Do Problem 26 for a rectangular membrane.

Klein-Gordon equation is given as below.

${\nabla }^{2}u=\left(\frac{1}{v^2}\right)\frac{{\partial }^{2}u}{\partial t^2}+{\lambda }^{2}u$

Any disturbance or energy transfer from one location to another is referred to as a wave. Each wave is guided by a mathematical formula. A wave can be either standing or stationary.

Consider the Klein Gordon equation.

${\nabla }^{2}u=\left(\frac{1}{c^2}\right)\frac{{\partial }^{2}u}{\partial t^2}+\left(\frac{m^2{c}^2}{{\hslash }^2}\right)u$

The above equation of the form $w(r,t)=e^{i(kr-\omega t)}$ where $w\in R$and $k\in R^3$.

Use the above given terms to write the dispersion relation.

$-{\left|k\right|}^2+\frac{{\omega }^2}{c^2}=\frac{m^2{c}^2}{{\hslash }^2}$ ….. (1)

Write the givenKlein-Gordon equation is as given below.

${\nabla }^{2}u=\left(\frac{1}{v^2}\right)\frac{{\partial }^{2}u}{\partial t^2}+{\lambda }^{2}u$

Compare the above equation with equation (1).

${\lambda }^2=\frac{m^2{c}^2}{{\hslash }^2}$

Here, the velocity c=v,

Put the value of${\lambda }^2$and c in equation (1).

$-{\left|k\right|}^2+\frac{{\omega }^2}{v^2}={\lambda }^2$

$v^2\left({\lambda }^2+k^2\right)={\omega }^2$

….. (2)

The lengths of the rectangular membrane is given by,

.$\left(k_1,k_2\right)=\left(\frac{n\pi }{a},\frac{m\pi }{b}\right)$

Use equation (2) to find the frequency.

$$\begin{gathered} {\omega }^2=v^2\left({\lambda }^2+k_1^2+k_2^2\right) \\ (2\nu \pi {)}^2=v^2\left({\lambda }^2+k_1^2+k_2^2\right) \\ {\nu }^2=\frac{v^2}{4\pi }\left({\lambda }^2+{\left(\frac{n\pi }{a}\right)}^2+{\left(\frac{m\pi }{b}\right)}^2\right) \\ \nu =\frac{v}{2}\sqrt{{\left(\frac{n}{a}\right)}^2+{\left(\frac{m}{b}\right)}^2+{\left(\frac{\lambda }{\pi }\right)}^2} \end{gathered}$$

Hence the frequency for a rectangular membrane is $\nu =\frac{v}{2}\sqrt{{\left(\frac{n}{a}\right)}^2+{\left(\frac{m}{b}\right)}^2+{\left(\frac{\lambda }{\pi }\right)}^2}$.