 Solution

Twice differentiating the proposed normal mode givesIt therefore solves the two-dimensional wave equation precisely whenThe fixed-edge boundary conditions require $z(x,0,t)=z(x,b,t)=0$. The first is automatic, and the second requires $\sin (k_{y}b)=0$. The nonzero normal modes are thereforeThe 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 $\cos (\omega t-k_{x}x)$ convention describes propagation in $+x$. Each positive integer $n$ labels a transverse standing wave across the strip.