= Solution
Twice differentiating the proposed <normal mode> gives
$$
\frac{\partial^2z}{\partial x^2}=-k_x^2z,\qquad
\frac{\partial^2z}{\partial y^2}=-k_y^2z,\qquad
\frac{\partial^2z}{\partial t^2}=-\omega^2z.
$$
It therefore solves the two-dimensional <wave equation> precisely when
$$
\omega^2=v^2(k_x^2+k_y^2).
$$
The fixed-edge <boundary conditions> require $z(x,0,t)=z(x,b,t)=0$. The first is automatic, and the second requires $\sin(k_yb)=0$. The nonzero <normal modes> are therefore
$$
\boxed{k_y=\frac{n\pi}{b},\quad n=1,2,\ldots;\qquad
k_x=\pm\sqrt{\frac{\omega^2}{v^2}-\frac{n^2\pi^2}{b^2}}.}
$$
The apparent $n=0$ choice gives identically zero <displacement> for this sine solution. There are no imposed end conditions in $x$, so $k_x$ is continuous rather than quantized by a second length. For a propagating mode it is real; the plus sign in the stated $\cos(\omega t-k_xx)$ convention describes propagation in $+x$. Each positive integer $n$ labels a transverse <standing wave> across the strip.
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