= Solution
Let the layer occupy $0<z<h$, with a free surface at $z=0$ and rigid base at $z=h$. Write an <SH-wave> as $u_y=Y(z)e^{i(kx-\omega t)}$. The equation and boundary conditions give
$$
Y''+q^2Y=0,\qquad q^2=\frac{\omega^2}{\beta'^2}-k^2,
\qquad Y'(0)=0,\quad Y(h)=0.
$$
Thus $Y=A\cos(qz)$ and the <guided SH modes between rigid and free planes> have
$$
q_n=\frac{\chi_n}{h},\quad\chi_n=(n+\tfrac12)\pi,\quad n=0,1,2,\ldots,
\qquad\omega^2=\beta'^2k^2+\omega_n^2,\quad\omega_n=\frac{\beta'\chi_n}{h}.
$$
The propagating branch has $\omega>\omega_n$ and $k=\sqrt{\omega^2-\omega_n^2}/\beta'$. Hence the <rigid-base approximation to Love waves> predicts
$$
\boxed{c_n(\omega)=\frac{\beta'}{\sqrt{1-\omega_n^2/\omega^2}},\qquad
U_n(\omega)=\beta'\sqrt{1-\omega_n^2/\omega^2},\qquad c_nU_n=\beta'^2}.
$$
At each threshold $c_n\to\infty$ and $U_n\to0$. As <frequency> rises, the <phase velocity> decreases towards $\beta'$ and the <group velocity> increases towards $\beta'$. <Frequencies> below $\omega_n$ have imaginary horizontal <wavenumber> and do not give propagating guided modes. The infinite threshold phase speed does not violate causality: the limiting group speed is zero. These quarter-wave thresholds differ from the actual <Love-wave cutoff frequencies> of an elastic substrate.
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