# When the oscillator is a system with an angle can we define the angular frequency to be the radians per unit time covered by the system itself?

I read on stackechange that in springs or any one dimensional oscillator the angular frequency is just describing a rate of angle change in the associated circle on which it's projected. Something like this :

My question is: suppose you have a pendulum as an oscillator. Would it be correct to say that the omega / angular frequency. Is a measure of radians per unit time that the pendulum itself going through. Or there is still some other hidden circle for which this is defined?

Edit: is the following definition possible? : we take the whole edge on which the pendulum is passing through and circulate it. Meaning, we make a closed circle out of it . Would then the radians in that circle can be thought of as the angular frequency? If it's correct, would that be correct for the line on which a linear spring is oscillating?

• It's still not radians per second where the radians are measuring angle in the pendulum's circle. One reason why it can't be is that the pendulum doesn't move through angles in this circle at a constant rate: at the top of its swing its instantaneously stationary. – jacob1729 Jun 1 '18 at 20:09
• OK. But I am not talking about velocity. I am ignoring anything else when I want to talk about angular frequency. Since each covering of the area is a repetition can't we take the total edge and make a circle out of it. So that each cycle is a period. And so 2pi devided by period would yeald frequency. What bothers you that the velocity isn't constant? Frequency doesn't care about velocity, it's just a matter of period and radians. All what matters is repetitions. – bilanush Jun 1 '18 at 20:28
• Hmmm ... $\omega$ does not point in the direction of the red arrow. $v_t$ points that way, but $\omega$ points out of the page. – dmckee --- ex-moderator kitten Jun 2 '18 at 0:55
• Possible duplicate of Is angular frequency the same as angular velocity or are they different? posted by the same user. – sammy gerbil Jun 2 '18 at 23:49

Would it be correct to say that the omega / angular frequency. Is a measure of radians per unit time that the pendulum itself going through.

If $\theta(t)$ is the (small, time varying) angle the pendulum makes with the vertical axis, then

$$\theta(t) = \theta_0 \cos\left(\phi(t)\right)$$

where $\theta(t)$ is the angular position of the pendulum and $\phi(t)$ is the phase (or phase angle) of the simple harmonic motion given by

$$\phi(t) = \omega t + \phi_0$$

where $\omega = \dot{\phi}$ is the (constant) angular frequency which is given by

$$\omega = \sqrt{\frac{g}{l}}$$

But the "radians per unit time that the pendulum itself [is] going through" isn't constant with time and isn't the angular frequency but is instead the angular velocity:

$$\dot{\theta}(t) = -\omega\,\theta_0\sin\left(\phi(t)\right)$$

So the answer to the quoted question is no, since the angular frequency of the pendulum is the time rate of change of the phase angle while the angular velocity of the pendulum is the time rate of change of the angular position.

• Looks like you were 1 minute faster than I was :) – Aaron Stevens Jun 1 '18 at 20:54
• Fine. But since in any case the angular frequency is pretty vague. Wouldn't my description give you a some sort of 'average velocity '? Can't we define the angular frequency to be the average angular velocity of the system? That makes sense why in constant velocity in circular motion both of them turn out to be identical because the instantaneous velocity is just equal to the average there. – bilanush Jun 1 '18 at 21:21
• @bilanush, the average angular velocity for the pendulum is zero (the average value of a sinusoid over a cycle is zero). – Hal Hollis Jun 1 '18 at 21:28
• I mean, you take the edge and you circulate it defining angular frequency on said circle would yeald the average of all velocities in all instants. – bilanush Jun 1 '18 at 21:34
• Hal can you please explain why does Wikimedia state that angular frequency and angular velocity are just the same? When in fact you got completely different results for each. One is a constant and one is a changing derivative. – bilanush Jun 2 '18 at 7:56

The construct that you are looking at is sometimes called "the reference circle" and a electrical engineer would call it a "phasor".

In either case the circle is not (necessarily1) a real thing, it is a mathematical construct used for several purposes

1. It gives a visual representation (and a simple addition scheme) to a mathematical construct that can be hard to to deal with intuitively. For instance comparing the behavior of out-of-phase oscillators algebraically involves many nasty trig identities, but in the phasor picture it involves only one angle.
2. It provides a way of understanding the system without calculus. (And has been used in many algebra/trig based textbooks for that reason, but PER people have been discouraging this practice).

You certainly can define the angular frequency of the oscillator as

the radians per unit time covered by the [phasor describing the] system

1 You can realize a phasor as part of a demonstration. Place a vertical pin on a turntable and shine a bright light on it from across a darkened room. The shadow of the pin will execute 1D harmonic motion on the wall.

For bonus points, rig a small angle pendulum with the same frequency just above the projection height of the pin, and set it in motion so that it matches the motion of the shadow. (This is hard but I've seen it done really well once and it is beautiful—even mesmerizing—to watch.)

With a pendulum and your defined circle, the rate at which the angle $\theta$ is changing $(\frac{d\theta}{dt})$ is not the same as the angular frequency ($\omega$).

Let's assume that our pendulum is only displaced by a small angle, so we can use the small-angle approximation and get a nice analytic solution of $\theta$ as a function of time. This solution could look something like $\theta(t)=Acos(\omega t)$.

Now, we see that the time derivative of this is $\frac{d\theta}{dt}=-A\omega sin(\omega t)$ which is not the same thing as just $\omega$. Therefore, the rate of change of the angle in this circle is not the same thing as the angular frequency of the oscillation.

We use the term "angular frequency" with sinusoidal oscillations since the argument to sinusoidal functions are angles. Even though in the example of the pendulum $\omega$ does not refer to an actual angle in our system, it is still called an angular frequency. It seems like you have a grasp on the actual physics involved, so I wouldn't get too hung up on definitions and semantics.

• Nice answer. But let me comment on your first paragraph . Correct me if I am wrong. You wrote "With a pendulum and your defined circle, the rate at which the angle θ is changing (dθ/dt) is not the same as the angular frequency (ω)." Isn't it different in your defined circle as well? Even in your system the(dθ/dt) is just a changing instantaneous velocity. While omega is a constant. In my defined circle I can say, that omega would be the average velocity of the system. Isn't it correct? – bilanush Jun 1 '18 at 21:30
• @bilanush What are you averaging over in your proposed averaging? – Aaron Stevens Jun 2 '18 at 0:22
• All the velocities in all instants. – bilanush Jun 2 '18 at 0:23
• Also, I don't understand your refutation. You refuted my claim by showing that the angular velocity doesn't equal angular frequency. But in your circles this wouldn't happen either. Suppose the velocity is changing but the period is constant. You would get two different results for omega and velocity. Even in circular motion if it doesn't move at constant velocity but period is constant the two would be different. Cuz angular frequency is constant and angular velocity is changing – bilanush Jun 2 '18 at 0:33
• @bilanush Here's the thing. In the simple pendulum system the angular velocity and the angular frequency as typically defined are not the same thing. If you want to define new things that are physically consistent then go for it. As for your averaging you aren't making sense. You cannot average over "all velocities in all instances" since this value would not converge to anything. You would need to define some finite time internal to average over. I'm not sure how this average would relate to the angular frequency, but like I said, you can define things however you want in your system. – Aaron Stevens Jun 2 '18 at 0:45