Consider the normal modes of vibration for a square membrane of side π (see Problem 6.3). Sketch the 2, 1 and 1, 2 modes. Show that the line $\mathit{y}\mathbf{=}\mathit{x}$ is a nodal line for the combination $\mathit{s}\mathit{i}\mathit{n}\left(x\right)\mathit{s}\mathit{i}\mathit{n}\left(2y\right)\mathbf{-}\mathit{s}\mathit{i}\mathit{n}\left(2x\right)\mathit{s}\mathit{i}\mathit{n}\left(y\right)$of these two modes. Thus find a vibration frequency of a membrane in the shape of a$\mathbf{45}\mathbf{°}$right triangle.

It has been asked to see problem 6.3 for reference.

Laplace’s equation in cylindrical coordinates is,

$$\begin{gathered} {\mathbf{\nabla }}^{\mathbf{2}}\mathit{u}\mathbf{=}\frac{\mathbf{1}}{\mathbf{r}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{r}}\left(r\frac{\partial u}{\partial r}\right)\mathbf{+}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{u}}{\mathbf{\partial }{\mathbf{\theta }}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{u}}{\mathbf{\partial }{\mathbf{z}}^{\mathbf{2}}} \\ \mathbf{=}\mathbf{0} \end{gathered}$$

And to separate the variable the solution assumed is of the form$\mathit{u}\mathbf{=}\mathit{R}\left(r\right)\mathit{\Theta }\left(\theta \right)\mathit{Z}\left(z\right)$.

For a rectangular membrane of sides a and b, the general solution for the displacement of vibration.

$z\left(x,y,t\right)=\sum _{m=1}^{\infty }\sin \left(\frac{m\pi }{a}x\right)\sin \left(\frac{n\pi }{b}x\right)\left[C_1\cos \left(k_{mn}vt\right)+C_2\sin \left(k_{mn}vt\right)\right]$

Here is a square of side$\mathrm{\pi }$, so$a=b=\pi .$

Nodal line for$m=2,n=1$

Now by setting$m=2,n=1$the displacement is

$z\left(x,y\right)=\sin \left(\frac{2\pi }{\pi }x\right)\sin \left(\frac{\pi }{\pi }y\right)$

Observe from the above expression that the displacement is 0 for $x=\frac{\pi }{2}$thus for the mode $m=2$and $n=1$the nodal line is placed at $x=\frac{\pi }{2}$as the following figure can be seen.

Follow the same process as before but for$m=1$and$n=2$the displacement for space variables is mentioned below.

$$\begin{gathered} z\left(x,y\right)=\sin \left(\frac{\pi }{\pi }x\right)\sin \left(\frac{2\pi }{\pi }y\right) \\ =\sin \left(x\right)\sin \left(2x\right) \end{gathered}$$

The displacement is 0 for $y=\frac{\pi }{2}$. Therefore for this model the nodal lines occur at $y=\frac{\pi }{2}$.

Consider the combination of$\sin x\sin 2y-\sin 2y\sin y,$write the displacement.

$z\left(x,y\right)=\sin x\sin 2y-\sin 2y\sin y,$

Observe that the only case where the displacement is 0 is if x=y, which sketch can be seen in the following figure.

The frequency of vibration of a membrane in the shape of a $45°$right triangle can be identified assuming a nodal line along x=y in a square membrane, which conveniently corresponds to the one found in the previous section.

The frequency for a square membrane.

${\nu }_{mn}=\frac{u}{2}\sqrt{{\left(\frac{n}{a}\right)}^2+{\left(\frac{m}{a}\right)}^2}$

From previous section that the displacement is a combination of modes.

$z\left(x,y\right)=\sin \left(x\right)\sin \left(2y\right)-\sin \left(2y\right)\sin \left(y\right)$

Which corresponds to m=2 and n=1 , or also m=1 and n=2. Thus, the frequency of vibration.

$$\begin{gathered} {\nu }_{mn}=\frac{u}{2}\sqrt{{\left(\frac{1}{a}\right)}^2+{\left(\frac{2}{a}\right)}^2} \\ =\frac{u}{2\pi }\sqrt{5} \end{gathered}$$

Hence, the figure has been drawn