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Period of a pendulum

  1. Jan 25, 2006 #1

    Hootenanny

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    I have already posted in the physics forum but its gone a bit quiet. Ive managed to derrive an equation up to this point. I have [tex]\frac{g}{L}\theta = \omega^2 \theta_{max} \sin (\omega t - \alpha) [/tex] and I need to prove that [tex] \omega = \sqrt{\frac{g}{L}} [/tex]. I'm stuped at this one. Thank's in advance for your help.
     
  2. jcsd
  3. Jan 25, 2006 #2
    well, you need to know that it's only an approximation. It is most accurate when sinx=x (in radians of course).
     
  4. Jan 26, 2006 #3

    Hootenanny

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    That doesnt help me much all that gives then is [tex]\frac{g}{L}\theta = \omega^2 \theta_{max}(\omega t - \alpha) [/tex]
     
  5. Jan 26, 2006 #4

    Hootenanny

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    Now I have [tex]\theta = \theta_{max} \sin(\sqrt{\frac{g}{L}} t - \alpha ) [/tex] How Can I remove the [itex]\alpha [/itex] ?
     
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