Behavior as $\alpha$ varies

For fixed wavenumbers $(k,m)$ and fixed $N$, increasing $\alpha$ increases the denominator $\alpha^2 k^2 + m^2$, which decreases $\omega^2$ and hence $\vert\omega\vert$. In particular, consider the limit of very short horizontal wavelength $k\to\infty$ (with $m$ fixed). Then

$$\omega^2 \to \frac{N^2 k^2}{\alpha^2 k^2} = \frac{N^2}{\alpha^2},\qquad \omega\to \pm \frac{N}{\alpha}.$$

Thus, as horizontal wavenumber increases the limiting frequency tends to $N/\alpha$. For $\alpha=1$ the limit is $N$, while for $\alpha> 1$ the limiting frequency is $N/\alpha< N$. That is, increasing $\alpha$ forces the high-$k$ limit to be a smaller value; the statement in the problem that “for $\alpha> 1$ the system approaches the limiting frequency of $N$ more rapidly than with $\alpha=1$” should be interpreted carefully: the limiting frequency (as $k\to\infty$) is $N/\alpha$ (not $N$) when $\alpha\neq 1$. What is likely intended in the textbook exercise is a comparison of how the dispersion curve $\omega(k)$ approaches the asymptotic bound ($N$ for conventional Boussinesq with $\alpha=1$) as $k$ increases: qualitatively, the presence of larger $\alpha$ suppresses vertical acceleration less or more depending on the precise nondimensionalization used in Vallis.

To visualize the dispersion relation, one can plot $\omega/N$ vs $k/m$ for several values of $\alpha$ (including 0 and 1). The algebraic form in nondimensional variables $\tilde{k}=k/m$ is

$$\left(\frac{\omega}{N}\right)^2 = \frac{\tilde{k}^2}{\alpha^2\tilde{k}^2 + 1}.$$

As $\tilde{k}\to 0$, $\omega\to 0$ for all $\alpha$. As $\tilde{k}\to \infty$, $(\omega/N)^2 \to 1/\alpha^2$.