91Ó°ÊÓ

It has been asked to see problem 6.3 for reference.

Laplace’s equation in cylindrical coordinates is,

$$\begin{gathered} {\mathbf{∇}}^{\mathbf{2}}\mathit{u}\mathbf{=}\frac{\mathbf{1}}{\mathbf{r}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{r}}\left(r\frac{∂u}{∂r}\right)\mathbf{+}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{{\mathbf{∂}}^{\mathbf{2}}\mathbf{u}}{\mathbf{∂}{\mathbf{θ}}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{∂}}^{\mathbf{2}}\mathbf{u}}{\mathbf{∂}{\mathbf{z}}^{\mathbf{2}}} \\ \mathbf{=}\mathbf{0} \end{gathered}$$

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

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

$z\left(x,y,t\right)=\underset{m=1}{\overset{∞}{∑}}\sin \left(\frac{m\mathrm{π}}{a}x\right)\sin \left(\frac{n\mathrm{π}}{b}x\right)\left[C_1\cos \left(k_{mn}vt\right)+C_2\sin \left(k_{mn}vt\right)\right]$

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

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

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

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

Observe from the above expression that the displacement is 0 for $x=\frac{\mathrm{Ï€}}{2}$thus for the mode $m=2$and $n=1$the nodal line is placed at $x=\frac{\mathrm{Ï€}}{2}$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.

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

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

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.

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

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.

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

Hence, the figure has been drawn