Question

Repeat Problem 17 for a membrane in the shape of a circular sector of angle$\mathbf{60}\mathbf{°}$.

Short answer

The lowest frequencies are as given below.

$$\begin{gathered} {\nu }_{13}=2.65{\nu }_{10,} \\ {\nu }_{23}=4.06{\nu }_{10}, \\ {\nu }_{4.13}=5.4{\nu }_{10} \end{gathered}$$

Step-by-step solution

Given Information:

It has been asked to repeat problem 17.

Definition of Laplace’ equation:

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$\mathit{u}\mathbf{=}\mathit{R}\left(r\right)\mathit{\Theta }\left(\theta \right)\mathit{Z}\left(z\right)$.

General solution for the wave equation:

Take the$\left(x,y\right)$plane to be the plane of the circular support and take the origin at its center. Let$z\left(x,y,t\right)$be the displacement of the membrane from the$\left(x,y\right)$plane.

Then z satisfies the wave equation.

${\nabla }^{2}z=\frac{1}{v^2}\frac{{\partial }^{2}z}{\partial t^2}$

Put$z=F\left(x,y\right)T\left(t\right)$ into the previous differential equation a space differential equation is obtained and a time differential equation. Since the question is solving the problem for a circular membrane write${\nabla }^2$ in polar coordinates and then separate the propose a solution following the separation of variables for the space equation. So, the general solution is $z=\cos \left(n\theta \right)\cos \left(kvt/a\right)$the circular membrane the frequencies are obtained as mentioned below.