Harmonic excitation of One-Degree-of-Freedom Systems

Forced vibrations of a one-degree-of-freedom system occur when work is being done on the system while the vibrations occur. For example forced excitation includes ground motion during an earthquake or the motion caused by an unbalanced reciprocating component.

A mechanical or structural system is said to undergo forced vibration whenever external energy is supplied to the system during the vibration. External energy can be supplied through either an applied force or an imposed displacement excitation. The applied force or displacement excitation may be harmonic, non-harmonic but periodic, non-periodic, or random in nature. The response of a system to a harmonic excitation is called harmonic response. The non-periodic excitation may have a long or short duration. The response of a dynamic system to suddenly applied non-periodic excitations is called transient response.

Equations of motion

Let’s examine the system which consists of a block with mass m, spring with stiffens k and the viscous damper with damping coefficient c. The force F(t) acts on the body and pulls it from its equilibrium position.

FIGURE 3.1 - A spring-mass-damper system

FIGURE 3.2 - Free Body Diagram a) External forces b) Effective forces

So now you can apply the Free Body Diagram Method or the Equivalent System Method. In this case we will apply the Free Body Diagram Method and by applying the Newton’s second law we will get the differential equation which describes the motion (vibrations) of the system.

$$m\ddot{x}+c\dot{x}+kx=F(t)$$

Since this equation is nonhomogeneous, its general solution x(t) is given by the sum of the homogeneous solution, and the particular solution. The homogeneous solution, which is the solution of the homogeneous equation

$$m\ddot{x}+c\dot{x}+kx=0$$

represents the free vibration of the system. As previously discussed the free vibrations dies out with time under each of three possible conditions of damping (under damping, critical damping and over damping) and under all possible initial conditions.

With time the general solution of the non-homogenous differential equation reduces to the particular solution which represents the steady-state vibration. The steady-state motion is present as long as the force or forcing function is present.

The Study of Vibrations - Examples

Example 1.1.

A spring is stretched 100 mm by a 10 kg block. If the block is displaced 50 mm downward from its equilibrium position and given a downward velocity of 3 m/s determine the differential equation which describes the motion. Assume that positive displacement is downward. Also, determine the position of the block when t=4 s. 

Figure 1.1 – a) Model of a system, b) Free body diagrams for external and effective forces

Solution: From previous figure we can derive the differential equation.

$$\begin{align} & +\downarrow \sum{{{F}_{y}}}=m{{a}_{y}} \\ & mg-k\left( y+{{y}_{st}} \right)=m\ddot{y} \\ \end{align}$$

where 

 $$k{{y}_{st}}=mg$$

Now we need to substitute the previous equation into the differential equation. With this operation we get: 

$$\begin{align} & mg-ky-mg=m\ddot{y}, \\ & \ddot{y}+\frac{k}{m}y=0, \\ \end{align}$$

Before we can determine  the natural frequency of the system we need to determine the stiffness of the spring. We know that the force the stretch the spring is proportion to the stiffness of the spring and the elongation of the spring. In this case we don’t have a force which acts on spring but we have a block which is attached to the sprig. So we will determine the stiffness of the spring by following expression

$$\begin{align} & F=ky,F=G=mg \\ & k=\frac{G}{y}=\frac{mg}{y}=\frac{10\cdot 9.81}{0.1}=981\text{ N/m}{{\text{m}}^{2}} \\ \end{align}$$

The next step is to determine the natural frequency and we will do that by following expression:

$$\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{981}{10}}=9.9045\text{ }{{\text{s}}^{-1}}$$

 

When we have determined the natural frequency we can insert it into differential equation. Whit this operation we get:

$$\begin{align} & \ddot{y}+{{(9.9045)}^{2}}y=0, \\ & \ddot{y}+98.1y=0, \\ \end{align}$$

All that is left is to apply the boundary conditions. At

$$t=0,y=0,05$$

By applying this boundary conditions we get:

$$\begin{align} & 0.05=A\sin 0+b\cos 0 \\ & b=0.05 \\ \end{align}$$

At

$$t=0,{{v}_{0}}=3m/s$$ $$\begin{align} & {{v}_{0}}=A\omega \cos 0-0=A \\ & A=\frac{{{v}_{0}}}{\omega }=\frac{3}{9.9045}=0.30289 \\ \end{align}$$

Example 1.2.

A 3 kg block is suspended from a spring having a stiffness of k=200N/m. If the block is pushed 50 mm upward from its equilibrium position and then released from rest, determine the equation that describes the motion. What are the amplitude and the frequency of the vibration? Assume that the positive displacement is downward.

From the given data we can determine the frequency of the vibrations.

$$\begin{align} & \omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{200}{3}}=8.165 \\ & f=\frac{\omega }{2\pi }=\frac{8.165}{2\pi }=1.299=1.3Hz \\ & x=A\sin \omega t+B\cos \omega t, \\ \end{align}$$

Boundary conditions are:

$$\begin{align} & t=0,x=-0.05, \\ & t=0,v=0 \\ \end{align}$$

By applying the boundary condition to the solution of the differential equation we get:

$$\begin{align} & 0.05=A\sin 0+B\cos 0, \\ & b=-0.05. \\ & v=A\omega \cos \omega t-B\omega \sin \omega t, \\ & 0=A\omega \cos 0-B\omega \sin 0 \\ & A=0 \\ \end{align}$$

In order to determine the amplitude of the system we will apply the following equation.

$$C=\sqrt{{{A}^{2}}+{{B}^{2}}}=\sqrt{2.5\cdot {{10}^{-3}}}=0.05m=50mm$$

Example 1.3

A spring has a stiffness of 800 N/m. If a 2 kg block is attached to the spring, pushed 50 mm above its equilibrium position, and released from rest, determine the equation that describes the block’s motion. Assume that positive displacement is downward.

$$\begin{align} & \omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{800}{2}}=20{{s}^{-1}}, \\ & y=A\sin \omega t+B\cos \omega t, \\ \end{align}$$

For t=0, y=-0.05 m

$$\begin{align} & -0.05=A\sin 0+B\cos 0, \\ & B=-0.05 \\ \end{align}$$ $$\begin{align} & y=A\sin \omega t+B\cos \omega t, \\ & v=A\omega \cos \omega t-B\omega \sin \omega t, \\ \end{align}$$

For t=0, v=0

$$\begin{align} & v=A\omega \cos \omega t-B\omega \sin \omega t, \\ & 0=A\omega \cos 0-B\omega \sin 0, \\ & A=0 \\ \end{align}$$

Thus we get:

$$y=-0.05\cos \left( 20t \right)$$

Example 1.4.

A spring is stretched 200 mm by a 15 kg block. If the bloc is displaced 100 mm downward from its equilibrium position and given downward velocity of 0.75 m/s determine the equation which describes the motion. What is a phase angle? Assume that positive displacement is downward

$$\begin{align} & F=ky\Rightarrow k=\frac{F}{y}=\frac{15\cdot 9.81}{0.2}=735.75N/m \\ & \omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{735.75}{15}}=7{{s}^{-1}} \\ & y=A\sin \omega t+B\cos \omega t \\ \end{align}$$

When t=0 then y=0.1 m and

When t=0 then v=0.75 m/s

$$\begin{align} & y=A\sin \omega t+B\cos \omega t \\ & 0.1=A\sin 0+B\cos 0 \\ & B=0.1m \\ \end{align}$$ $$\begin{align} & v=A\omega \cos \omega t-B\omega \sin \omega t \\ & 0.75=A\omega \cos 0-B\omega \sin 0 \\ & 0.75=A\omega \\ & A=\frac{0.75}{7}=0.107 \\ \end{align}$$

Now we can insert the values of A and B into the solution.

$$\begin{align} & y=0.107\sin (7t)+0.100\cos (7t) \\ & \phi =arctan\left( \frac{B}{A} \right)=\arctan \left( \frac{0.100}{0.107} \right)={{43.0}^{\circ }} \\ \end{align}$$

Free vibration of single degree of freedom systems

Free vibrations are oscillations about a system’s equilibrium position that occur in the absence of an external excitation. Free vibrations are result of a kinetic energy imparted to the system or of a displacement from the equilibrium position that leads to a difference in potential energy from the system’s equilibrium position.

Let’s consider a model of one degree of freedom system which is shown in Fig 2.1. This system consists of a block of mass m and the spring with stiffens k. When block is displaced a distance a from its equilibrium position, a potential energy ka^2/2 is developed in the spring. When the system is released from equilibrium, the spring force draws the block toward the system’s equilibrium position, with the potential energy being converted to kinetic energy. When the block reaches its equilibrium position the kinetic energy reaches a maximum and the motion continues. The kinetic energy is converted to potential energy until the spring is compressed a distance a. This process of transfer of potential energy to kinetic energy and vice versa is continual in the absence of non-conservative forces.

Remainder:

            Conservative force is a force with the property that the work done in moving a particle between two points is independent of the taken path. For example when you lift a book, the work that you do against gravity in lifting is stored and is available for kinetic energy of the book once you let go. Some conservative forces are: Gravity, Elastic (Hooke’s Law), electric etc.

            Non-conservative force is a force with the property that the work done in moving a particle between two points is dependent on the taken path. For example friction is a non-conservative force. When you move an object the work that you do “against friction” is apparently lost-it is certainly not available to the object as kinetic energy. In general any friction type force, like air resistance is non-conservative force.

            Examples of free vibrations of systems that can be modeled using one degree of freedom include the oscillations of a pendulum about a vertical equilibrium position, the motion of a recoil mechanism of a firearm once it has been fired, and the motion of a vehicle suspension system after the vehicle encounters a pothole.

            Vibrations of the system with one degree of freedom can be described with the second order differential equation. The independent variable is time, while the dependent variable is the chosen generalized coordinate. The chosen generalized coordinate represents the displacement of a particle in the system or an angular displacement and is measured from the system’s equilibrium position.        

            We can derive differential equation by applying one of two methods and these methods are:

1)      The Free Body Diagram Method or

2)      Equivalent System Method

The general solution of the differential equation is a linear combination of two linearly independent solutions. The arbitrary constants, called constants of integration, are uniquely determined on application of two initial conditions. The necessary initial conditions are values of the generalized coordinate and its first time derivative at specified time, usually t = 0.

     The form of the solution of the differential equation depends on system parameters. That means that solution of system with no damping will be different from the solution with viscous damping.