Consider an elastic sheet with mass density $\sigma$[kg m$^{-2}$] we have
$$ \nabla^2 f = \frac{\sigma}{T} \frac{\partial^2 f}{\partial t^2} $$
where $f(x,y,t)$ is the displacement, $T$[N m$^{-1}$] plays the role of the n and $\nabla^2$ is the Laplacian operator
$$ \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \frac{\sigma}{T} \frac{\partial^2 f}{\partial t^2} $$
$$ \begin{aligned} \frac{\text d^2 f_x(x)}{\text d x^2} &= -k_x^2 f_x(x) \\ \frac{\text d^2 f_y(y)}{\text d y^2} &= -k_y^2 f_y(y) \\ \frac{1}{v^2}\frac{\text d^2 f_t(t)}{\text d t^2} &= -k^2 f_t(t) \end{aligned} $$
$$ \begin{aligned} f_x(x) &= A_x \cos k_x x + B_x \sin k_x x \\ f_y(y) &= A_y \cos k_y y + B_y \sin k_y y \\ f_t(t) &= A_t \cos \omega t + B_t \sin \omega t \end{aligned} $$
where $\omega^2 = k^2 v^2$
$$ f(x,y,t)=f_x(x) f_y(y) f_t (t) $$
$$ \begin{aligned} f(0,y,t) &= 0 \quad f(a,y,t)=0 \\ f(x,0,t) &= 0 \quad f(x,a,t)=0. \end{aligned} $$
Initial shape and velocity are:
$$ \begin{aligned} f(x,y,0) &\equiv f(x,y) \\ \frac{\partial f}{\partial t}(x,y,0) &\equiv g(x,y) \end{aligned} $$
$$ \frac{\text d^2 f_x}{\text dx^2} = - k_x^2 f_x\quad ; \quad f_x(0)=f_x(a)=0 $$
General solution: $f_x(x) = A \cos k_x x + B \sin k_x x.$ then using the boundary conditions we get
$$ f_x(x) = B \sin \frac{n \pi x}{a} \quad n=\Z $$
Same process for $z$ and $t$