# Why are angular frequencies $\omega=2\pi f$ used over normal frequencies $f$?

When we first studying vibrations in crystals we begin by studying the monoatomic chain, and then go onto the diatomic chain with a series of alternating masses. In studying these we look to calculate the dispersion relation, which is the angular frequency as a function of the wave vector.

For example, in the monoatomic chain we can derive the dispersion relation as $$\omega=\sqrt{\frac{4C}{M}}\sin^2\Big(\frac{ka}{2}\Big),$$ where $C$ is a 'spring' constant inherent in the crystal structure, $M$ is the mass of the atoms on the chain, $k$ is the wave vector and $a$ is the atomic spacing in the chain.

When studying the diatomic chain, we get two solutions corresponding to the optical (diatomic only) and acoustic (diatomic and monoatomic) waves.

What I don't understand is exactly why we are concerned with an angular frequency. What has the property of angular frequency? As far as I know there is no rotational motion, and the intrinsic frequency of a wave is surely more useful?

In addition to this question, how can we calculate the frequency, $f$ of, say, an optical wave of a diatomic chain given the angular frequency from the dispersion relation, $\omega$?

You are right in noting that $f$ is the more "physically intuitive" quantity, and at the end of the day measurements typically done in $f$, not $\omega$. However, the relationship between $f$ and $\omega$ is always $2\pi f = \omega$, so it is a very simple conversion to the point where people operationally don't really think of them as different. The reason $\omega$ is typically preferred over $f$ is because it is more convenient to write in equations: $\sin(2\pi f t)$ is much more cumbersome to write than $\sin (\omega t)$. This essentially has to do with the fact that sinusoids have a period of $2\pi$, not $1$. For similar reasons, people tend to use $\hbar$ and not $h$ in many equations.

Mainly because

$$\frac{\rm{d}}{\rm{d}t}\sin\left(\omega t\right) = \omega\cos\left(\omega t\right)$$

but

$$\frac{\rm{d}}{\rm{d}t}\sin\left(2 \pi f t\right) = 2 \pi f\cos\left(2 \pi f t\right)$$

All those pre-factors every time you take a derivative or an integral get to be a pain to keep track of.

As you get even more mathematical in your physics, you might prefer to work with complex exponentials, $e^{i\omega t}$ instead of sine and cosine, because that makes dealing with differential equations even easier (no flipping back and forth between sine and cosine every time you take a derivative).