## 4 1/3 Subharmonic Resonance  
In this section, to analyze the 1/3 subharmonic resonance of the system, we let:  
\[\Omega=3\omega_{0}+\varepsilon\sigma. \tag{31}\]  
Figure 5: The amplitude-frequency curve of the superharmonic resonance.  
Figure 6: The response-excitation amplitude curve of the superharmonic resonance.  
Figure 7: The time history curves of the response of the controlled beams.  
To eliminate the secular terms in Equation (18), we put:  
\[i\omega_{0}(2A_{n}^{\prime}+\mu_{n}A_{n})+6\Gamma_{mmm}A_{n}\Lambda_ {n}^{2}+3\Gamma_{mmn}A_{n}^{2}\bar{A}_{n}+3\Gamma_{mmn}\Lambda_{n}\bar{A}_{n}^{2 }\exp(i\sigma T_{1})\] \[+k_{ann}A_{n}\exp(-i\omega_{0}\tau)=0, \tag{32}\]  
Substituting \(A_{n}=a_{n}\exp(i\beta_{n})/2\) into Equation (32) and separating the result into real and imaginary parts, we have:  
\[a_{n}^{\prime}=-\mu_{ne}a_{n}-\frac{3\Gamma_{mmn}\Lambda_{n}}{4\omega_{0}}a_{n }^{2}\sin\gamma_{n}, \tag{33}\]  
\[a_{n}\gamma_{n}^{\prime}=\left(\sigma_{e}-\frac{9\Gamma_{mmn}\Lambda_{n}^{2}} {\omega_{0}}\right)a_{n}-\frac{9\Gamma_{mmn}}{8\omega_{0}}a_{n}^{3}-\frac{9 \Gamma_{mmn}\Lambda_{n}}{4\omega_{0}}a_{n}^{2}\cos\gamma_{n}, \tag{34}\]  
where \(\gamma_{n}=\sigma T_{1}-\beta_{n},\mu_{ne}=\mu_{n}/2-k_{ann}\sin(\omega_{0} \tau)/(2\omega_{0})\) and \(\sigma_{e}=\sigma-3k_{ann}\cos(\omega_{0}\tau)/(2\omega_{0})\). Therefore, for the first approximation:  
\[q_{n}=\frac{1}{2}a_{n}\cos[\frac{1}{3}(\Omega t-\gamma_{n})]+f_{n}(\omega_{0}^ {2}-\Omega^{2})^{-1}\cos\Omega t+O(\varepsilon). \tag{35}\]  
The steady-state motions correspond to \(a_{n}^{\prime}=\gamma_{n}^{\prime}=0\), that is they correspond to the solutions of:  
\[-\mu_{ne}a_{n}=\frac{3\Gamma_{mmn}\Lambda_{n}}{4\omega_{0}}a_{n}^{2}\sin\gamma _{n} \tag{36}\]  
\[\left(\sigma_{e}-\frac{9\Gamma_{mmn}\Lambda_{n}^{2}}{\omega_{0}}\right)a_{n}- \frac{9\Gamma_{mmn}}{8\omega_{0}}a_{n}^{3}=\frac{9\Gamma_{mmn}\Lambda_{n}}{4 \omega_{0}}a_{n}^{2}\cos\gamma_{n} \tag{37}\]  
Squaring and adding these equations leads to the frequency-response equation:  
\[\left[9\mu_{ne}^{2}+\left(\sigma_{e}-\frac{9\Gamma_{mmn}\Lambda_{n}^{2}}{ \omega_{0}}-\frac{9\Gamma_{mmn}}{8\omega_{0}}a_{n}^{2}\right)^{2}\right]a_{n} ^{2}=\frac{81\Gamma_{mmn}^{2}\Lambda_{n}^{2}}{16\omega_{0}^{2}}a_{n}^{4}. \tag{38}\]  
Equation (38) shows that either \(a_{n}=0\) or:  
\[9\mu_{ne}^{2}+\left(\sigma_{e}-\frac{9\Gamma_{mmn}\Lambda_{n}^{2}}{\omega_{0}} -\frac{9\Gamma_{mmn}}{8\omega_{0}}a_{n}^{2}\right)^{2}=\frac{81\Gamma_{mmn}^{ 2}\Lambda_{n}^{2}}{16\omega_{0}^{2}}a_{n}^{2}, \tag{39}\]  
which is quadratic in \(a_{n}^{2}\). Its solution is:  
\[a_{n}^{2}=\
u\pm(\
u^{2}-\iota)^{1/2}, \tag{40}\]  
where:  
\[\
u=\frac{8\omega_{0}\sigma_{e}}{9\Gamma_{mmn}}-6\Lambda_{n}^{2}\quad\text{ and}\quad\iota=\frac{64\omega_{0}^{2}}{81\Gamma_{mmn}^{2}}\left[9\mu_{ne}^{2}+ \left(\sigma_{e}-\frac{9\Gamma_{mmn}\Lambda_{n}^{2}}{\omega_{0}}\right)^{2} \right], \tag{41}\]  
We note that \(\iota\) is always positive, and thus, nontrivial free-oscillation amplitudes occur only when \(\
u>0\) and \(\
u^{2}\geq\iota\). These conditions demand that:  
\[\Lambda_{n}^{2}<\frac{4\omega_{0}\sigma_{e}}{2\Gamma_{mmn}},\frac{\Gamma_{mmn }\Lambda_{n}^{2}}{\omega_{0}}\left(\sigma_{e}-\frac{63\Gamma_{mmn}\Lambda_{n}^ {2}}{8\omega_{0}}\right)-2\mu_{ne}^{2}\geq 0. \tag{42}\]  
is follows that \(\Gamma_{mmn}\) and \(\sigma_{e}\) must have the same sign.  
It follows from Equation (42) that, for a given \(\Lambda_{n}\), nontrivial solutions can exist only if:  
\[\Gamma_{mmn}\sigma_{e}\geq\frac{2\mu_{ne}^{2}\omega_{0}}{\Lambda_{n}^{2}}+ \frac{63\Gamma_{mmn}\Lambda_{n}^{2}}{8\omega_{0}} \tag{43}\]  
It is worth noting that the amplitude can be effectively suppressed by adjusting the control gain \(k_{a11}\) and time delay.