# Study on Band Gap and Sound Insulation Characteristics of an Adjustable Helmholtz Resonator
## 3 Modeling and Analysis of Low-Frequency Band Gap Formation Mechanism
### Establishment of Equivalent Model  
In order to explore the mechanism of band gap generation, the equivalent method of \"electro-mechanical-acoustic analogy\" was used to establish the \"spring-oscillator\" model of the structure. By establishing the equivalent model, the upper and lower limits of band gaps of the structure were calculated, and the influencing factors of the band gap were quantitatively analyzed. Firstly, the structure of the cell is divided into five parts, namely outer cavity, cavity II, the tube of cavity II, cavity I and the tube of cavity I. As a Helmholtz resonator, the structure of the cell can be hypothesized as follows:  
1. The volume of tubes is much smaller than that of cavity I and cavity II;
2. The linearity of the cell is much smaller than the wavelength of low-frequency sound waves;
3. The wall of the cavity does not deform when the air in the inner cavity is compressed and expanded. For the air in the tubes of cavity I and cavity II, the wavelength of low-frequency sound waves is much longer than the length of the tube, so each part of the air in tube can be thought to vibrate the same. Therefore, the air in the tube can be regarded as a vibrating oscillator, the mass of the equivalent oscillator of air in the tube of cavity I and tube of cavity II is \(m_{1}\) and \(m_{2}\) respectively, and if the height of the cell is set to 1, the expression is:  
\[m_{1}=m_{2}=\rho_{air}l_{1}S \tag{7}\]  
where \(\rho_{air}\) is the air density, \(l_{1}\) is the length of the tube and \(S\) is the cross-sectional area of the tube. For the air in the cavity II, cavity I and outer cavity, when the air vibrates in the tube, the wall of the cavity does not deform, so the air in cavity II, cavity I and outer cavity will compress and expand with the vibration of air in the tube. These parts of air act like an air spring and apply an elastic force to the air in the tube, and the expression of the elastic force is as follows:  
\[F=-pS=-\frac{\rho_{air}c^{2}S^{2}}{V}\xi \tag{8}\]  
where \(\xi\) is the displacement of air in the tube. The air in cavity II, cavity I and outer cavity can be equivalent to three springs, whose equivalent stiffness is \(k_{1}\) (cavity I), \(k_{2}\) (cavity II) and \(k_{3}\) (outer cavity), respectively, and the expressions are as follows:  
\[k_{1}=\frac{\rho_{air}c^{2}S^{2}}{V_{1}} \tag{9}\]  
\[k_{2}=\frac{\rho_{air}c^{2}S^{2}}{V_{2}} \tag{10}\]  
\[k_{3}=\frac{\rho_{air}c^{2}S^{2}}{V_{3}} \tag{11}\]  
where \(V_{1}\), \(V_{2}\) and \(V_{3}\) are the volumes of cavity I, cavity II and outer cavity, respectively. According to the sound pressure fields at each point, the vibration modes were analyzed, and the corresponding equivalent model was established. For point A and point C,it can be seen from the above analysis that the vibration modes of point A and point C are similar. As shown in Figure 7a, the sound pressure field of the outer cavity, cavity I and the tube of cavity I at point A is almost 0, so the effect of air in the outer cavity, cavity I and the tube of cavity I can be ignored. The two regions of cavity II and the tube of cavity II are considered. According to the principle of mechanical-acoustic analogy, the long and narrow labyrinth tube can be equivalent to an oscillator, while the air in the cavity II can be equivalent to an air spring. The established \"spring-oscillator\" model of point A is shown in Figure 8a. Similarly, as shown in Figure 7c, the sound pressure field of the outer cavity, cavity II and the tube of cavity II at point C is 0, so these parts of air can be ignored. Therefore, the spring-oscillator model of point C is shown in Figure 8b.