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Homework Help: Angular frequency - how is it derived?

  1. Apr 10, 2005 #1
    How is the formula for angular derived?

    [tex]\omega=\sqrt{\frac{g}{L}}[/tex]

    my book has these equations:

    [tex]F= -mg sin\theta = m \frac{d^2s}{dt^2}[/tex]

    [tex]\frac{d^2 \theta}{dt^2}=-\frac{g}{L}sin\theta[/tex]

    [tex]\frac{d^2 \theta}{dt^2}=-\frac{g}{L}\theta[/tex]

    [tex]\omega=\sqrt{\frac{g}{L}}[/tex]

    what exactly is m [tex]\frac{d^2s}{dt^2}[/tex] and [tex]\frac{d^2 \theta}{dt^2}[/tex]
     
  2. jcsd
  3. Apr 10, 2005 #2

    dextercioby

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    [tex]\mbox{s} [/itex] is arc length...Because the movement is on a circle

    [tex]s=L\theta[/tex]

    ,where [itex] \theta[/itex] is the angle at the center of the circle...

    Daniel.
     
  4. Apr 10, 2005 #3

    Galileo

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    s is the position of the particle, so the second derivative wtr time denotes the acceleration 'a' of the particle. (Hence F=ma in the first equation)

    Solve the D.E. and check what that you get a periodic function. What is its angular frequency?
     
  5. Apr 10, 2005 #4

    dextercioby

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    Instead of applying Newton's second law for the translation movement [itex] \frac{d\vec{\mbox{p}}}{dt}=\sum_{k} \vec{\mbox{F}}_{k} [/itex],try to apply it for the rotation movement [itex] \frac{d\vec{\mbox{L}}}{dt}=\sum_{k} \vec{\mbox{M}}_{k} [/itex]


    Daniel.
     
  6. Apr 10, 2005 #5
    okay, I understand up to here:
    [tex]\frac{d^2 \theta}{dt^2}=-\frac{g}{L}\theta[/tex]


    I think the next step is this...
    [tex]\sqrt{\frac{d^2 \theta}{dt^2}}=\sqrt{-\frac{g}{L}\theta}[/tex]
    [tex]\sqrt{\frac{d^2 \theta}{dt^2}}=\frac{d \theta}{dt}= \omega[/tex]
    [tex]\omega = \sqrt{\frac{g}{L}[/tex]

    but I dont think I'm correct
     
  7. Apr 10, 2005 #6

    dextercioby

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    Of course not.

    It's a definition

    [tex] \frac{g}{l}=:\omega_{0}^{2} [/tex]

    That's all to it.

    Daniel.
     
  8. Apr 10, 2005 #7
    but how do I get [tex]\frac{d^2 \theta}{dt^2}=-\frac{g}{L}\theta[/tex] to go to [tex]\omega=\sqrt{\frac{g}{L}}[/tex]
     
  9. Apr 10, 2005 #8

    jtbell

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    How much have you studied about solving differential equations? Here in the USA, very few people study differential equations at the K-12 level.

    It turns out that a general solution of the differential equation

    [tex]\frac {d^2 \theta} {d t^2} = - \frac {g}{l} \theta[/tex]

    is

    [tex]\theta = A \sin \left(\sqrt {\frac {g}{l}} t + \theta_0\right)[/tex]

    where [itex]A[/itex] and [itex]\theta_0[/itex] are arbitrary constants. You can verify this by working out the second derivative and plugging it back into the differential equation. A general form of a sinusoidal function is

    [tex]x = A \sin (\omega t + \theta_0)[/itex]

    where [itex]\omega[/itex] is the angular frequency. Matching up the preceding two equations gives you

    [tex]\omega = \sqrt {\frac {g}{l}}[/tex]
     
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