I can't seem to find a satisfactory answer on stack exchange for this question, so I will present an example which I would appreciate some clarification on. Let's say we have a pendulum with mass $m$ attached to it, and it goes through one full oscillation in period $T$. Now, say we divide the arc s which the mass goes through in half a period into 10 pieces. So, a full oscillation corresponds to 20 of those pieces. I will call each of those small pieces x. Say the mass is at rest, and we just release it. It will first go through a distance x in time t1, then through another x through time t2. I think it's obvious that t1 is greater than t2. So, it's pretty obvious that the angular velocity is greater for the second x that is traversed. Now, the angular frequency is $\frac{2\pi}{T}$. The way I interpret this is that the mass goes through $\frac{2\pi}{T}$ cycles per second. But I don't see how that makes sense if the mass is traversing different "parts" of the cycle (i.e, the small pieces x which each correspond to $\frac{1}{20}$th of a cycle) in different amounts of time.
2 Answers
The angular frequency in this case does not correspond to the angle the pendulum makes with the vertical. It corresponds to the "angle" of the argument of your sinusoidal functions. This question could also be seen from a different view by considering a block of mass $m$ oscillating on a spring with spring constant $k$. The natural angular frequency of this system is $\omega=\sqrt{k/m}$, but there isn't any sort of relevant physical angle to be found here.
The solutions of SHM have the form $$x(t)=A\sin(\omega t+\phi)$$
Where $A$ is an amplitude and $\phi$ determines the phase of the oscillation. The angular frequency $\omega$ describes the rate of change of the argument $\omega t+\phi$ which is an angle. Hence "angular frequency".
Of course, I agree that this gets confusing because we use also can talk about angular velocity, which is also usually denoted with the symbol $\omega$. You just need to keep everything straight in your head. You are correct that the angular velocity of the pendulum about its pivot does change throughout the oscillation process, but the angular frequency will remain constant.
The problem is that you are mixing two models.
On the one hand, you can regard a pendulum as a circular motion.
On the other hand, you can see it as a harmonic motion.
You have to choose your model. If the oscillations are small enough ($\theta<14º$), the movement can be seen as a harmonic motion. That is an approximation. In that approximation, the movement is considered to be straight and oscillating, in a harmonic way. That is, we say "a pendulum and a spring is the same", and we describe the movement with an amplitude and a frequency (or period).
In that case, $\omega=\dfrac{2\pi}{T}=\sqrt{\dfrac{k}{m}}$
Mainly because you solve a differential equation, for which the movement is
$$ A\cdot \sin(\omega t + \phi_0)$$
And the reason we call it "omega" is because a big analogy with circular motion. It's an analogy, but not the same.
Note: actually, $\omega$ is the angular velocity of the phasor which is solution to the complex differential equation. That's another reason to call it $\omega$ as well.
On the other hand, if you want to describe it as a circular motion (or a motion around a circular arc), then it does not make sense to talk abour angular frequency because it is not a harmonic motion anymore.
Nevertheless, you could describe it as a non-uniformly accelerated circular motion, with varying acceleration, and varying angualr velocity $\omega$, but $\omega$ would be non-constant.
So, the thing is to be aware that you are choosing a model, and not to mix models.
