I know there are duplicates. But the answers seem to disagree and also I have more specific questions related to this title.

First of all, most questions on this site which ask this question have at least two answers of which one is at variance with the other.

Some say they are the same thing, just that one is scalar 'magnitude ' and one is a vector with 'direction'. On the other hand you have some answers saying they are completely different things apart from the case of complete cyclical motions of constant velocity.

Don't these answers contradict each other? You have these types of answers posted on a few of such questions.

Wikipedia too, in 'angular frequency ' states that it's just the same as angular velocity. How is that possible? Isn't it wrong? At least in oscillating systems which don't have complete circular motion with constant velocity. The angular velocity is just the derivative of theta while angular frequency is omega.

I am very confused. When I am asked about an angular velocity of an oscillator, does that mean omega? Or does that mean derivative of theta?
The two don't seem to be identical at all. For omega you get a constant which is dependent on some given quantities. While the derivative of theta turns out to be completely different and is dependent on t=time and has a cos or sin functions for it's own function.

To sum it up. There is a lot of confusion here , and seemingly contradictionary material as well. Can someone please make it clear?-

  • $\begingroup$ Duplicate and not contradictory? physics.stackexchange.com/q/281408/104696 $\endgroup$
    – Farcher
    Jun 1, 2018 at 5:00
  • $\begingroup$ This is a completely different question. I never asked about the meaning of angular frequency or velocity in linear oscillators. I know the meaning of it. I am asking what is the angular velocity as compared with angular frequency. Wikipedia states that they are the same thing. In reality, if you take the derivative of theta it doesn't get anywhere near the constant omega in many problems. $\endgroup$
    – bilanush
    Jun 1, 2018 at 8:16

2 Answers 2


In a simple harmonic oscillator, or any vibrating system, the angular frequency is a scaling of the frequency by $2\pi$, e.g. $\sin(2\pi ft)$, $\omega = 2\pi f$.

This ensures that the sine function will complete one cycle in one period $T = 1/f$.

Angular velocity is a definition related to angular motion. For an object rotation about an axis $\omega = (\theta_f - \theta_i)/(t_f - t_i)$.

An object rotating with a constant angular speed (or velocity) we have $\theta(t) = \omega t + \omega_0$. The Cartesian coordinates of the point are related to the angular variable by $x = r\cos(\omega t + \omega_0)$ etc. So you see that the individual coordinates of the "rotating" object appear to move as simple harmonic oscillators.

From a purely physical point of view I would say these two quantities are manifestly different because the basic definitions are different. Once does not need "rotation" to define a periodic motion. However they can be related. Keep in mind that the function $\sin(2\pi ft)$ can also describe an oscillating voltage or other quantity that has not physical relation to rotation or vibration.

From a mathematical point of view one may see them as the same. This equivalence can be seen from the complex representation of $\sin(x)$ as $\operatorname{Im}(e^{ix})$. In the complex plane $e^{i2\pi ft}$ can be seen as "rotating" about the origin of the complex plane. Sometimes when two quantities have the same units it hints at a deep connection, sometimes not.

My personal perspective is that they are different quantities for the reasons described above, but there are interesting and useful connections between the two depending on use and mathematical representation.

I hope that lightens your confusion.

  • $\begingroup$ Thank! Still, this isn't completely clear to me . In rotating objects with constant velocity aren't the two definitions turn out to be identical? Since the angular velocity is just the same as rate of radian change. So that angular frequency and angular velocity overlap. In oscillating systems when I am asked about the angular velocity of something should I answer with omega or with the derivative of theta? The derivative is certainly a correct answer. But it's not clear to me what is the exact physical meaning of omega in this case. $\endgroup$
    – bilanush
    Jun 1, 2018 at 0:01
  • $\begingroup$ In oscillators. If I were to take the average of the angular velocity (or something like that) can I possibly get the value of the angular frequency? Can omega be thought of as a some general averaged velocity of the system and hance the overlap in circular constant velocity? $\endgroup$
    – bilanush
    Jun 1, 2018 at 0:06
  • $\begingroup$ Only in the case of rotation do they turn out to be related. Forget about motion. The 2*pi*f could be frequency of A.C. current in a wire, or some other quantity. There is no rotation in this case. $\endgroup$
    – user196418
    Jun 1, 2018 at 1:24
  • $\begingroup$ They are always related. In rotation with constant velocity they are identical. What I am asking is whether or not omega can be thought of as a some sort of average of angular velocity. $\endgroup$
    – bilanush
    Jun 1, 2018 at 1:35
  • $\begingroup$ @bilanush, angular velocity is simply the time rate of change of the angular displacement of something for which that is defined. A mass-spring system doesn't have an angular displacement nor an angular velocity but, since the motion is sinusoidal, there is a period $T$ and an associated angular frequency $\omega = \frac{2\pi}{T}$ which is simply defined as the time rate of change of the angular argument of the sin/cos function $\endgroup$ Jun 1, 2018 at 1:35

Circular Motion

In the case of circular motion, the angular velocity $\vec \omega$ is the velocity in terms of angular displacement, i.e.

$$ \vec \omega = \frac{d \vec \theta}{dt}. $$

Here the vector indicates the direction of rotation -- whether it is clockwise or counter-clockwise with respect to some axis -- as given by the right-hand rule.

Note that the magnitude of this quantity, the angular speed

$$ \omega = \frac{d\theta}{dt} $$

is synonymous with the term "angular frequency" because it quantifies how many radians the motion completes in a given cycle time. In other words, it can be converted into an absolute frequency via

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

where $f$ indicates how many cycles/oscillations/circular-trips occur in a given time period.


Take a spring-mass system. It moves up and down in a periodic manner. The time to complete one oscillation is the period $T$, and thus the frequency is $f = \frac{1}{T}$.

However, its up-and-down oscillatory behavior is sinusoidal, and thus it is related to the unit circle.

enter image description here

The rotating arm in the circle of the above GIF, you would agree, has an angular "speed" because it moves in a circle. However, note the blue dot on the y-axis -- it moves like a spring, so it has both an angular speed and angular frequency. This angular speed/frequency $\omega$ corresponds to the absolute frequency $f$ by

$$ \omega = 2\pi f, $$

as also given above. Here, $\omega$ does not mean the $\frac{d\theta}{dt}$ of the mass itself, but of the unit$^1$ circle corresponding to its motion. This is highly related to phasor diagrams, which may be of some interest to you. They are applied extensively in the analysis of AC electricity.

Bottom line

Angular speed and angular frequency are equivalent. However, the angular velocity is a vector quantity.

In different physical situations, it might be more evocative to have a preference for either term (e.g., for angular speed in rotating systems). However, they are equivalent due to their intimate relationship through trigonometry: the oscillations that are experienced in spring-mass-type oscillators are described by sinusoidal functions, and these functions are defined via unit circles.


  1. It's not actually a unit circle because the sine wave will generally have an amplitude that is not equal to 1. But the concept still applies.
  • $\begingroup$ Thank. A very beautiful answer. Still not perfectly clear. Two main things which I didn't understand in your answer . 1) you say. "This angular speed corresponds to the frequency by ω=2πf" I don't understand. Isn't it a relation between the angular frequency to the normal frequency f? Or please, can you can elaborate more explicitly what would be the angular velocity as compared with frequency. 2) You say in the end 'Angular speed and angular frequency are equivalent' how are they equivalent? You have shown their equivalency in circle only. $\endgroup$
    – bilanush
    Jun 1, 2018 at 8:01
  • $\begingroup$ Also, you have described the difference between circular motion to oscillators only with the geometrical difference of their motion. Isn't the cause of the difference the constant velocity in circular motion as opposed to oscillator? For example, if in circular motion there wouldn't be constant velocity the terms angular velocity and angular frequency would certainly not correspond. $\endgroup$
    – bilanush
    Jun 1, 2018 at 8:11
  • $\begingroup$ @bilanush Thanks for your comments. I have edited it to make my language more explicit regarding $\omega = 2\pi f$. For (2), they are equivalent because a circular motion is an oscillation. Look at the GIF above. Both the x- and y-components, taken independently, act merely like spring-mass oscillators. Neither component has a constant speed, despite the fact that $v_x^2 + v_y^2 = v^2$ is constant. $\endgroup$
    – zh1
    Jun 1, 2018 at 11:42
  • $\begingroup$ OK. Again very nice. If I understand you correctly (correct me if I am wrong ) we can discuss angular velocity and frequency in whatever system by looking at the corresponding circle. Then you say, that once it's defined on the circle. .. we will find out that omega turns out to be just equal to velocity d0/dt. Is it correct? If it's correct. If so, is it true even if the speed isn't constant? Or can you please explain how is it possible that omega which in many problems is just a constant would be equal to the rate of angle change which is dependent of time? $\endgroup$
    – bilanush
    Jun 1, 2018 at 12:25
  • $\begingroup$ @bilanush Yes, that is correct. For objects moving in a circle, $v = \omega R$. Thus if $v$ is time-dependent, $\omega$ will be as well (for constant $R$). After all, if the speed is changing, so is your rotational period, and thus the absolute and angular frequencies must change too. $\endgroup$
    – zh1
    Jun 1, 2018 at 23:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.