# Theoretical Optimization of Trapped-Bubble-Based Acoustic Metamaterial Performance
## 3 Results and Discussion  
### Metascreen Modes  
To determine the resonant frequencies of the array, we should first know the resonant frequencies of the individual scatterer (trapped bubble). Instead of commonly used Minnaert frequency, we employ an approach proposed in [19] and developed in our previous work [18]. The modes of the interface vibrations read as [18]:  
\[\omega_{b}=\frac{\sqrt{\kappa_{mn}^{3}\sigma J_{-n}(\kappa_{mn}a)(\rho_{1}+ \rho_{2})-\kappa_{mn}^{4}\eta^{2}}}{\rho_{1}+\rho_{2}}, \tag{16}\]  
where \(J_{n}(\kappa_{mn}a)\) is the Bessel function of the first type of the order \(n\), with \(\kappa_{mn}\) being mode-dependent radial wave number. Here, we only account for the viscous damping. It is well known that for the clamped membrane multiple modes can be excited. However, we only pick the modes \((m,1)\) for our calculations. Once the resonant frequency of the individual scatterer was identified, we apply all of the analysis [13, 14, 20] to make the corrections to the resonant frequency of the system due to the strong coupling between the bubbles that allows shifting the resonant frequencies of the system. The general approach to find resonant frequencies of the metamaterial includes two steps: (1) find \(\Omega\) from extremal problem for energy transmission coefficient; and, (2) find \(\omega\) from Equation (7). Let us first consider an array of bubbles positioned in the liquid bulk. Solving Equation (9), we arrive at:  
\[\Omega_{01}=0, \tag{17}\]  
or in terms of resonant frequency:  
\[\omega_{01}=\frac{\omega_{b}}{\sqrt{1-A}}, \tag{18}\]  
where \(A=\frac{2\sqrt{\pi}a}{d}\), \(a\), and \(d\) stand for the bubble radius and distance between the centers of adjacent bubbles. Hereafter, we will use the notions of \"negative\" and \"positive\" branches of the solutions. These branches correspond to the values of \(\Omega_{i}\) that has \"\(-\)\" and \"\(+\)\" in front of the \"square root\" terms. For instance, in Equation (19) \(\Delta_{0}-\sqrt{\delta^{2}+\Delta_{0}^{2}}\) is a \"negative\" branch of the solution, whereas \(\Delta_{0}+\sqrt{\delta^{2}+\Delta_{0}^{2}}\) is a \"positive\" one. The transmission dip given by Equation (18) is always present in the system, regardless of its complexity. Next, let us assume the presence of the interface where the acoustic impedance changes abruptly from one side of the bubble array and bulk fluid from the other side. The interface can be formed by two media with different acoustic properties (e.g., water-air interface). Subsequently, solving Equation (18), we get:  
\[\Omega_{1,2}=\Delta_{0}\pm\sqrt{\delta^{2}+\Delta_{0}^{2}}, \tag{19}\]  
where \(\Delta_{0}=\left(B(1+h^{2}k^{2})+2 hk\delta\right)/2\). Here, \(B=\frac{4\pi a\,h}{d^{2}}\) and \(h\) is the distance between the bubble's center and adjacent interface. Subsequently, the resonant frequencies of the system read as:  
\[\omega_{01}=\frac{\omega_{b}}{\sqrt{1-A+\Delta_{0}+\sqrt{\delta^{2}+\Delta_{0 }^{2}}}},\quad\omega_{02}=\frac{\omega_{b}}{\sqrt{1-A+\Delta_{0}-\sqrt{\delta^ {2}+\Delta_{0}^{2}}}} \tag{20}\]  
Now, let us assume that \(hk\ll 1\). Subsequently, Equation (19) becomes:  
\[\Omega_{1,2}=\frac{1}{2}\Big{(}B\pm\sqrt{B^{2}+4\delta^{2}}\Big{)} \tag{21}\]  
and the resonant frequencies of the system read as:  
\[\omega_{01}=\frac{\omega_{b}}{\sqrt{1-A+B/2+\sqrt{(B/2)^{2}+\delta^{2}}}}, \quad\omega_{02}=\frac{\omega_{b}}{\sqrt{1-A+B/2-\sqrt{(B/2)^{2}+\delta^{2}}}} \tag{22}\]  
Finally, if both \(hk\ll 1\) and \(\delta\ll 1\), the solution reduces to:  
\[\Omega_{1}=0,\quad\Omega_{2}=B, \tag{23}\]  
that results in two resonant frequencies of the system:  
\[\omega_{01}=\frac{\omega_{b}}{\sqrt{1-A}},\quad\omega_{02}=\frac{\omega_{b}}{ \sqrt{1-A+B}} \tag{24}\]  
Similarly, if there are interfaces with abrupt acoustic impedance change from both sides of the bubble array, we obtain from Equation (13):  
\[\Omega_{1,2}=\Delta_{1}\pm\sqrt{\delta^{2}+\Delta_{1}^{2}}, \tag{25}\]  
where \(\Delta_{1}=\left(B+4 hk\delta\right)/4\).