# Proof Of Theory Of Helmholtz Resonance

## Introduction

When you blow on the mouth of an empty bottle, you may hear a “boing” sound. This phenomenon is called Helmholtz resonance. In this article, we will explain the mechanism of Helmholtz resonance.

The frequency of the “boo-boo” sound (resonance frequency) produced by Helmholtz resonance can be expressed as follows.

Resonance frequency of Helmholtz resonance

If the speed of sound is $V$, the volume of the body of the bottle is $V_0$, and the cross-sectional area and length of the neck are $S, L$, respectively, the resonance frequency $f_{\rm H}$ of the Helmholtz resonance is as follows

\begin{eqnarray*}
f_{\rm H} = \frac{v}{2\mathrm{\pi}}\sqrt{\frac{S}{V_0 L}}\tag{☆}
\end{eqnarray*}

Let us derive this equation.

## How the sound works

The mechanism of Helmholtz resonance is relatively simple and is as follows.

(1) When you blow into the mouth of the bottle, air from the neck is pushed into the bottle.

(2) Pressure inside the bottle increases and pushes back air.

The repetition of (1) ⇄ (2) causes the air to vibrate and sound is produced.

To derive (☆), let us first consider the air in the neck as a piston of mass $m$. If the cross-sectional area and length of the neck section are $S, L$, respectively, and the density of air is $\rho$, then

\begin{eqnarray}
m \approx \rho S L
\end{eqnarray}

Now, with no force applied to the piston, the pressure of the gas inside is $P_0$, equal to the outside pressure, and its volume is $V_0$.

Suppose now that the piston is pushed in by $x$ and the pressure changes to $P$ and the volume to $V$. The piston is subjected to the following restoring forces

\begin{eqnarray}
F = -S(P – P_0)
\end{eqnarray}

Let us also assume that the expansion and compression of gas due to the pushing of the piston is an adiabatic operation. Then, from Poisson’s relation

\begin{eqnarray}
PV^{\gamma} = P_0V_0^{\gamma}
\end{eqnarray}

where $\gamma$ is the specific heat ratio. Also, the following equation holds

\begin{eqnarray}
V = V_0 – Sx
\end{eqnarray}

Substituting (4) into (3), we obtain

\begin{eqnarray*}
&P&(V_0 – Sx)^{\gamma} = P_0V_0^{\gamma} \\
\\
\therefore\ &P& = P_0V_0^{\gamma} \cdot (V_0 – Sx)^{-\gamma} \\
&&= P_0V_0^{\gamma} \cdot V_0^{-\gamma} \left(1 – \frac{Sx}{V_0}\right)^{-\gamma} \\
&&= P_0 \left(1 – \frac{Sx}{V_0}\right)^{-\gamma} \\
\end{eqnarray*}

Here, given that $\frac{Sx}{V_0}$ is a minute quantity, using the relation: $(1 + \varepsilon)^{\alpha} \approx 1 + \alpha\varepsilon$ which holds when $|\varepsilon| \ll 1$.

\begin{eqnarray}
\begin{split}
P &=& P_0 \left(1 – \frac{Sx}{V_0}\right)^{-\gamma} \\
&\approx& P_0 \left(1 + \gamma \frac{Sx}{V_0}\right)
\end{split}
\end{eqnarray}

Therefore, substituting (5) into (2) yields

\begin{eqnarray*}
F &=& -S\left\{ P_0 \left(1 + \gamma \frac{Sx}{V_0}\right) – P_0\right\} \\
&=& – \gamma \frac{P_0S^2}{V_0} \cdot x
\end{eqnarray*}

Thus, the equation of motion followed by the air in the neck section of mass $m$ is

\begin{eqnarray*}
m \frac{\mathrm{d^2}x}{\mathrm{d}t^2} =\, – \gamma \frac{P_0S^2}{V_0} \cdot x
\end{eqnarray*}

This can be viewed as a single oscillation with spring constant $k = \gamma\frac{P_0S^2}{V_0}$.

Therefore, the frequency of air vibration, or resonance frequency $f_{\rm H}$, is

\begin{eqnarray*}
f_{\rm H} &=& \frac{1}{2\pi} \sqrt{\frac{k}{m}} \\
&=& \frac{1}{2\pi} \sqrt{\frac{\gamma P_0S^2}{mV_0}}
\end{eqnarray*}

Substituting (1): $m \approx \rho S L$ and the relationship between the speed of sound $v$: $v = \sqrt{\frac{\gamma P_0}{\rho}}$, we obtain

\begin{eqnarray*}
f_{\rm H} = \frac{v}{2\mathrm{\pi}}\sqrt{\frac{S}{V_0 L}}
\end{eqnarray*} Yukkuri Machine Learning
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