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Time period of a simple gravity pendulum

  1. Nov 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Put of curiosity I have tried to derive a formula for the time period of a pendulum that oscillates under gravity. Below you can find my workings, which I have checked on wikipedia and the workings are practically identical. However I got stuck at one point, and this is where internet pages show the further workings. However I do not understand how they have been obtained. Any help would be appreciated.

    2. The attempt at a solution

    [tex] F_{net}=-mgsin\theta [/tex]

    The overall net force will pull towards the equilibrium position. The force which causes the pendulum to oscillate is due to gravity. Tension/other force all result because of the force of gravity.

    Assume [tex]\theta\ll1[/tex] then [tex]sin\theta\approx\theta[/tex]

    This small angle approximation is necessary as the resultant upcoming differential equation would not reduce to an appropriate solution. As long as the angular displacement is small enough, the solution will hold true.

    [tex]
    \therefore F_{net}=-mgsin\theta and a_{net}=-g\theta
    \\
    \\
    [/tex]

    [tex]
    s=l\theta
    [/tex]
    where s = arc length and l = radius from centre

    [tex]
    \\
    \\v=\dfrac{ds}{dt}=l\dfrac{d\theta}{dt}[/tex] and [tex]a=\dfrac{d^{2}s}{dt^2}=l\dfrac{d^{2}\theta}{dt^2}
    \\
    \\ l\dfrac{d^{2}\theta}{dt^2}=g\theta
    \\
    \\ \dfrac{d^{2}\theta}{dt^2}-\dfrac{g\theta}{l}=0
    [/tex]

    (From this point onewards I don't understand. Please explain it to me like I'm 5 if possible.)

    [tex]
    \\ \theta=\theta_{max}sin\sqrt{\dfrac{g}{l}}t
    \\
    \\ \omega=\sqrt{\dfrac{g}{l}}
    \\
    \\ f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}
    \\
    \\ f^{-1}=(\dfrac{1}{2\pi}\dfrac{g}{l})^{-1}=T=2\pi\sqrt{\dfrac{l}{g}}
    [/tex]
     
    Last edited: Nov 8, 2013
  2. jcsd
  3. Nov 8, 2013 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    I guess it's simply a case of this sinusoid being known (by mathematicians) to be a solution to that second order D.E. that you arrived at above. You can show this yourself by differentiating it to get theta dot, then differentiating it again to get theta double dot, and substituting those values back into your second order D.E.

    This is a sinewave of maximum value theta max, and of frequency (in radians/sec) given by that square root term.

    P.S. you are missing the square root sign in the line: ω = g/l
     
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