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Explaining Pendulum

  1. Feb 16, 2006 #1
    Why is the period of the pendulum proportional to the square root of the length?
     
  2. jcsd
  3. Feb 16, 2006 #2

    arildno

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    Well, what physical parameters would you think might affect the period?
    How can you combine these parameters in order to get the physical dimension of "time" (i.e, the dimension of the period)
     
  4. Feb 16, 2006 #3

    Tide

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    The vertical displacement of a pendulum during its swing is approximately [itex]L \theta^2 / 2[/itex] (using the small angle approximation for the cosine). Multiply by m g to get for the available potential energy for its motion.

    Also, the speed of the pendulum as it passes its lowest point will be about

    [tex]v \sim \frac {L \theta}{T}[/tex]

    Now set

    [tex]g L \theta^2 / 2 \sim v^2/2 \sim \frac {1}{2} \left(\frac {L \theta}{T}\right)^2[/tex]

    to find [itex]T \sim \sqrt {L/g}[/itex].
     
    Last edited: Feb 16, 2006
  5. Feb 16, 2006 #4
    erm..can i say..i have no idea what you just said..physics isnt my strong point :cry:
     
  6. Feb 16, 2006 #5

    Tide

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    Well, you did ask a physics question in a physics forum so ... ! :)
     
  7. Feb 17, 2006 #6
    Okay, forget the why question

    What if the simple forumla showing that the period of the pendulum proportional to the square root of the length?
     
  8. Feb 17, 2006 #7

    Tide

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    Okay,

    [tex]T = 2 \pi \sqrt { \frac {L}{g}}[/tex]
     
  9. Feb 17, 2006 #8
    Ah-ha. I thought so. Well i knew there was a 2pi in there somewhere..
    I dont need to say why its like that....thankfully.
    Hopefully that should be all the help i need...
     
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