Answer in two parts:
(a) for circular motion, the angular speed \(\omega\) measures the number of radians per second. So how long does it take to do a full revolution ((2\pi\) radians)? Answer: \(2\pi/\omega\). Right?
(b) For simple harmonic motion, the general solution is something like \(u(t) = A\cos(\omega t)+B\sin(\omega t)\) - this usually has nothing to do with circular motion, so \(\omega\) is not \(v/r\). But the period of \(u(t)\) is what? Again, it is \(2\pi/\omega\). Because \(u(t+2\pi/\omega)=u(t),\) right?
The similarity is why the same letter is traditionally used for angular speed and (angular) frequency.