Prismatic horn

A prismatic horn is one whose cross-sectional dimensions are constant in the principal direction of vibration.

The following discussion is for a longitudinally resonant thin (slender) prismatic horn. (A thin horn is one whose transverse dimensions (e.g., diameter) are sufficiently small at a specified frequency that any energy due to transverse motion due to Poisson coupling can be neglected. zzz Poisson = 0? no xverse motion. One-dimensional wave eqn.)

In such a horn the amplitude is cosinually distributed along the horn's length —

\begin{align} \label{eq:10701a} u(x) &= U_p \cos \left( {\frac{2\pi}{\lambda_{tw}}} \, x \right) \end{align}

where —

\( x \)	= Distance, starting from a free end (antinode)
\( u \)	= Amplitude distribution
\( U_p \)	= Maximum (peak) amplitude
\( \lambda_{tw} \)	= Thin-wire wavelength at the resonant frequency

The quantity in ( ) is in radians. The amplitude \( u(x) \) is maximum at the antinodes (the free end and every half wavelength (\( \lambda_{tw}/2 \)) thereafter). \( u(x) \) is zero at the nodes (\( \lambda_{tw}/4 \) and every \( \lambda_{tw}/2 \) thereafter).

The stress is the product of Young's modulus \( E \) and the strain \( \epsilon \), where the strain is the derivative (slope) of the amplitude curve —

\begin{align} \label{eq:10702a} {\sigma}(x) &= E \, \left[ U_p \left( {\frac{2\pi}{\lambda_{tw}}} \right) \sin \left( {\frac{2\pi}{\lambda_{tw}}} \, x \right) \right] \end{align}

The stress \( {\sigma}(x) \) is zero at the antinodes (the free end and every half wavelength (\( \lambda_{tw}/2 \)) thereafter). \( {\sigma}(x) \) is maximum at the nodes (\( \lambda_{tw}/4 \) and every \( \lambda_{tw}/2 \) thereafter).

The thin-wire wavelength can be expressed as —

\begin{align} \label{eq:10703a} \lambda_{tw} = \frac{c_{tw}}{f} \end{align}

where —

\( c_{tw} \)	= Thin-wire wave speed = \( \sqrt{\frac{E}{\rho}} \)
\( f \)	= frequency

Then, rather than expressing equations \eqref{eq:10701a} and \eqref{eq:10702a} in terms of the wavelength, they may be more conveniently expressed in terms of the frequency and wave speed —

\begin{align} \label{eq:10704a} u(x) &= U_p \cos \left( {\frac{2\pi \, f}{c_{tw}}} \, x \right) \end{align}

\begin{align} \label{eq:10705a} {\sigma}(x) &= E \, \left[ U_p \left( {\frac{2\pi \, f}{c_{tw}}} \right) \sin \left( {\frac{2\pi \, f}{c_{tw}}} \, x \right) \right] \end{align}

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Appendix A. Energy stored in a thin longitudinally resonant member

\( \dot{u} \) For a thin longitudinally resonant member the kinetic energy stored in any slice \( dx \) is —

\begin{align} \label{eq:10721a} dW &= \zsfrac{1}{2} \, v^2 \, dm \\[0.7em]%eqn_interline_spacing &= \zsfrac{1}{2} \, v^2 \, \left( \rho \, A \, dx \right) \nonumber \end{align}

The energy of a quarter wave is then —

\begin{align} \label{eq:10722a} W &= \zsfrac{1}{2} \, \int_{0}^{\lambda/4}{v^2 \, \left( \rho \, A \, dx \right) } \end{align}

For a thin prismatic resonator the amplitude distribution (from the free end) is —

\begin{align} \label{eq:10723a} u &= U_p \cos \left( {\frac{2\pi}{\lambda}} \, x \right) \, \sin \left( 2\pi f \, t \right) \end{align}

The velocity is the rate of change of the amplitude. Thus, differentiating equation \eqref{eq:10723a} with respect to time gives the velocity distribution —

\begin{align} \label{eq:10724a} v &= \frac{du}{dt} \\ &= \left( 2\pi f \right) U_p \, \cos \left( {\frac{2\pi}{\lambda}} \, x \right) \, \cos \left( 2\pi f \, t \right) \nonumber \\ &= V_p \, \cos \left( {\frac{2\pi}{\lambda}} \, x \right) \, \cos \left( 2\pi f \, t \right) \nonumber \end{align}

Substituting equation \eqref{eq:10724a} into equation \eqref{eq:10722a} and ignoring the time varying component (to get the maximum energy during the cycle)  —

\begin{align} \label{eq:10725a} W &= \zsfrac{1}{2} \, \int_{0}^{\lambda/4}{\left[ U_p \left( 2\pi f \right) \cos \left( {\frac{2\pi}{\lambda}} \, x \right) \right]^2 \, \left( \rho \, A \, dx \right) } \end{align}

For a prismatic resonator the area \( A \) is constant. Also, neither the peak amplitude \( U_p \) nor the frequency \( f \) depend on \( x \). Assume that the density \( \rho \) does not vary with \( x \). Then all of these can be taken outside the integral —

\begin{align} \label{eq:10726a} W &= \zsfrac{1}{2} \left( \rho \, A\right)\, U_p \left( 2\pi f \right) \int_{0}^{\lambda/4}{\left[ \cos \left( {\frac{2\pi}{\lambda}} \, x \right) \right]^2 \, dx } \end{align}

Integrating yields —

\begin{align} \label{eq:10727a} W &= \zsfrac{1}{2} \left( \rho \, A\right)\, U_p \left( 2\pi f \right) {\left[ \frac{x}{2} + \frac{1}{4} \, \sin \left( {\frac{4\pi }{\lambda}} \, x \right) \right]}_0 ^{\lambda/4} \end{align}

When equation \eqref{eq:10727a} is evaluated for \( x \) between the limits of 0 and \( {\lambda}/4 \), the sin term is 0 at both limits and so drops out. Then equation \eqref{eq:10727a} is —

\begin{align} \label{eq:10728a} W &= \zsfrac{1}{2} \left( \rho \, A\right)\, U_p \left( 2\pi f \right) {\left[ \frac{x}{2} \right]}_0 ^{\lambda/4} \\[0.7em]%eqn_interline_spacing &= \zsfrac{1}{2} \left( \rho \, A \right)\, U_p \left( 2\pi f \right) {\left[ \frac{\lambda}{8} \right]} \nonumber \\[0.7em]%eqn_interline_spacing &= \zsfrac{1}{4} \left( \rho \, A \, \frac{\lambda}{4} \right)\, U_p \left( 2\pi f \right) \nonumber \end{align}

The first factor in ( ) is just the mass of the quarter wave section (\( m_{\lambda/4} \)) so equation \eqref{eq:10728a} can be written as —

\begin{align} \label{eq:10729a} W &= \zsfrac{1}{2} \left( \zsfrac{m_{\lambda/4}}{2} \right) \, U_p \left( 2\pi f \right) \end{align}

Thus, the energy that is cosinually distributed in a prismatic member is the same as if zzz

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