axial mode width

Submitted by dougkeenan on Sun, 02/18/2018 - 00:23

\begin{equation}f_{resonance} = \frac{v}{2} \sqrt{ (\frac{w}{W})^2}\end{equation}

where W is the width (east-west) of the RESONANCE CHAMBER.

\begin{equation} f_{resonance} = \frac{v}{2} \sqrt{ (\frac{w}{10.472})^2 } \end{equation}

The lowest mode is for w = 1.

\begin{equation} f_{resonance} = \frac{v}{2} \sqrt{ (\frac{1}{10.472})^2  } \end{equation}

\begin{equation} f_{resonance} = 0.04775 v \end{equation}

\begin{equation} f = \frac{1}{\lambda} v \end{equation}

\begin{equation} \lambda = 20.94m \end{equation}

For air in the chamber $v = 330m/s$ therefore \begin{equation} f_{air} = 15.8 Hz \end{equation}

For hydrogen in the chamber $v = 1280m/s$ therefore \begin{equation} f_{H_2} = 61.1 Hz \end{equation}

For steam in the chamber $v = 500m/s$ therefore \begin{equation} f_{H_2} = 23.9 Hz \end{equation}

\begin{equation} f_{em} = 3\e{8}m/s / 7.31 m = 14.3 MHz \end{equation}