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Is angular SHM also related to UCM ?

  1. Jan 10, 2009 #1
    Hi all,

    I have learned that linear SHM is just projection of UCM onto one axis. Is angular SHM also related to UCM ?

    Moreover, I have also learned that theta = theta0 * sin(wt+P) where theta is angular displacement, theta0 is angular amplitude and P is phase constant. Where does this equation come from, or how can I derive it ?
  2. jcsd
  3. Jan 10, 2009 #2
    Well, I somehow get the idea that they are related..

    Thus instead, "how they are related" would be a better question actually.
  4. Jan 10, 2009 #3
    I guess I need to be more precise. Assume we have a simple pendulum. We say that w = sqrt(g/L). Moreover, we say T = 2PI / w

    It is quite a lot confusing for me to say period is equal to 2PI / w for a simple pendulum. First of all, if the angular displacement was 2PI and w was constant; then it would make sense to say T = 2PI / w . However, w is not constant and the angular displacement is not 2PI..

    Well, if motion is analog to a uniform circular motion in which w is constant and always equal to sqrt(g/L); then it seems to be fine. But I'm not sure about that, too.

    I seriously need some help here.
  5. Jan 10, 2009 #4

    Doc Al

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    Staff: Mentor

    Don't confuse the angular frequency of the pendulum's simple harmonic motion (ω), with the angular speed of the pendulum itself.
  6. Jan 10, 2009 #5
    Hmm, so as far as I understand, angular freq. is constant and angular speed (of pendulum) is variable.

    saying w = sqrt(g/L) ; w is here angular freq. I guess.

    Can we say that angular frequency is actually angular velocity of a UCM(not the pendulum) analog of our SHM in this case ?
  7. Jan 10, 2009 #6

    Doc Al

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    Staff: Mentor

    I don't see why not.

    [tex]x = x_0\sin \omega t[/tex]

    is analogous to

    [tex]\theta = \theta_0\sin \omega t[/tex]

    (You'd have to have the radius of the circular motion represent θ0.)
  8. Jan 10, 2009 #7
    Thanks a lot!

    Now, I would like to prove that

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

    I have tried:

    [tex] Ftan = - mg \sin\theta [/tex]

    (for simple pendulum)


    [tex] a = g \sin\theta[/tex]

    [tex] V = \int g\sin\theta dt[/tex] 0 to t


    [tex] X = \int V dt [/tex] 0 to T/2.


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

    Then; using [tex] w = \frac {2PI} {T} [/tex]

    I would obtain a formula for w -- which hopefully would be equal to

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

    So I guess now the problem is to calculate

    [tex] V = \int g \sin\theta dt[/tex]

    Well, you say that

    [tex]\theta = \theta_0\sin \omega t [/tex]

    but I don't really know why this equation is correct.
    Last edited: Jan 10, 2009
  9. Jan 10, 2009 #8

    Doc Al

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    Staff: Mentor

    That's for a mass on a spring, not a pendulum.

    Why the tan?

    Just: F = -mg sinθ

    Then use a small angle approximation and Newton's 2nd law to set up (then solve) the differential equation.
  10. Jan 10, 2009 #9
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