= Solution
Take <shear-horizontal waves> with displacement $\boldsymbol u=(0,0,w(x,y,t))$. This is divergence-free, so the elastic equation reduces to $\rho w_{tt}=\mu(w_{xx}+w_{yy})$. Rigid boundaries require $w=0$ at $y=0,h$, giving modes $w=A\sin(n\pi y/h)\cos(kx-\omega t)$, $n\ge1$. With shear speed $c_s=\sqrt{\mu/\rho}$, the <dispersion relation> is
$$
\boxed{\omega^2=c_s^2\left(k^2+\frac{n^2\pi^2}{h^2}\right),\qquad \omega_{c,n}=\frac{n\pi c_s}{h}.}
$$
For $k>0$, the <phase velocity> is $v_p=\omega/k>c_s$ and the <group velocity> is $v_g=d\omega/dk=c_s^2k/\omega<c_s$, with $v_pv_g=c_s^2$.
The <kinetic energy density> is $T=\rho w_t^2/2$. The only strains are $e_{xz}=w_x/2$ and $e_{yz}=w_y/2$, together with their symmetric counterparts; thus the <isotropic linear-elastic energy density> is $W=\mu(w_x^2+w_y^2)/2$. Average over one temporal period and integrate across the layer. The sine- and cosine-squared transverse integrals are both $h/2$, giving
$$
\boxed{\left\langle\int_0^hT\,dy\right\rangle=\frac{\rho A^2\omega^2h}{8}
=\frac{\mu A^2h}{8}\left(k^2+\frac{n^2\pi^2}{h^2}\right)
=\left\langle\int_0^hW\,dy\right\rangle.}
$$
The <guided shear-horizontal mode> has average equipartition after both operations; a time average at a fixed transverse point alone need not give equality.
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