Q:

What does it mean when frequency has dimension of (1/cm)?

I have to solve this problem.

Let WDOS (weighted DOS) be
$$ \rho (\omega) = A \omega e^{- \omega / \omega _0}\ \ \ \ \ \ (1) $$
where $\omega _0 = 30\ \color{red}{\text{cm}^{-1}}$ and $A$ is a constant.
The amplitude of fluctuation $D$ is given as
$$ D^2 (T) = \int _0 ^\infty \rho (\omega) \omega ^2 \text{coth} \left( \frac{\hbar \omega}{2 k_B T} \right) d\omega .\ \ \ \ \ \ (2)$$

Numerically determine the value of $A$, assuming $D = 100\ \text{cm}^{-1}$ when $T = 300\ \text{K}$.

... (A middle level problem.)

... (A higher level problem.)

I don't even understand the problem. If $\omega _0$ has dimension of cm${}^{-1}$, the frequency $\omega$ should have the same dimension to make $\omega / \omega _0$ dimensionless in eq.(1). If so, however, $\hbar \omega / 2 k_B T$ in eq.(2) does have dimension, which is so strange that we cannot determine dimension of $\text{coth} \left( \frac{\hbar \omega}{2 k_B T} \right)$.
Is the problem incorrect? But please note that, in the field of spectroscopy, sometimes it seems frequency has dimension of length. For example, see this plot from this paper.
How can we handle frequency whose dimension is cm${}^{-1}$?

For the future reader:
I found a related post Why does this paper use 1/cm for units of frequency?

A:

It means that the units of $\hbar$ have also been adjusted to match.
Multiply your frequency by $c$ to get a conventional angular frequency in rad/s, or multiply $\hbar \mapsto\hbar c$ to get the appropriate units for $\hbar$.

