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Angular Frequency of Pendulum.

  1. Oct 11, 2013 #1
    1. The problem statement, all variables and given/known data
    How does a simple pendulum have a constant ω if it's velocity changes?


    2. Relevant equations
    (1) v=ωR
    (2) [itex]\omega=\stackrel{\sqrt{l}}{\sqrt{g}}[/itex]

    3. The attempt at a solution

    Very confused about this. The pendulum bob is at a fixed distance from where it's attached (i.e R is constant). It seems to me that if the velocity changes, from the equations above, the angular frequency has also got to change. I thought perhaps that the angular frequency derived from SHM was perhaps an average and not instantaneous? But I'm not so sure.
     
  2. jcsd
  3. Oct 11, 2013 #2

    jfizzix

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    A simple pendulum's velocity does change with time, but it changes periodically with the angular frequency [itex]\omega[/itex]. The acceleration of the pendulum as well as the posiiton of the pendulum also change with time, but periodically with angular frequency [itex]\omega[/itex].

    If you think of the position function with time as being a sine wave

    [itex]x(t) = X_{0}Sin[\omega t + \phi][/itex]

    the time derivative of this (i.e. the velocity) will also be a sine wave (a cosine wave has the same shape and behavior in any case)

    [itex]v(t)= \dot{x}(t) = \omega X_{0}Cos[\omega t + \phi][/itex]

    Similarly, differentiating again gives us the acceleration which also varies sinusoidally with the same frequency [itex]\omega[/itex]

    [itex]a(t) \ddot{x}(t) = -\omega^{2}X_{0}Sin[\omega t + \phi][/itex]
    Note that here we can write the acceleration in terms of the position, and see that sine waves are solutions to the simple harmonic oscillator equation

    [itex]\ddot{x}(t) = -\omega^{2}x(t)[/itex]

    So in short, the constant [itex]\omega[/itex] does not refer to the instantaneous angular velocity, but it gives how rapidly the instantaneous angular velocity oscillates with time.

    In particular, if you look at the time averages of [itex]x(t)^{2}[/itex] and [itex]v(t)^{2}[/itex], you can show that

    [itex]\omega = \sqrt{\frac{<v^{2}>}{<x^{2}>}}[/itex]
     
  4. Oct 11, 2013 #3
    I know what you mean. Its natural to get confused, but ω represents different things in both equations.

    In ##v=ωr##, ω isn't constant because gravity applies torque. : here ω means angular velocity

    in ## ω= \sqrt{\frac{g}{l}}##, ω is constant- its a value associated with Simple harmonic motion, which jfizzix has explained above. : here ω means angular frequency
     
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