There is a simple way to see why the natural frequencies of a rectangular plate with simply-supported edges give rise to a modal density that is constant on average. Equation (6) of section 3.2.3 gave natural frequencies

$$\omega_{nm} = \sqrt{\frac{EK}{\rho h}} \left[\frac{n^2 \pi^2}{a^2}+\frac{m^2 \pi^2}{b^2} \right] \tag{1}$$

which we can write in the form

$$\omega_{nm} = \sqrt{\frac{EK}{\rho h}} \left[k_x^2+k_y^2 \right] \tag{2}$$

where $k_x$ and $k_y$ are the components of *wavenumber* in the $x$ and $y$ directions respectively. The solutions allow $n$ and $m$ to take any integer values 1,2,3…, so if we plot these in the wavenumber plane we obtain a regular grid, sketched in Fig. 1.

If we now ask how many modes have natural frequencies below some fixed value $\Omega$, we have to count the number of points on this wavenumber grid lying inside a quarter-circle with radius $r$ satisfying

$$\sqrt{\frac{EK}{\rho h}} r^2 = \Omega \tag{3}$$

such as the one plotted in blue in Fig. 1. The points are uniformly distributed over the plane, so this mode count is proportional to the area inside the blue curve. But the radius $r$ is proportional to $\sqrt{\Omega}$, and of course the area is proportional to $r^2$, so we deduce that the count of natural frequencies below $\Omega$ is proportional to $\Omega$. This is another way of saying that the average modal density is a constant, independent of $\Omega$.

To obtain an explicit formula, we first note that the area of one “box” of the wavenumber grid is $(\pi/a) \times (\pi/b)$. The area of the quarter-circle is

$$A_c = \dfrac{\pi r^2}{4} = \dfrac{\pi}{4} \Omega \sqrt{\dfrac{\rho h}{EK}} \tag{4}$$

so using the value of $K$ from eq. (4) of section 3.2.3 we can deduce that the number of “boxes” needed to fill this area, and hence the mode count $N(\Omega)$, is given by

$$N \approx \dfrac{ab}{\pi^2} \dfrac{\pi}{4} \Omega \sqrt{\dfrac{12 \rho (1-\nu^2)}{Eh^2}} = \Omega \dfrac{ab}{2 \pi h} \sqrt{\dfrac{3 \rho (1-\nu^2)}{E}} . \tag{5}$$

Thus the modal density $n$ is

$$n=\dfrac{dN}{d \Omega} = \dfrac{ab}{2 \pi h} \sqrt{\dfrac{3 \rho (1-\nu^2)}{E}} . \tag{6}$$

This argument has been based on the special case of the rectangular plate, with simply-supported edges. If the shape and/or the boundary conditions are changed in some continuous way, the points in the wavenumber plot will move around continuously, but they never appear or disappear. So the average density of points in the plane remains fairly uniform, and the argument given above still applies. Of course, the density of points and hence the numerical value of the modal density may change, for example if the plate is made smaller, larger or thicker. But it will remain the case that the average modal density will remain roughly constant.