## 3 Key Components Design  
### Vibration Mechanism  
#### 3.1.1 Dynamic Model of Vibration Harvesting  
At present, there are three main methods for picking winter jujubes: Hand picking, suction picking, and vibration picking. Among them, hand picking requires human beating of fruits with high labor intensity and low efficiency, and it is easy to cause damage to the jujubes. The suction picking needs complicated equipment, and the technology is not yet mature. The vibration picking device is simple, easy to operate, and the acquisition rate of fruit is high. Therefore, a small eccentric vibration picking mechanism of winter jujubes was designed as shown in Figure 2, and it consists of eccentric blocks [12], direct current motor (DC motor), fixation bracket, etc. The forced vibration of jujube tree under external harmonic excitation can be simplified to a second-order linear mass-spring system. The motion differential equation of this system can be obtained by Equation (1) [13]. \[m\ddot{x}+c\dot{x}+kx=F(t)=F_{0}\sin(\omega t) \tag{1}\]  
where, \(m\) is the mass of the vibration system, kg; \(k\) is the stiffness coefficient, N/m; \(c\) is the damping factor, N-s/m; \(x\) is the vibration displacement, mm; \(F\) is the exciting force, N; \(F_{0}\) is the maximum exciting force, N; \(t\) is the time, s; \(\omega\) is the frequency, Hz. All the solutions of the differential equation \(x\) include a general solution \(x_{1}\) and a particular solution \(x_{2}\), and \(x=x_{1}+x_{2}\). In the case of small damping, the general solution \(x_{1}\) is not considered, and the special solution is set as:  
\[x_{2}=X\sin(\omega t-\varphi) \tag{2}\]  
where, \(X\) is the amplitude of forced vibration, mm; \(\varphi\) is the phase angle, rad. Equation (2) was substituted into Equation (1), and we can get,  
\[\big{(}k-m\omega^{2}\big{)}X\sin(\omega t-\varphi)+\mathrm{c}\omega X\mathrm{ cos}(\omega t-\varphi)=F_{0}\sin(\omega t) \tag{3}\]  
Rewrite \(F_{0}\sin(\omega t)\) at the right end of the above expression as,  
\[F_{0}\sin(\omega t)=F_{0}\sin[(\omega t-\varphi)+\varphi]=F_{0}\cos\varphi\sin (\omega t-\varphi)+F_{0}\sin\varphi\cos(\omega t-\varphi) \tag{4}\]  
Substitute Equation (4) into Equation (3), and the equation can be obtained:  
\[\big{[}\big{(}k-m\omega^{2}\big{)}X-F_{0}\cos\varphi\big{]}\sin(\omega t- \varphi)+(\omega X-F_{0}\sin\varphi)\cos(\omega t-\varphi)=0 \tag{5}\]  
Figure 2: Vibration mechanism diagram: (1) Eccentric block; (2) direct current motor; (3) fixation bracket; (4) rubber pad; (5) fixed belt. This equation is equal to 0 for any time \(t\), therefore,  
\[\left(k-m\omega^{2}\right)X-F_{0}\text{cos}\ \varphi=0 \tag{6}\]  
\[c\omega X-F_{0}\text{sin}\ \varphi=0 \tag{7}\]  
Thus,  
\[X=\frac{F_{0}}{\sqrt{\left(k-m\omega^{2}\right)^{2}+\left(c\omega\right)^{2}}} \tag{8}\]  
\[\tan\varphi=\frac{c\omega}{k-m\omega^{2}} \tag{9}\]  
Equations (8) and (9) divide the numerator and denominator by \(k\), and we can get:  
\[x_{2}=X\sin(\omega t-\varphi) \tag{2}\]0 \[x_{2}=X\sin(\omega t-\varphi) \tag{2}\]1  
where, \(\omega_{n}=\sqrt{\frac{k}{n}}\), \(\zeta=\frac{c}{c_{c}}\), \(c_{c}=2m\omega_{n}\). The \(\omega/\omega_{n}\) is called frequency ratio. The main function of the conversion here is to simplify the formula form and make it more tidy. From the above analysis, the specific solution of the equation can be written:  
\[x_{2}=X\sin(\omega t-\varphi) \tag{2}\]2  
In this mechanism, two eccentric blocks with identical shape and size are installed on both sides of the support symmetrically. Under the same angular velocity, the excitation force changes in sin form, and the direction of the force is perpendicular to the connection of the rotating center of two eccentric blocks. \[x_{2}=X\sin(\omega t-\varphi) \tag{2}\]3  
where, \(m_{e}\) is the mass of eccentric block, kg; \(R\) is the eccentricity of the eccentric block, m, \(R=\frac{4r}{3\pi}\); \(\omega\) is the angular velocity of the eccentric block, rad/s; \(\omega=\frac{\pi n}{60}\), where \(n\) is the motor speed, rev/min.