= Solution
Represent a <time-harmonic wave> by $A\cos(kx-\omega t+\delta)$, with $k,\omega>0$. The <angular frequency> $\omega$ is the rate at which the <wave phase> changes at a fixed position, measured in radians per unit time. The <wavenumber> $k$ is its spatial rate of change, measured in radians per unit length. One full spatial cycle defines the <wavelength> $\lambda$, and one temporal cycle defines the <wave period> $T$, so
$$
\boxed{k=\frac{2\pi}\lambda,\qquad\omega=\frac{2\pi}T.}
$$
The ordinary <frequency> is $f=1/T=\omega/(2\pi)$; the paper's symbol $\omega$ denotes <angular frequency>.
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