Question

Use Problem 7.16 to find the characteristic vibration frequencies of sound in a spherical cavity.

Short answer

The characteristic vibration frequencies of sound in a spherical cavity is$\nu =\frac{{\lambda }_{1}v}{2\pi a}$ .

Step-by-step solution

Given information:

Spherical harmonics $Y_l^m\left(\theta ,\varphi \right)=P_l^m\left(\cos \theta \right)e^{\pm im\varphi }$.

Definition of Wave

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.

Use the Wave equation:

The wave equation is as given below.

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

Write the wave equation in spherical coordinates in the form of spherical harmonics.

$u\left(r,\theta ,\Phi ,t\right)=\left(\begin{array}{c}{j}_1\left(kr\right)\\ y_1\left(kr\right)\end{array}\right)P_l^m\cos \theta e^{\pm im\Phi }\left(\begin{array}{c}\cos \left(kvt\right)\\ sin\left(kvt\right)\end{array}\right)$

Neglect $y_1\left(kr\right)$as inside the spherical cavity, it has no effect.

$u\left(r,\theta ,\Phi ,t\right)=\left\{j_1\left(kr\right)\right\}{P}_l^m\cos \theta e^{\pm im\Phi }\left\{\cos \left(kvt\right)\right\}$

Apply the boundary condition.

If $ka={\lambda }_1$then the Bessel function equal to zero.

$\Rightarrow j_1\left(ka\right)=0$

Step 4:Find the vibration frequency

The frequency for normal mode is$\omega =2\pi \nu$ .

Write the equation in terms of v.

$\nu =\frac{\omega }{2\pi }$

Replace w by kv in the above equation.

$\nu =\frac{kv}{2\pi }$ ….. (1)

Use the relation written below.

${\lambda }_1=ka$

$k=\frac{{\lambda }_1}{a}$ ….. (2)

Put equation (2) in (1).

$$\begin{gathered} \nu =\frac{\frac{{\lambda }_1}{a}v}{2\pi } \\ =\frac{{\lambda }_{1}v}{2\pi a} \end{gathered}$$

Hence the characteristic vibration frequencies of sound in a spherical cavity is $\nu =\frac{{\lambda }_{1}v}{2\pi a}$.