# Relation between angular velocity and linear velocity in a pendulum

For a pendulum in simple harmonic motion, I understand that the period is independent of amplitude, and I know that the angular velocity is inversely proportional to the period. However, doesn't the angular velocity change as linear speed changes?

• Congratulations! This is the three hundred thousandth post (#300000) ever to be posted on this site. Dec 21, 2016 at 18:42

Yes there is a very simple relationship between the angular velocity and the tangential velocity, given by $v_t = \omega r$, where $v_t$ is the tangential velocity, $\omega$ is the angular velocity and $r$ is the radius of the pendulum.

There are two different concepts of angular velocity associated with a pendulum, and you have mixed up the two.

The displacement of the pendulum is given by something like:

$$x = A \sin(\omega t) \tag{1}$$

where $\omega$ is a constant called the angular velocity and related to the frequency and period by:

$$\omega = 2\pi f = \frac{2\pi}{\tau}$$

However there also the angluar velocity of the pendulum bob around the pivot point which is given by:

$$\Omega = \frac{v}{r}$$

and if we differentiate equation (1) to get the velocity we get:

$$\Omega = \frac{A\omega}{r} \cos(\omega t) \tag{2}$$

so this second type of angular velocity is not a constant. Your confusion arises because you have mixed up the constant $\omega$ and the variable $\Omega$.

• Thank you for your answer! So if the variable angular velocity is the bob around the pivot, where does the constant angular velocity come from? How would you measure that? I'm a bit confused why there are two. Dec 22, 2016 at 7:06
• @johnsmith4725: it's because simple harmonic motion is closely connected to circular motion. Have a look through these related videos. The constant angular velocity is the angular velocity of the associated circular motion. Dec 22, 2016 at 8:28