*3.1. The Pendulum*

Two cases are presented below in deriving the equations of motion of the pendulum. Case I follows the standard models using Newton's second law of motion to derive the pendulum with constant length equations. The works existing in literature use this type of pendulum to act as the system inputs. In Case II of the present work, we used a vertically excited parametric pendulum with variable length as the system input. Hence, more frequent and faster oscillations can be achieved. In this case, the Euler–Lagrange equation is utilized to obtain the governing equations.

Case I: The energy needed for starting the pumping process is initiated by swinging the pendulum with minimum effort [20,21]. The pendulum model showing the free body diagram is shown in Figure 4, and its equation of motion is derived below. The force response does not transmit a moment around the position 0, where the rotation takes place. Therefore, the sum of the moments is 0 about the attachment point.

**Figure 4.** The pendulum free body diagram.

Summing all moments of forces, the following equation is found:

$$
\sum M\_{\mathbf{f}} = l\_0 \ddot{\boldsymbol{\varphi}}\_{\mathbf{\prime}} \tag{3}
$$

$$
\sum M\_{\rm t} = -(m+m\_{\rm b})gl\sin\varphi - T\_{\rm v} + T\_{\rm hv} \tag{4}
$$

where: *mb*—pendulum's bar mass (kg), *m*—the pendulum body mass (kg), *l*—length to the pivot of mass of the end weight (m), *Tv*—moment of viscous friction force (N·m), *Th* excitation moment (N·m), *ϕ*—pendulum angular displacement (rad), *I*0—mass moment of inertia (kg·m2), and *<sup>g</sup>*—acceleration due to gravity (m/s2).

Simplifications: (i) the pendulum mass is concentrated in the center point of the end mass; (ii) sin *ϕ* ≈ *ϕ*, so we assume small angles of rotation in this case.

The mass moment of inertia *I*<sup>0</sup> is given as:

$$I\_0 \approx (m + m\_b)l^2,\tag{5}$$

For a more accurate approach, we consider the body's relative mass of the system to be situated at the system's pivot of mass, which is:

$$l\_{\mathcal{S}} = \frac{(ml + 0.5m\_{\mathbb{b}}l)}{(m + m\_{\mathbb{b}})},\tag{6}$$

so the mass moment of inertia yields:

$$I\_0 = \frac{m\_b l^2}{3} + ml^2.\tag{7}$$

The moment of viscous friction *Tv* = *<sup>b</sup>ϕ*˙, where *<sup>b</sup>* (N·s·m−1) states the coefficient of viscous damping.

Equation (4) becomes:

$$-\left(m+m\_{b}\right)g l\_{\mathcal{S}} \sin \varphi - b\phi + T\_{h} = I\_{0} \dot{\varphi},\tag{8}$$

and at the assumed simplifications:

$$
\ddot{\varphi} + \frac{b}{I\_0} \dot{\varphi} + \frac{(m + m\_b) g l\_\mathcal{g} q}{I\_0} = T\_{h\prime} \tag{9}
$$

where: *Th* =*f*<sup>0</sup> cos *ωt*, *f*0—amplitude of forcing excitation (N·m).

The linearized method has the configuration of a typical unconstrained differential equation of the second-order. Complementing Equation (9) to a canonical form, one finds

$$
\ddot{\varphi} + 2\zeta\omega\_n \dot{\varphi} + \omega\_n^2 \dot{\varphi} = T\_{h\nu} \tag{10}
$$

where *<sup>ω</sup><sup>n</sup>* is the excitation frequency (rad·s−1), *<sup>ζ</sup>* is the damping coefficient (N·s·m−1), 2*ζω<sup>n</sup>* = *<sup>b</sup> I*0 , *ω*<sup>2</sup> *<sup>n</sup>* <sup>=</sup> (*m*+*mb*)*glg<sup>ϕ</sup> <sup>I</sup>*<sup>0</sup> . Later, we will show how numerous variables of the system affect the independent response of the pendulum coupling. For the simplified pendulum having its mass concentrated at its end, we find:

$$
\omega\_n \approx \sqrt{\frac{\mathcal{S}}{l}}.\tag{11}
$$

Case II: As shown in Figure 5, Lagrange's equation is utilized to derive the governing equation of the vertically excited parametric pendulum with variable length [16] using the sine function as input for the angular position [16].

**Figure 5.** Schematic diagram of the vertical parametric pendulum with variable-length.

It is possible to remodel the overall length of the pendulum *l*<sup>03</sup> by readjusting the location of the telescopic bar, *l*02. As the bar is recanted, the original length of the pendulum *l*<sup>01</sup> (*l*<sup>03</sup> = *l*01) is regained. The centre of rotation is stimulated vertically as a result of excised motion *I* = *I*(*t*). Using the Euler–Lagrange equation in two degrees of freedom, the unconventional system is expressed in Equation (12)

$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right) - \frac{\partial L}{\partial q} + \frac{\partial D}{\partial \dot{q}} = 0,\tag{12}$$

where: *ϕ*(*t*)—angular displacement defined from the downward inclination point, *Ek* kinetic energy equation of the system, *Ep*—potential energy, and *D*—dissipative energy of the system:

$$\begin{split} E\_k &= \frac{1}{2}m\left( (\dot{\mathfrak{x}}\_{\varepsilon}(t)\sin\varrho(t) + \cos\varrho(t)(l\_{01} + l\_{02} + \mathfrak{x}\_{\varepsilon})\dot{\varrho}(t)) \right)^2 + \\ &\left( -f\_0\omega\sin\left(\omega t\right) + \dot{\mathfrak{x}}\_{\varepsilon}(t)\cos\varrho(t) - \sin\varrho(t)(l\_{01} + l\_{02} + \mathfrak{x}\_{\varepsilon})\dot{\varrho}(t) \right)^2 \end{split} \tag{13}$$

$$E\_p = \frac{1}{2}kx\_\varepsilon^2(t) - gm(f\_0 \cos\left(\omega t\right) + \cos\left(t\right)(l\_{01} + l\_{02} + x\_\varepsilon(t))),\tag{14}$$

$$D = \frac{1}{2} c\_p \dot{\varphi}^2(t). \tag{15}$$

where: *m*—the mass of the pendulum body (a bob), *k*—spring stiffness, *cp*—viscous damping coefficient of the pendulum. Putting Equations (13)–(15) into Equation (12) and carrying out the analogous derivatives, the equation of motion is obtained as presented in Equations (16) and (17):

$$\begin{split} \ddot{\boldsymbol{\varphi}} &= \left( \frac{1}{m(l\_{01} + l\_{02} + \mathbf{x}\_{\varepsilon})(l\_{01} + l\_{02} + \mathbf{x}\_{\varepsilon})} \right) \Big( -m(l\_{01} + l\_{02}) \Big( \mathbf{g} + f\_{0} \boldsymbol{\omega}^{2} \cos \left( \omega t \right) \Big) \sin \boldsymbol{\varphi}(t) - \\ & \boldsymbol{c}\_{p} \boldsymbol{\varphi}(t) - m \mathbf{x}\_{\varepsilon}(t) \Big( \mathbf{g} + f\_{0} \boldsymbol{\omega}^{2} \cos \left( \omega t \right) \sin \boldsymbol{\varphi}(t) \Big) - m(l\_{01} + l\_{02} + \mathbf{x}\_{\varepsilon}(t)) (2 \dot{\boldsymbol{x}}\_{\varepsilon}(t) \boldsymbol{\varphi}(t)) \Big), \end{split} \tag{16}$$

$$\begin{split} \ddot{\mathbf{x}}\_{\varepsilon}(t) &= \mathbf{g} \cos \varphi(t) + f\_0 \omega^2 \cos \left(\omega t\right) \cos \varphi(t) - \frac{k \mathbf{x}\_{\varepsilon}(t)}{m} - \frac{c \dot{\mathbf{x}}\_{\varepsilon}(t)}{m} + l\_{01} \dot{\boldsymbol{\phi}}^2(t) + \\ &l\_{02} \dot{\boldsymbol{\phi}}^2(t) + \mathbf{x}\_{\varepsilon}(t) \, \dot{\boldsymbol{\phi}}^2(t). \end{split} \tag{17}$$

The motion considered here is the enforce motion existing as a sinusoidal wave of the conformation *I*(*t*) = *f*<sup>0</sup> cos (*ωt*), where *f*<sup>0</sup> is the excitation amplitude.

The above equation regains the established governing equation of the vertical parametric pendulum when *l*<sup>03</sup> is varied due to the extension *xe*. It is assumed that an external

force causes the excitation. Thus, we assumed the almost ideal case where an external force could partially compensate for dissipation of mechanical energy through a frictional contact from the human effort [16,22].

### *3.2. The Lever*

The lever amplifies the pendulum force and transmits the force to the piston through the bracket foundation link. To understand the lever model fully, we start with a simple model, as can be seen in Figure 6, which shows the lever with the pivot in the middle of the lever or center of gravity (CG), assuming that the CG is at the middle of the lever, that is, *l*<sup>1</sup> is equal to *l*2. Figure 7 represents the lever component showing the pendulum's connection and then to the piston through two different bracket foundation links, respectively. From Newton's second law, the equation of motion of the lever is obtained, where *ϕ* is denoted as the input for the lever model, which is the output from the pendulum model.

**Figure 6.** Schematic diagram of the lever device.

**Figure 7.** Schematic diagram of the lever device and connecting components like spring and damper.

Analyzing Figure 6, one writes: *x*<sup>1</sup> = *l*<sup>1</sup> sin *ϕ*, *x*<sup>2</sup> = *l*<sup>2</sup> sin *ϕ* and *<sup>x</sup>*<sup>1</sup> *<sup>x</sup>*<sup>2</sup> <sup>=</sup> *<sup>l</sup>*<sup>1</sup> *<sup>l</sup>*<sup>2</sup> = *<sup>L</sup>*. In addition, *<sup>v</sup>*<sup>1</sup> *<sup>v</sup>*<sup>2</sup> <sup>=</sup> *<sup>l</sup>*<sup>1</sup> *<sup>l</sup>*<sup>2</sup> states a ratio of lever arms: where *l*1—distance from *f*<sup>1</sup> and position *p*, *l*2—distance from *f*<sup>2</sup> and position *p*, *x*<sup>1</sup> and *x*<sup>2</sup> are the displacements at both ends of the lever. With small-angle approximation and the motion at the lever end to be purely translational in '*x*' direction, the relationship between the forces and displacements is carried out by summing the torques *T* around the pivot point (about fulcrum), i.e.,

$$\sum\_{ccw} T\_{ccw} - \sum\_{cw} T\_{cw} = 0,\tag{18}$$

where the subscript *ccw*—counter-clockwise direction, and *cw*—clockwise direction. With the relations: *<sup>f</sup>*1*l*<sup>1</sup> <sup>−</sup> *<sup>f</sup>*2*l*<sup>2</sup> <sup>=</sup> 0 (balance of momentums), *<sup>f</sup>*1*l*<sup>1</sup> <sup>=</sup> *<sup>f</sup>*2*l*<sup>2</sup> or *<sup>f</sup>*<sup>1</sup> *<sup>f</sup>*<sup>2</sup> <sup>=</sup> *<sup>l</sup>*<sup>2</sup> *<sup>l</sup>*<sup>1</sup> <sup>=</sup> <sup>1</sup> *L* (conservation of power in levers, see [23]); *f*<sup>1</sup> and *f*<sup>2</sup> (N) are the forces on both ends of the lever.

It can be noted that the force relationship with the velocity relationship is given as *v*1 *<sup>v</sup>*<sup>2</sup> <sup>=</sup> *<sup>l</sup>*<sup>1</sup> *<sup>l</sup>*<sup>2</sup> = *<sup>L</sup>*.

Therefore, *<sup>v</sup>*<sup>1</sup>

$$\frac{v\_1}{v\_2} \frac{f\_1}{f\_2} = \frac{l\_1}{l\_2} \frac{l\_2}{l\_1} = 1,\tag{19}$$

where we find that the power is preserved if *v*<sup>1</sup> *f*<sup>1</sup> = *v*<sup>2</sup> *f*2, so it is the same on both sides of the lever.

In Figure 7, one observes that the lever arms are unequal; one side will experience a higher force and velocity than the other side. We assume some definitions *Fk* = −*kx*, *Fc* <sup>=</sup> <sup>−</sup>*cx*, *<sup>F</sup>* <sup>=</sup> *mx*¨, and *<sup>l</sup>*<sup>3</sup> <sup>=</sup> <sup>1</sup> <sup>3</sup> *l*2, which is the distance from *CG* to *p*, *k* stiffness of the stiffness (N·m<sup>−</sup>1), and *<sup>c</sup>*—viscous damping coefficient (N·s·m<sup>−</sup>1).

Using Newton's second law, the lever governing equation is obtained as follows:

$$m\_l \ddot{\mathbf{x}} l\_3 + c \dot{\mathbf{x}} l\_2 + k \mathbf{x} l\_2 = f\_1(\mathbf{t}) l\_1. \tag{20}$$

where *ml*—is the total mass of the lever, *x*—is the displacement of the lever. *f*1(*t*) = sin *ϕ*.
