﻿ 纵向换能器的频率方程
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 哈尔滨工程大学学报  2019, Vol. 40 Issue (7): 1245-1250  DOI: 10.11990/jheu.201805106 0

### 引用本文

MO Xiping, LIU Yongping. Frequency equations for a longitudinal transducer[J]. Journal of Harbin Engineering University, 2019, 40(7), 1245-1250. DOI: 10.11990/jheu.201805106.

### 文章历史

1. 哈尔滨工程大学 水声工程学院, 黑龙江 哈尔滨 150001;
2. 中国科学院 声学研究所, 北京 100190;
3. 哈尔滨工程大学 水声技术重点实验室, 黑龙江 哈尔滨 150001

Frequency equations for a longitudinal transducer
MO Xiping 1,2, LIU Yongping 1,2,3
1. College of Underwater Acoustic Engineering, Harbin Engineering University, Harbin 150001, China;
2. Institute of Acoustics, Chinese Academy of Sciences, Beijing 100190, China;
3. Acoustic Science and Technology Laboratory, Harbin Engineering University, Harbin 150001, China
Abstract: Currently used frequency equations for a longitudinal transducer are described as two segments from an unknown node, which must employ the mathematical iterative method to calculate the resonant frequency of the entire transducer. Therefore, frequency equations are inconvenient and have adaptability problems on transducers with different cross-sectional shapes. In this study, we obtained the simplified electromechanical equivalent model on the basis of the one-dimensional vibration theory and energy method and deduced the parametric expression of equivalent elements through the equivalent relationship between kinetic and potential energies, considering the parameters of cross sections. Finally, on the basis of the mechanical resonance condition, we derived the parameterized mathematical expression of frequency equations for the entire transducer, which is precisely deduced through the mathematical method. This parameterized mathematical expression can be widely used to calculate the resonant frequency and vibrational node of longitudinal transducers with unequal cross-sectional parts or different cross-sectional shapes and sizes. The calculated results are consistent with those of the finite element method.
Keywords: longitudinal transducer    energy method    equivalent parameter    resonant frequency    vibrational node    finite element method

1 基于能量法的纵向换能器等效模型

1.1 纵向换能器等效模型

 ${\xi _{\rm{e}}} = {A_{\rm{e}}}\sin \left( {{k_{\rm{e}}}z} \right)$

 ${\xi _1} = {A_1}\sin \left( {{k_1}z + {\varphi _1}} \right)$ (1)

 $\xi_{2}=A_{2} \frac{1}{z+F l_{2}-l_{\mathrm{e}_{2}}} \sin \left(k_{2} z+\varphi_{2}\right)$ (2)

 $h=\frac{d_{T}}{d_{T}-d_{2}} l_{2}$

 $h=(F+1) l_{2}$

 $\left\{\begin{array}{l}{\left.\frac{\partial \xi_{1}}{\partial z}\right|_{z=-l_{1}-l_{e_{1}}}=0} \\ {\left.\frac{\partial \xi_{2}}{\partial z}\right|_{z=l_{e_{2}}+l_{2}}=0}\end{array}\right.$

 $\left\{ {\begin{array}{*{20}{l}} {\cos \left( {{k_1}\left( { - {l_1} - {l_{{{\rm{e}}_1}}}} \right) + {\varphi _1}} \right) = 0}\\ {\cos \left( {{k_2}\left( {{l_{{{\rm{e}}_2}}} + {l_2}} \right) + {\varphi _2} + {\varphi _0}} \right) = 0} \end{array}} \right.$

 $\varphi_{0}=\arcsin \frac{1}{\sqrt{k_{2}^{2} l_{2}^{2}(F+1)^{2}+1}}$ (3)

 $\left\{ {\begin{array}{*{20}{l}} {{\varphi _1} = - \frac{{\rm{ \mathsf{ π} }}}{2} + {k_1}\left( {{l_1} + {l_{{{\rm{e}}_1}}}} \right)}\\ {{\varphi _2} = \frac{{\rm{ \mathsf{ π} }}}{2} - {k_2}\left( {{l_2} + {l_{{{\rm{e}}_2}}}} \right) - {\varphi _0}} \end{array}} \right.$

φ1φ2代入式(1)、(2)，得到位移函数表达式：

 $\left\{ \begin{array}{l} {\xi _e} = {A_e}\sin \left( {{k_{\rm{e}}}z} \right)\\ {\xi _1} = - {A_1}\cos \left( {{k_1}\left( {z + {l_1} + {l_{{{\rm{e}}_1}}}} \right)} \right)\\ {\xi _2} = {A_2}\frac{1}{{z + F{l_2} + {l_{{{\rm{e}}_2}}}}}\cos \left( {{k_2}\left( {z - {l_2} - {l_{{{\rm{e}}_2}}}} \right) - {\varphi _0}} \right) \end{array} \right.$

 $\left\{ \begin{array}{l} {A_{\rm{e}}} = \frac{{\cos \left( {{k_1}{l_1}} \right)}}{{\sin \left( {{k_{\rm{e}}}{l_{{{\rm{e}}_1}}}} \right)}}{A_1}\\ {A_2} = \frac{{\cos \left( {{k_1}{l_1}} \right)}}{{\sin \left( {{k_{\rm{e}}}{l_{{{\rm{e}}_1}}}} \right)}} \cdot \frac{{\sin \left( {{k_{\rm{e}}}{l_{{{\rm{e}}_2}}}} \right)}}{{\cos \left( {{k_2}{l_2} + {\varphi _0}} \right)}}F{l_2}{A_1} \end{array} \right.$

 Download: 图 2 纵向换能器集总参数机电等效电路 Fig. 2 The lumped parameter equivalent diagram of longitudinal transducer

1.2 尾端支路等效参数

 $\frac{1}{2}{D_{{{\rm{e}}_1}}}A_1^2 = \frac{1}{2}\int\limits_{ - {l_{{{\rm{e}}_1}}}}^0 {{E_{\rm{e}}}{S_{\rm{e}}}{{\left( {\frac{{\partial {\xi _{\rm{e}}}}}{{\partial z}}} \right)}^2}{\rm{d}}z}$

 ${D_{{{\rm{e}}_1}}} = {\left( {\frac{{\cos \left( {{k_1}{l_1}} \right)}}{{\sin \left( {{k_{\rm{e}}}{l_{{{\rm{e}}_1}}}} \right)}}} \right)^2}{E_{\rm{e}}}{S_{\rm{e}}}{k_{\rm{e}}}\left[ {\frac{{{k_{\rm{e}}}{l_{{{\rm{e}}_1}}}}}{2} + \frac{1}{4}\sin \left( {2{k_{\rm{e}}}{l_{{{\rm{e}}_1}}}} \right)} \right]$

 $\frac{1}{2}{m_{{{\rm{e}}_1}}}{\omega ^2}A_1^2 = \frac{1}{2}\int\limits_{ - {l_{{{\rm{e}}_1}}}}^0 {{\rho _{\rm{e}}}S{\omega ^2}\xi _{\rm{e}}^2{\rm{d}}z}$

 ${m_{{{\rm{e}}_1}}} = {\left( {\frac{{\cos \left( {{k_1}{l_1}} \right)}}{{\sin \left( {{k_{\rm{e}}}{l_{{{\rm{e}}_1}}}} \right)}}} \right)^2}\frac{{{\rho _{\rm{e}}}{S_{\rm{e}}}}}{{{k_{\rm{e}}}}}\left[ {\frac{{{k_{\rm{e}}}{l_{{{\rm{e}}_1}}}}}{2} - \frac{1}{4}\sin \left( {2{k_{\rm{e}}}{l_{{{\rm{e}}_1}}}} \right)} \right]$

 $\frac{1}{2}{D_1}A_1^2 = \frac{1}{2}\int\limits_{ - \left( {{l_{{{\rm{e}}_1}}} + {l_1}} \right)}^{ - {l_{{{\rm{e}}_1}}}} {{E_1}{S_1}{{\left( {\frac{{\partial {\xi _1}}}{{\partial z}}} \right)}^2}{\rm{d}}z}$

 ${D_1} = {E_1}{S_1}{k_1}\left[ {\frac{{{k_1}{l_1}}}{2} - \frac{1}{4}\sin \left( {2{k_1}{l_1}} \right)} \right]$

 $\frac{1}{2}{m_1}{\omega ^2}A_1^2 = \frac{1}{2}\int\limits_{ - \left( {{l_{{{\rm{e}}_1}}} + {l_1}} \right)}^{ - {l_{{{\rm{e}}_1}}}} {{\rho _1}{S_1}{\omega ^2}\xi _1^2{\rm{d}}z}$

 $m_{1}=\frac{\rho_{1} S_{1}}{k_{1}}\left[\frac{k_{1} l_{1}}{2}+\frac{1}{4} \sin \left(2 k_{1} l_{1}\right)\right]$
1.3 前端支路等效参数

 $\frac{1}{2}{D_{{{\rm{e}}_2}}}A_1^2 = \frac{1}{2}\int\limits_0^{{l_{{{\rm{e}}_2}}}} {{E_{\rm{e}}}{S_{\rm{e}}}{{\left( {\frac{{\partial {\xi _{\rm{e}}}}}{{\partial z}}} \right)}^2}{\rm{d}}z}$

 ${D_{{{\rm{e}}_2}}} = {\left( {\frac{{\cos \left( {{k_1}{l_1}} \right)}}{{\sin \left( {{k_{\rm{e}}}{l_{{{\rm{e}}_1}}}} \right)}}} \right)^2}{E_{\rm{e}}}{S_{\rm{e}}}{k_{\rm{e}}}\left[ {\frac{{{k_{\rm{e}}}{l_{{{\rm{e}}_2}}}}}{2} + \frac{1}{4}\sin \left( {2{k_{\rm{e}}}{l_{{{\rm{e}}_2}}}} \right)} \right]$

 $\frac{1}{2}{m_{{{\rm{e}}_2}}}{\omega ^2}A_1^2 = \frac{1}{2}\int\limits_0^{{l_{{{\rm{e}}_2}}}} {{\rho _{\rm{e}}}{S_{\rm{e}}}{\omega ^2}\xi _{\rm{e}}^2{\rm{d}}z}$

 ${m_{{{\rm{e}}_2}}} = {\left( {\frac{{\cos \left( {{k_1}{l_1}} \right)}}{{\sin \left( {{k_{\rm{e}}}{l_{{{\rm{e}}_1}}}} \right)}}} \right)^2}\frac{{{\rho _{\rm{e}}}{S_{\rm{e}}}}}{{{k_{\rm{e}}}}}\left[ {\frac{{{k_{\rm{e}}}{l_{{{\rm{e}}_2}}}}}{2} - \frac{1}{4}\sin \left( {2{k_{\rm{e}}}{l_{{{\rm{e}}_2}}}} \right)} \right]$

 $\frac{1}{2}{D_2}A_1^2 = \frac{1}{2}\int\limits_{{l_{{{\rm{e}}_2}}}}^{{l_{{{\rm{e}}_2}}} + {l_2}} {{E_2}{{\left( {\frac{{z + F{l_2} - {l_{{{\rm{e}}_2}}}}}{{F{l_2}}}} \right)}^2}{S_2}{{\left( {\frac{{\partial {\xi _2}}}{{\partial z}}} \right)}^2}{\rm{d}}z}$

 $\begin{array}{l} \frac{1}{2}{D_2}A_1^2 = \frac{1}{2}\int\limits_{{l_{{{\rm{e}}_2}}}}^{{l_{{{\rm{e}}_2}}} + {l_2}} {{E_2}{{\left( {\frac{{z + F{l_2} - {l_{{{\rm{e}}_2}}}}}{{F{l_2}}}} \right)}^2}{S_2}A_2^2} \times \\ \;\;\;\;\;\;\left( {\frac{1}{{{{\left( {z + F{l_2} - {l_{{{\rm{e}}_2}}}} \right)}^2}}}\cos \left( {{k_2}\left( {z - {l_2} - {l_{{{\rm{e}}_2}}}} \right) - {\varphi _0}} \right.} \right) + \\ \;\;\;\;\;\;\frac{{{k_2}}}{{z + F{l_2} - {l_{{{\rm{e}}_2}}}}}\sin {\left( {{k_2}\left( {z - {l_2} - {l_{{{\rm{e}}_2}}}} \right) - {\varphi _0}} \right)^2}{\rm{d}}z \end{array}$

 $y = z + F{l_2} - {l_{{{\rm{e}}_2}}}$

 $z - {l_2} - {l_{{{\rm{e}}_2}}} = y - \left( {F + 1} \right){l_2}$

 $\begin{array}{*{20}{c}} {{D_2} = \frac{{A_2^2{E_2}{S_2}}}{{A_1^2{F^2}l_2^2}}\int\limits_{F{l_2}}^{\left( {F + 1} \right){l_2}} {\left( {\frac{1}{y}\cos \left( {{k_2}\left( {y - \left( {F + 1} \right){l_2}} \right) - {\varphi _0}} \right) + } \right.} }\\ {{{\left. {{k_2}\sin \left( {{k_2}\left( {y - \left( {F + 1} \right){l_2}} \right) - {\varphi _0}} \right)} \right)}^2}{\rm{d}}y} \end{array}$

 $\theta = 2\left( {F + 1} \right){k_2}{l_2} + 2{\varphi _0}$

 $\begin{array}{l} I = \int\limits_{F{l_2}}^{\left( {F + 1} \right){l_2}} {\left( {\frac{1}{{2{y^2}}}\cos \left( {2{k_2}y - \theta } \right) + \frac{{{k_2}}}{y}\sin \left( {2{k_2}y - \theta } \right)} \right){\rm{d}}y} - \\ \;\;\;\;\;\;\;\frac{{{k_2}}}{4}\left[ {\sin \left( {2\left( {{k_2}{l_2} + {\varphi _0}} \right)} \right) - \sin \left( {2{\varphi _0}} \right)} \right] + \frac{1}{2}k_2^2{l_2} + \\ \;\;\;\;\;\;\;\frac{1}{{2F\left( {F + 1} \right){l_2}}} \end{array}$

 $\begin{array}{l} \frac{1}{{2{y^2}}}\cos \left( {2{k_2}y - \theta } \right) + \frac{{{k_2}}}{y}\sin \left( {2{k_2}y - \theta } \right) = \\ \;\;\;\;\;\;\;\;\;\;\frac{{\cos \theta }}{2}\left[ {\frac{1}{{{y^2}}}\cos \left( {2{k_2}y} \right) + \frac{{2{k_2}}}{y}\sin \left( {2{k_2}y} \right)} \right] + \\ \;\;\;\;\;\;\;\;\;\;\frac{{\sin \theta }}{2}\left[ {\frac{1}{{{y^2}}}\sin \left( {2{k_2}y} \right) - \frac{{2{k_2}}}{y}\cos \left( {2{k_2}y} \right)} \right] \end{array}$

 $\int\left(\frac{\sin (\alpha x)}{x^{2}}-\alpha \frac{\cos (\alpha x)}{x}\right) \mathrm{d} x=-\frac{\sin (\alpha x)}{x}$
 $\int\left(\frac{\cos (\alpha x)}{x^{2}}+\alpha \frac{\sin (\alpha x)}{x}\right) \mathrm{d} x=-\frac{\cos (\alpha x)}{x}$

 $\begin{array}{l} I = \frac{1}{2}k_2^2{l_2} + \frac{1}{{2F{l_2}}}\cos \left( {2{k_2}{l_2} + 2{\varphi _0}} \right) - \\ \;\;\;\;\;\;\frac{{{k_2}}}{4}\sin \left( {2\left( {{k_2}{l_2} + {\varphi _0}} \right)} \right) + \frac{1}{{2F(F + 1){l_2}}} + \frac{{{k_2}}}{4}\sin {(2\varphi )_0} - \\ \;\;\;\;\;\;\frac{1}{{2(F + 1){l_2}}}\cos \left( {2{\varphi _0}} \right) = \frac{1}{2}k_2^2{l_2} + \\ \;\;\;\;\;\;\frac{1}{{2F{l_2}}}\cos \left( {2\left( {{k_2}{l_2} + {\varphi _0}} \right)} \right) - \frac{{{k_2}}}{4}\sin \left( {2\left( {{k_2}{l_2} + {\varphi _0}} \right)} \right) + \\ \;\;\;\;\;\;\frac{1}{{2F(F + 1){l_2}}} + \frac{1}{{2(F + 1){l_2}}} - \frac{{{k_2}}}{4}\sin \left( {2{\varphi _0}} \right) \end{array}$

 $\begin{array}{l} {D_2} = \frac{{A_2^2{E_2}{S_2}{k_2}}}{{A_1^2{F^2}l_2^2}}\left( {\frac{1}{2}{k_2}{l_2} + \frac{1}{{2F{k_2}{l_2}}}\cos \left( {2\left( {{k_2}{l_2} + {\varphi _0}} \right)} \right) - } \right.\\ \;\;\;\;\;\;\;\;\frac{1}{4}\sin \left( {2\left( {{k_2}{l_2} + {\varphi _0}} \right)} \right) + \frac{1}{{2F\left( {F + 1} \right){k_2}{l_2}}} + \\ \;\;\;\;\;\;\;\;\left. {\frac{1}{{2\left( {F + 1} \right){k_2}{l_2}}} - \frac{1}{4}\sin \left( {2{\varphi _0}} \right)} \right) \end{array}$

 $\frac{1}{2}{m_2}{\omega ^2}A_1^2 = \frac{1}{2}\int\limits_{{l_{{{\rm{e}}_2}}}}^{{l_{{{\rm{e}}_2}}} + {l_2}} {{\rho _2}{{\left( {\frac{{z + F{l_2} - {l_{{{\rm{e}}_2}}}}}{{F{l_2}}}} \right)}^2}{S_2}{\omega ^2}\xi _2^2{\rm{d}}z}$

 $\begin{array}{l} {m_2} = \frac{{A_2^2{\rho _2}{S_2}}}{{A_1^2{F^2}{k_2}l_2^2}}\left[ {\frac{{{k_2}{l_2}}}{2} + \frac{1}{4}\sin \left( {2\left( {{k_2}{l_2} + {\varphi _0}} \right)} \right) - } \right.\\ \;\;\;\;\;\;\;\;\frac{1}{4}\sin \left( {2{\varphi _0}} \right)] \end{array}$
2 纵向换能器的频率方程

 $\omega {m_{{{\rm{e}}_1}}} + \omega {m_1} - \frac{1}{\omega }{D_{{{\rm{e}}_1}}} - \frac{1}{\omega }{D_1} = 0$

ω=v1k1=vekeω2=E1/ρ1=Ee/ρe代入上式，得：

 $\begin{array}{l} {\rho _1}{v_1}{S_1}\left[ {\frac{{{k_1}{l_1}}}{2} + \frac{1}{4}\sin \left( {2{k_1}{l_1}} \right)} \right] + {\rho _{\rm{e}}}{v_{\rm{e}}}{S_{\rm{e}}}{\left( {\frac{{\cos \left( {{k_1}{l_1}} \right)}}{{\sin \left( {{k_{\rm{e}}}{l_{{{\rm{e}}_1}}}} \right)}}} \right)^2} \cdot \\ \;\;\;\;\;\;\;\;\;\left[ {\frac{{{k_{\rm{e}}}{l_{{{\rm{e}}_1}}}}}{2} - \frac{1}{4}\sin \left( {2{k_{\rm{e}}}{l_{{{\rm{e}}_1}}}} \right)} \right] - \\ \;\;\;\;\;\;\;\;\;{\rho _1}{v_1}{S_1}\left[ {\frac{{{k_1}{l_1}}}{2} - \frac{1}{4}\sin \left( {2{k_1}{l_1}} \right)} \right] - \\ \;\;\;\;\;\;\;\;\;{\rho _{\rm{e}}}{v_{\rm{e}}}{S_{\rm{e}}}{\left( {\frac{{\cos \left( {{k_1}{l_1}} \right)}}{{\sin \left( {{k_{\rm{e}}}{l_{{{\rm{e}}_1}}}} \right)}}} \right)^2}\left[ {\frac{{{k_{\rm{e}}}{l_{{{\rm{e}}_1}}}}}{2} + \frac{1}{4}\sin \left( {2{k_{\rm{e}}}{l_{{{\rm{e}}_1}}}} \right)} \right] = 0 \end{array}$

 $\tan \left( {{k_{\rm{e}}}{l_{{{\rm{e}}_1}}}} \right) = \frac{{{\rho _{\rm{e}}}{v_{\rm{e}}}{S_{\rm{e}}}}}{{{\rho _1}{v_1}{S_1}}}\cot \left( {{k_1}{l_1}} \right)$ (4)

 $\omega {m_{{{\rm{e}}_2}}} + \omega {m_2} - \frac{1}{\omega }{D_{{{\rm{e}}_2}}} - \frac{1}{\omega }{D_2} = 0$

 $\begin{array}{l} \omega {\left( {\frac{{\cos \left( {{k_1}{l_1}} \right)}}{{\sin \left( {{k_{\rm{e}}}{l_{{{\rm{e}}_1}}}} \right)}}} \right)^2}\frac{{{\rho _{\rm{e}}}{S_{\rm{e}}}}}{{{k_{\rm{e}}}}}\left[ {\frac{{{k_{\rm{e}}}{l_{{{\rm{e}}_2}}}}}{2} - \frac{1}{4}\sin \left( {2{k_{{{\rm{e}}_2}}}{l_{{{\rm{e}}_2}}}} \right)} \right] + \\ \;\;\;\omega \frac{{A_2^2{\rho _2}{S_2}}}{{A_1^2{F^2}{k_2}l_2^2}}\left[ {\frac{{{k_2}{l_2}}}{2} + \frac{1}{4}\sin \left( {2\left( {{k_2}{l_2} + {\varphi _0}} \right)} \right) - \frac{1}{4}\sin \left( {2{\varphi _0}} \right)} \right] - \\ \;\;\;\frac{1}{\omega }{\left( {\frac{{\cos \left( {{k_1}{l_1}} \right)}}{{\sin \left( {{k_{\rm{e}}}{l_{{{\rm{e}}_1}}}} \right)}}} \right)^2}{E_{\rm{e}}}{S_{\rm{e}}}{k_{\rm{e}}}\left[ {\frac{{{k_{\rm{e}}}{l_{{{\rm{e}}_2}}}}}{2} + \frac{1}{4}\sin \left( {2{k_{\rm{e}}}{l_{{{\rm{e}}_2}}}} \right)} \right] - \\ \;\;\;\frac{1}{\omega }\frac{{A_2^2{E_2}{S_2}{k_2}}}{{A_1^2{F^2}l_2^2}}\left( {\frac{1}{2}{k_2}{l_2} + \frac{1}{{2F{k_2}{l_2}}}\cos \left( {2\left( {{k_2}{l_2} + {\varphi _0}} \right)} \right) - } \right.\\ \;\;\;\frac{1}{4}\sin \left( {2\left( {{k_2}{l_2} + {\varphi _0}} \right)} \right) + \frac{1}{{2F(F + 1){k_2}{l_2}}} + \\ \;\;\;\frac{1}{{2(F + 1){k_2}{l_2}}} - \frac{1}{4}\sin \left( {2{\varphi _0}} \right)) = 0 \end{array}$

 $\tan \left( {{k_{\rm{e}}}{l_{{{\rm{e}}_{\rm{2}}}}}} \right) = \frac{{{\rho _{\rm{e}}}{v_{\rm{e}}}{S_{\rm{e}}}}}{{{\rho _2}{v_2}{S_2}}}\frac{1}{{\tan \left( {{k_2}{l_2} + {\varphi _0}} \right) - \frac{1}{{F{k_2}{l_2}}}}}$ (5)

 $\begin{array}{l} \cot \left( {{k_{\rm{e}}}\left( {{l_{{{\rm{e}}_1}}} + {l_{{{\rm{e}}_2}}}} \right)} \right) = \frac{1}{{\tan \left( {{k_{\rm{e}}}{l_{{{\rm{e}}_1}}}} \right) + \tan \left( {{k_{\rm{e}}}{l_{{{\rm{e}}_2}}}} \right)}} - \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{{\cot \left( {{k_{\rm{e}}}{l_{{{\rm{e}}_1}}}} \right) + \cot \left( {{k_{\rm{e}}}{l_{{{\rm{e}}_2}}}} \right)}} \end{array}$

 $\begin{array}{*{20}{c}} {\cot \left( {{k_{\rm{e}}}{l_{\rm{e}}}} \right) = \frac{{F{k_2}{l_2}\tan \left( {{k_2}{l_2} + {\varphi _0}} \right) - 1}}{{\frac{{{\rho _{\rm{e}}}{v_{\rm{e}}}{S_{\rm{e}}}}}{{{\rho _{\rm{1}}}{v_{\rm{1}}}{S_{\rm{1}}}}}F{k_2}{l_2}\cot \left( {{k_1}{l_1}} \right)\tan \left( {{k_2}{l_2} + {\varphi _0}} \right) - \frac{{{\rho _{\rm{e}}}{v_{\rm{e}}}{S_{\rm{e}}}}}{{{\rho _{\rm{1}}}{v_{\rm{1}}}{S_{\rm{1}}}}}\cot \left( {{k_1}{l_1}} \right) + \frac{{{\rho _{\rm{e}}}{v_{\rm{e}}}{S_{\rm{e}}}}}{{{\rho _{\rm{2}}}{v_{\rm{2}}}{S_{\rm{2}}}}}F{k_2}{l_2}}} - }\\ {\frac{{F{k_2}{l_2}}}{{\frac{{{\rho _{\rm{1}}}{v_{\rm{1}}}{S_{\rm{1}}}}}{{{\rho _{\rm{e}}}{v_{\rm{e}}}{S_{\rm{e}}}}}F{k_2}{l_2}\tan \left( {{k_1}{l_1}} \right) + \frac{{{\rho _{\rm{2}}}{v_{\rm{2}}}{S_{\rm{2}}}}}{{{\rho _{\rm{e}}}{v_{\rm{e}}}{S_{\rm{e}}}}}F{k_2}{l_2}\tan \left( {{k_2}{l_2} + {\varphi _0}} \right) - \frac{{{\rho _{\rm{2}}}{v_{\rm{2}}}{S_{\rm{2}}}}}{{{\rho _{\rm{e}}}{v_{\rm{e}}}{S_{\rm{e}}}}}}}} \end{array}$ (6)

3 纵向换能器谐振频率计算分析

 Download: 图 3 多边形前盖板示例 Fig. 3 The example of polygon front cover

4 结论

1) 从一维纵振动理论出发，建立纵向换能器的最简机电等效模型，利用能量法推导等效模型和元件参数表达式，进一步推导得到频率方程。

2) 通过与有限元方法相比，验证了频率方程对纵向换能器基频谐振特性描述的准确性，可用于部件不等截面情形和多种截面形状纵向换能器谐振频率的计算，具有更广的适用性。

3) 由于该方程直接给出了参数化表达式，方便直接解算频率参数，不需要预先假定节点位置的逼近处理。

4) 在前盖板截面尺寸与压电堆相比差异悬殊时，计算结果偏差增大，其原因主要是本文的推导仍然基于一维振动模型的假设，而随着前盖板截面尺寸与压电堆相比变大时，这一假设偏离实际情况就越远。

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