## 3 Dynamic Analysis  
### Primary Frequency--Amplitude Relation  
Considering that the load plate is acted by a harmonic force with the amplitude \(F_{\rm e}\) and frequency \(\omega\), the motion of the load plate is expressed as following:  
\[M\ddot{u}+C\dot{u}-\left(a_{1}u+a_{2}u^{2}+a_{3}u^{3}+a_{4}u^{4}+a_{5}u^{5} \right)=F_{e}\cos(\omega t) \tag{12}\]  
By introducing parameters as following:  
\[ \begin{array}{l}\hat{u}=\frac{u}{u_{a}};\ \omega_{n}^{2}=-\frac{a_{1}}{M}; \ \Omega=\frac{\omega}{\omega_{n}};\ \tau=\omega_{n}t;\ \xi=\frac{C}{2\omega_{n}M};\\ \alpha_{n}=-\frac{a_{n}u^{n-1}}{M\omega_{n}^{2}};\ u_{o}=\left.\frac{F_{\rm e} }{-a_{1}}\right|_{\omega=0\alpha_{\hat{u}}=0(i=2,3,\ldots,5)}\end{array} \]  
In dimensionless form, Equation (12) is rewritten as below:  
\[\hat{u}^{\prime\prime}+2\xi\hat{u}^{\prime}+\hat{u}+\sum_{n=2}^{5}\alpha_{n} \hat{u}^{n}=cos(\Omega\tau) \tag{13}\]  
where the number of primes denote the order of differentiation with respect to dimensionless time \(\tau\)  
In order to find the solution in the neighborhood of the equilibrium position \(\hat{u}=0\), we let \(\hat{u}=\varepsilon\hat{x}\)  
\[\overline{x}^{\prime\prime}+2\varepsilon\hat{x}^{\prime}\overline{x}+\overline {x}+\sum_{n=2}^{5}\alpha_{n}\varepsilon^{n-1}\overline{x}^{\prime\prime}= \varepsilon^{4}\overline{k}\cos(\Omega\tau) \tag{14}\]  
herein \(\varepsilon<<1\), and \(\varepsilon^{4}\overline{\xi}\overline{x}^{\prime}=\xi\hat{x}^{\prime};\ \cos( \Omega\tau)=\varepsilon^{4}k\cos(\Omega\tau)\). Figure 7: Stiffness curve for different values of \(\mu\) given in the panel of figure: (**a**) Quasi-zero stiffness at position (\(\hat{u}=0.008;\ 0.021;\ 0.046;\ 0.083\)); (**b**) Arbitrary stiffness. In addition, the differentiation of \(a\) and \(\beta\) with respect to dimensionless time \(\tau\)_is_ presented as below:  
\[ \begin{array}{ll}\dot{a}^{\prime}=\frac{da}{d\tau}=\varepsilon D_{1}\dot{a}+ \varepsilon^{2}D_{2}\dot{a}+\varepsilon^{3}D_{3}\dot{a}+\varepsilon^{4}D_{4} \dot{a}\\ \beta^{\prime}=\frac{d\beta}{d\tau}=\varepsilon D_{1}\beta+\varepsilon^{2}D_{2 }\beta+\varepsilon^{3}D_{3}\beta+\varepsilon^{4}D_{4}\beta\end{array}  \tag{22}\]  
From Equation (21), Equation (22) is rewritten as the following:  
\[ \begin{array}{ll}\dot{a}^{\prime}&=-\widetilde{\xi}\epsilon^{4}\dot{a}+ \frac{1}{2}\epsilon^{4}\widetilde{K}\sin(\sigma T_{2}-\beta)\\ \dot{a}\beta^{\prime}&=\frac{3a_{2}}{8}\varepsilon^{2}\dot{a}^{3}-\frac{10a_ {2}^{2}}{24}\varepsilon^{2}\dot{a}^{3}+\left(\frac{10a_{1}^{2}}{6}-\frac{3a_ {2}}{2}\right)^{2}\frac{1}{2}\epsilon^{4}\dot{a}^{5}\\ &+\epsilon^{4}\alpha_{1}\left(\frac{26a_{1}a_{2}}{12}-\frac{172a_{1}^{3}}{216 }-\frac{28a_{2}}{3}\right)\frac{1}{2^{5}}\epsilon^{4}\dot{a}^{5}\\ &+\alpha_{2}\left(\frac{1926a_{1}^{2}}{216}+\frac{3a_{2}}{8}\right)\frac{1}{2 ^{5}}\epsilon^{4}\dot{a}^{5}+\frac{56a_{1}a_{3}}{3}\frac{1}{2^{5}}\epsilon^{4 }\dot{a}^{5}\\ &+\frac{10a_{1}}{2^{5}}\epsilon^{4}\dot{a}^{5}-\epsilon^{4}\frac{\widetilde{K }}{2}\cos(\sigma T_{2}-\beta)\end{array}  \tag{23}\]  
Recalling \(\dot{u}=\varepsilon\dot{a}\), and by letting \(\gamma=\sigma T_{2}-\beta\), Equation (23) is expressed with respect to \(\dot{a}_{u}\) as below:  
\[ \begin{array}{ll}\dot{a}_{u}^{\prime}=-\dot{\varsigma}\dot{a}_{u}+\frac{1}{ 2}K\sin\gamma\\ \dot{a}_{u}\gamma^{\prime}=\dot{a}_{u}\sigma-G(\alpha_{1},\alpha_{2})\dot{a}_ {u}^{3}-F(\alpha_{1},\alpha_{2})\dot{a}_{u}^{5}+\frac{K}{2}\cos\gamma\end{array}  \tag{24}\]  
with \(\dot{a}_{u}=\varepsilon a\), \(G(\alpha_{1},\alpha_{2})=\left(\frac{3a_{2}}{8}-\frac{10a_{1}^{2}}{24}\right)\);  
\[F(\alpha_{1},\alpha_{2})=\left( \begin{array}{ll}-\left(\frac{10a_{1}^{2}}{6} -\frac{3a_{2}}{2}\right)^{2}\frac{1}{2^{5}}+\epsilon^{4}\alpha_{1}\left(\frac {26a_{1}a_{2}}{12}-\frac{1172a_{1}^{3}}{216}-\frac{28a_{3}}{3}\right)\frac{1}{ 2^{5}}\\ +\alpha_{2}\left(\frac{1926a_{1}^{2}}{216}+\frac{3a_{2}}{8}\right)\frac{1}{2^{ 5}}+\frac{56a_{1}a_{3}}{3}\frac{1}{2^{5}}+\frac{10a_{1}}{2^{5}}\end{array}