass m is suspended in a car that is traveling with constant speed v around a circle of radius R. If the pendulum undergoes small oscillations in a radial direction about its equilibrium position, what will its frequency of oscillation be?

    I don’t know how to post diagrams, so...
    The negative x direction is to the left, toward the center of the circle of radius R. [itex]\theta[/itex] is the angle L makes with the vertical.
    The forces on m are the gravitational force and the tension along length L. The forces result in SHM for small values of [itex]\theta[/itex].

    [itex]F_{x}=-kx=-mgsin\theta -m\frac{v^{2}}{R}sin\theta[/itex]

    For small values of [itex]\theta[/itex], [itex]\theta \approx sin\theta[/itex].
    [itex]s=L\theta[/itex]. Where s is the arc length L sweeps through angle [itex]\theta[/itex].
    For small values of [itex]\theta[/itex], [itex]s \approx x[/itex]

    Using these two approximations, the top equation becomes
    [itex]kx=mg\frac{x}{L} +m\frac{v^{2}x}{RL}[/itex]
    After cancelling the x's and rearranging, I get
    [itex]\frac{k}{m}=\frac{g+\frac{v^{2}}{R}}{L}[/itex]
    For frequency f and period T, [itex]f=\frac{1}{T}[/itex], [itex]T=2 \pi \sqrt{\frac{m}{k}}[/itex],
    [itex]f=\frac{1}{2 \pi}\sqrt{\frac{g+ \frac{v^{2}}{R}}{L}}[/itex]
    Unfortunately, the answer in the back of the book is
    [itex]f=\frac{1}{2 \pi}\sqrt{\frac{\sqrt{g^{2}+\frac{v^{4}}{R^{2}}}}{L}}[/itex]

    Thanks for any help.
     
    Last edited: Sep 27, 2011
  2. jcsd
  3. Sep 28, 2011 #2

    ehild

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    The pendulum will oscillate around an angle θ0 because of the centrifugal force acting in the car as rotating frame of reference: tan(θ0)=v2/(Rg). The deviation from this angle can be considered small, but not the angle with respect to vertical.

    ehild
     
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