Why Does a Simple Pendulum Have a Constant ω If Its Velocity Changes?

[bananabandana]

Homework Statement

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

Homework Equations

(1) v=ωR
(2) \omega=\stackrel{\sqrt{l}}{\sqrt{g}}

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.

 

[jfizzix]

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

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

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

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)

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

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

a(t) \ddot{x}(t) = -\omega^{2}X_{0}Sin[\omega t + \phi]
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

\ddot{x}(t) = -\omega^{2}x(t)

So in short, the constant \omega 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 x(t)^{2} and v(t)^{2}, you can show that

\omega = \sqrt{\frac{<v^{2}>}{<x^{2}>}}

 

[NihalSh]

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