Angular Frequency of Pendulum.

  • #1

Homework Statement


How does a simple pendulum have a constant ω if it's velocity changes?


Homework Equations


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

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.
 

Answers and Replies

  • #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]
 
  • #3
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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.

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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