Electric Circuit Analogs

The vibrating system and its characteristics can be represented by an electric circuit. Now let’s consider the electrical circuit which is shown in the next figure.

 

Figure 1 - Electric Circuit

The system consists of inductor L, a resistor R and a capacitor C. When the voltage E(t) is applied to the system, it caused a current of magnitude i to flow through the circuit. As the current flows past the inductor the voltage drop is L(di/dt). When the current flows through the resistor the voltage drop is Ri, and when it arrives at the capacitor the drop is 

$$\frac{1}{C}\int{idt}$$

Since the current cannot flow past a capacitor, it’s only possible to measure the charge q acting on the capacitor. The charge can be related to the current by the equation

$$i=\frac{dq}{dt}$$

Thus the voltage drops, which occur across the inductor, resistor, and capacitor becomes

$$L\frac{{{d}^{2}}q}{d{{t}^{2}}},R\frac{dq}{dt},\frac{q}{C}$$

By applying the Kirchhoff’s law which states that the applied voltage balances the sum of the voltage drops around the circuit. Therefore:

$$L\frac{{{d}^{2}}q}{d{{t}^{2}}}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)$$

As you can see the previous differential equation which represents the change of voltage in electrical system is similar to the differential equation which describes the motion of Viscous Dampers Forced Vibration system.

By comparing these to differential equation we can see that these equation have the same form, and hence mathematically the procedure of analyzing an electric circuit is the same as that of analyzing a vibration mechanical system

$$m\frac{{{d}^{2}}x}{d{{t}^{2}}}+c\frac{dx}{dt}+kx=F(t)$$

From this we can derive the analogs between two equations and we will show them in the following table.

Electrical		Mechanical
Electric Charge	q	Displacement	x
Electric Current	i	Velocity	dx/dt
Voltage	E(t)	Applied Force	F(t)
Inductance	L	Mass	m
Resistance	R	Viscous damping coefficient	c
Reciprocal of capacitance	1/C	Spring stiffness	k

Free Body Diagram Method

The system consisting of a body with mass m and the spring of stiffens k is shown in the next figure. The body is displaced form equilibrium position and released.

Figure 2.2 a) Mass-spring system b) Free-body diagrams at an arbitrary instant. Directions of external and effective forces are consistent with positive direction of generalized coordinate x.

So now we will derive the differential equation of motion for this system by applying the FBD method. First we need to draw this body without connections. On this body we will add the weight force in downward directions and the force of the spring in upward direction.

Now we need to choose the generalized coordinate. Let’s say that x(t) is a generalized coordinate and the positive values of that coordinate is in the downward direction. Since the body is pulled down from equilibrium position and then released we need to add the distance from the point of release and the equilibrium position. 
  $$\begin{align} & \sum{{{F}_{ext}}=\sum{{{F}_{eff}}}} \\ & mg-k\left( x+{{\Delta }_{st}} \right)=m\ddot{x}, \\ & mg-kx-k{{\Delta }_{st}}=m\ddot{x}, \\ & mg=k{{\Delta }_{st}}, \\ & mg-kx-mg=m\ddot{x}, \\ & m\ddot{x}+kx=0. \\ \end{align}$$

After we add these forces to the body we can sum forces and derive the differential equation.

In general the procedure of FBD method is: 1)      Choose the generalized coordinate. This variable should represent the displacement of a particle in the system. If we analyze the rotational motion then the generalized coordinate could represent an angular displacement. 2)      FBD diagrams are drawn showing the system at an arbitrary instant of time. For any system you need to make two Free Body Diagrams one with external forces and one with effective forces. 3)      The appropriate form of Newton’s law is applied to free-body diagrams. 4)      Applicable assumptions are used along with algebraic manipulation. The result is the governing differential equation which describes the motion of  the analyzed system.