# B Difference between angular frecuency and velocity pendulum

1. Aug 7, 2017

Hello! I hope someone could help me to solve mu doubt, I am very confused and I don't find answers in internet. My question is about pendulums. I know the angular frecuency of a pendulum is give by the equation w= sqrt(g/L). But also i know the angular velocity (also named with "omega") can be found from v=w.r . My question is, those w's have the same measure? or they differ?

This question arose from the following exercise:

A simple pendulum has a SHM, doing a maximum angle regarding to the vertical of 5°. Its period is 2.21s.
a) What is its lenght?
b) What is its maximum velocity?
c)What is its maximum angular velocity?

The book solution is this:
https://ibb.co/exBCuv

As you can see, the "w" is reckoned using v=w.r , but if I reckon the "w" with w= sqrt(g/L), the anwer is different and i don't know why.
Thanks in advance for you help

2. Aug 7, 2017

The pendulum has a motion that is sinusoidal, basically $x=A \cos(\omega t)$, with some amplitude $A$, to a very good approximation, = provided the oscillations are not too large. $\omega=2 \pi f=2 \pi /T =\sqrt{g/L}$. (with credit to @jbriggs444 for the last equality from his post below). (It can also have an arbitrary phase angle $\phi$). $\\$ If you want to consider circular motion about the point where it rotates, $x/L=tan(\theta) \approx \theta$ (measuring in radians, small angles), so that $\dot{\theta}=\dot{x}/L=-A \omega \sin(\omega t)/L$, (at least for small oscillations). This $\dot{\theta}$ is the angular velocity of the object in a rotation about the point where the pendulum is hung from. Velocity $v=L \dot{\theta}$.

Last edited: Aug 7, 2017
3. Aug 7, 2017

### jbriggs444

They differ.

[Note that you can find an "omega" symbol on these forums by clicking on the Σ sign above the editting panel and finding the "ω" character in the resulting list of special characters]

The omega for angular frequency relates to the tick rate of the pendulum. Two pi radians per back-and-forth cycle at a constant rate. You could think of this as if the position of the tip of the pendulum was matching the x coordinate on a blip moving in a small circle whose radius is equal to the amplitude of the pendulum swing. This is the $\omega=\sqrt{\frac{g}{L}}$

The omega for the angular velocity of the pendulum relates to the angle of the shaft of the pendulum as the tip cycles back and forth over the bottom part of a larger circle centered on the axis to which the pendulum is attached. This angular velocity will vary between a maximum at the bottom of the stroke and zero at the ends of the stroke. This is the $v=\omega r$.

4. Aug 7, 2017

### pixel

I'm not following what jbriggs444 is saying. The two angular velocities (radians per second) that you reference are the same. v is the velocity (meters per second) of a point along the pendulum and that will vary with distance, r, from the pivot point. That's reasonable as during the same time interval a point at the end of the pendulum travels further than a point at, say, the midpoint, hence it has higher velocity.

5. Aug 7, 2017

### jbriggs444

No, they are not.

The one is the notional angular velocity of an imaginary point going around an imaginary circle in lock-step with the ticks of the pendulum, one rotation per pendulum round trip. That angular velocity is roughly constant.

The other is the physical angular velocity of the pendulum shaft. That angular velocity is changing as the pendulum swings back and forth.

6. Aug 8, 2017

### pixel

Oh, I see. Thanks.