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Not so simple harmonic motion

  1. Oct 17, 2006 #1
    OK, the answer for this problem seems a bit high to me, so I'm going to ask if it all seems alright.

    You have a 1m rod of no mass, fixed so that it may rotate about it's center. At the top of the rod is a mass m1 (0.5kg), and the bottom is a mass m2 (1.0kg). Find the angular frequency assuming a small amplitude of oscillations.

    I just stared at it for about an hour, trying to find a way to get it into the form

    [tex]\ddot{\psi} + \omega_0 ^2 \psi = 0[/tex]

    Eventually i went at it with torques, using the [tex]\tau = r x F[/tex] forumla.


    [tex] \tau = rm_1g \sin (180- \theta) - rm_2g \sin (\theta) [/tex]

    and using small angle approximation, both sins go to theta.

    [tex] \tau = rg (m_2 - m_1) \theta [/tex]
    [tex] \tau = I \ddot{\theta} = -rg (m_2 - m_1) \theta [/tex]
    [tex] I = \sum m r^2 = (m_1 + m_2) r^2 [/tex]

    [tex] \ddot{\theta} = \frac{-rg (m_2 - m_1)}{r^2 (m_1+m_2)}[/tex]
    [tex] \ddot{\theta} + \frac{rg (m_2 - m_1)}{r^2 (m_1+m_2)} = 0 [/tex]

    Which is in the right form. Meaning that the frequency is:
    [tex] \omega_0 = \sqrt{\frac{g(m_2-m_1)}{r(m_1+m_2)}} [/tex]

    Now, that may be ALL wrong, but it's what first came to mind. I end up with 2.6Hz for an answer which seems a bit off...

    Thanks for any redirections/confirmations/help you can give!

    (if the tex is messy i'm working on it still!)
    Last edited: Oct 17, 2006
  2. jcsd
  3. Oct 17, 2006 #2


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    You could check your answer using the standard physical pendulum equation


    You already have an expression for I, and the mass and g are easy. You just need to compute the distance from support to CM. I think you will like it. I did not check your computation, but the answer looks good.
  4. Oct 17, 2006 #3
    excellant! what a helpful script! The answer does end up being the same, so I'm going to assume my theory-work was good. It's not EXACTLY how we;ve been doing those questions (been using energies and such) but I found this one much easier to follow.

    Thanks a lot! (still open for comments though)
  5. Oct 17, 2006 #4


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    Well, if you look at the first part of the link, you wil find your

    [tex]\ddot{\psi} + \omega_0 ^2 \psi = 0[/tex]

    Just replace their variable with yours
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