## 5 Equivalent Circuit of the Coupled Mechanical Oscillators
### Equivalent Circuit Diagram of the Dry-Friction Term \(f_{\rm DFI}(z_{i})\)  
The Figure 11 depicts the equivalent circuit diagram of the force due to the dry-friction \(f_{\rm DFI}(z_{1})\) of the first oscillator. \(z_{1}\) is the applied input voltage. In order to obtain the expression of \(f_{\rm DFI}(z_{1})\), which is, \(f_{\rm DFI}(z_{1})=\frac{t_{1}z_{1}}{\big{(}z_{1}^{2}+\alpha_{1}^{2}\big{)}^{ 2}}\), using analog circuit components, we first make the square of \(z_{1}\). The multiplier \(M1\) does this operation. So, the voltage \(V_{15}\) in the Figure 11 is \(z_{1}^{2}\). After that, the voltage \(V_{15}\) is added with \(q_{1}\) (i.e., \(a_{1}^{2}\)) using an inverting adder, \(OA1\). So the output of \(OA1\) is the \(-\big{(}z_{1}^{2}+a_{1}^{2}\big{)}\). As, the output is negative, we now use a unity gain inverting op-amp \(OA2\) to get the expression \(\big{(}z_{1}^{2}+a_{1}^{2}\big{)}\).  
As the denominator of the second term of the expression of \(f_{\rm R1}(z_{1})\), is the square root of \(\big{(}z_{1}^{2}+a_{1}^{2}\big{)}\), we use the op-amp \(OA3\) to make the square-root of the output voltage of \(OA2\), which is \(V_{\rm OA2}\). For that, we have connected the \(OA2\) op-amp's output with the non-inverting input terminal of \(OA3\). The output of the \(OA3\) op-amp, \(V_{\rm OA3}\) is connected to the inverting input of the \(OA3\) op-amp through a multiplier \(M2\). The output of \(M2\) is the square of the output voltage of the op-amp \(OA3\), i.e., \(V_{\rm OA3}^{2}\). Due to the virtual connection concept of the two input terminals of an ideal op-amp, we can write, \(V_{\rm OA3}^{2}=V_{\rm OA2}\). Hence, \(V_{\rm OA3}=\sqrt{V_{\rm OA2}}=\sqrt{\big{(}z_{1}^{2}+a_{1}^{2}\big{)}}\).  
To get the numerator of the second term of \(f_{\rm R1}(z_{1})\), we have passed \(z_{1}\) through a proportional block of gain \(0.2413\) to achieve \(0.2413z_{1}\) at the output of the proportional block.  
The op-amp \(OA4\) is used to make the ratio of the two signals. The output voltage of the proportional block is directly connected to the non-inverting input of the op-amp \(OA4\). We have used the multiplier \(M3\) to multiply the output of the op-amp \(OA4\), i.e., \(V_{\rm OA4}\) with the \(V_{\rm OA3}\). So, the output of the multiplier \(M3\) is \(V_{\rm OA3}\cdot V_{\rm OA4}\). This is connected to the inverting input of the op-amp \(OA4\). Due to the virtual connection concept of an ideal op-amp, we can say that \(0.2413z_{1}=V_{\rm OA3}\cdot V_{\rm OA4}\), which makes \(V_{\rm OA4}=\frac{0.2413z_{1}}{V_{\rm OA3}}\). So, the output of the op-amp \(OA4\) provides the expression of the force due to the dry-friction, which is \(f_{\rm DFI}(z_{1})=V_{\rm OA4}=\frac{0.2413z_{1}}{\sqrt{\big{(}z_{1}^{2}+a_{1}^ {2}\big{)}}}\).  
Figure 11: The equivalent circuit diagram of the force due to the dry-friction \(f_{\rm DFI}(z_{1})\). The resistance values are: \(R_{104}=R_{105}=R_{106}=R_{1}=R_{2}=R_{3}=R_{4}=100\) k\(\Omega\), \(k=0.2413\). \(OA1\) to \(OA4\) are the Op-Amps, \(M_{1}\) to \(M_{3}\) are the multipliers, \(q_{1}\) is the dc voltage source with the value of \(a_{1}^{2}\). \(K\) is the gain of a proportional block. (Color Online).  
Similarly, using the same logic, we have constructed the equivalent circuit diagram of the non-dimensional dry-friction force term of the second oscillator. For the second mechanical oscillator, the gain of the proportional block is chosen as \(K=0.2880\). The circuit diagram is shown in the Figure 12. The output voltage of the op-amp OA8 provides the expression of \(F_{\text{DF2}}(z_{2})\).