 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 $T$, soThe ordinary frequency is $f=1/T=\omega /(2\pi )$; the paper's symbol $\omega$ denotes angular frequency.