# Angular Frequency of Pendulum.

## 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
Gold Member
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

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

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