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Finding the period of a simple pendulum given angular acceleration

  1. Dec 23, 2011 #1
    1. The problem statement, all variables and given/known data
    A physical pendulum oscillates through small angles about the vertical with the angle, measured in radians, obeying the differential equation d2θ/dt2 = -4πθ. What is the period of the oscillation?


    2. Relevant equations
    T = 2π √(L/g)


    3. The attempt at a solution
    I attempted to integrate twice which should give me θ. This did not work, though, because the acceleration is defined by θ.

    Thanks!
     
  2. jcsd
  3. Dec 23, 2011 #2

    BruceW

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    Homework Helper

    try to think of a function which you could use for theta. (i.e. a substitution)
     
  4. Dec 23, 2011 #3
    I am trying to think...
    θ = acos(H/L)
    But I don't think that's what you're thinking of.

    Another hint please?
     
  5. Dec 23, 2011 #4

    I like Serena

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    Homework Helper

    The differential equation for a pendulum (that belongs to your relevant equation) is:
    $${d^2\theta \over dt^2} + {g \over L}\theta=0$$
    (See http://en.wikipedia.org/wiki/Pendulum_(mathematics).)

    Perhaps you can match that with your differential equation?


    Otherwise you would need to learn how to solve the DE...
     
  6. Dec 23, 2011 #5
    So from there,
    [itex]-4\pi\theta + {g \over L}\theta=0[/itex]

    [itex]{L \over g}={1 \over 4\pi}[/itex]

    [itex]T=\sqrt{4\pi^{2} \over 4\pi}[/itex]

    [itex]T=\sqrt{\pi}[/itex]

    Which is the answer! Thanks!
     
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