Free vibration of an Undamped Translational System

As mentioned earlier we are going to use the Free body diagram to derive the second order differential equation which describes the motion of the system. The procedure can be summarized as follows:

1)      Select a suitable coordinate to describe the position of the mass or rigid body in the system. Use a linear coordinate to describe the linear motion of a point mass or the centroid of a rigid body, and an angular coordinate to describe the angular motion of a rigid body.

2)      Determine the static equilibrium configuration of the system and measure the displacement of the mass or rigid body from its static equilibrium position.

3)      Draw the free body diagram of the mass or rigid body when a positive displacement and velocity are given to it. Indicate all the active and reactive forces acting on the mass or rigid body.

4)      And finally apply Newton’s second law of motion to the mass or rigid body shown in free body diagram. Newton’s second law of motion can be state as follows :

Rate of change of momentum of a mass is equal to the force acting on it.

 If the mass of the system is displaced at a distance x(t) when acted upon by a resultant force in the same direction, Newton’s second law of motion gives

$$\overrightarrow{F}\left( t \right)=\frac{d}{dt}\left( m\frac{d\overrightarrow{x}\left( t \right)}{dt} \right)$$

If a mass m is constant, this equation reduces to:

$$\overrightarrow{F}\left( t \right)=m\frac{{{d}^{2}}\vec{x}\left( t \right)}{d{{t}^{2}}}=m\ddot{\vec{x}}$$

is the acceleration of the mass. Previous equation can be stated in words as

Resultant force on the mass = mass * acceleration

Where

$$\ddot{\vec{x}}=\frac{{{d}^{2}}\vec{x}\left( t \right)}{d{{t}^{2}}}$$

For a rigid body undergoing rotational motion Newton’s law gives

$$\vec{M}\left( t \right)=J\ddot{\vec{\theta }}$$

Where $$\vec{M}$$  is resultant moment acting on the body and 
$$\ddot{\vec{\theta }}={{{d}^{2}}\theta \left( t \right)}/{d{{t}^{2}}}\;$$  

  and θ are resulting angular displacement and angular acceleration respectively. Equation or represents the equation of motion of the vibrating system.

FIGURE 1 - SPRING MASS SYSTEM

The procedure is now applied to the undamped single-degree of freedom system shown in Fig 1. Here the mass is supported on frictionless rollers and can have translator motion in the horizontal direction. When the mass is displaced a distance +x from its equilibrium position the force in spring is kx, and the free-body diagram of the mass can be represented as shown in fig 1 the application of the 1 to mass m yields the equation of motion.

$$\begin{align} & F\left( t \right)=-kx=m\ddot{x}, \\ & m\ddot{x}+kx=0 \\ \end{align}$$

The initial conditions are:

$$\begin{align} & x(0)={{x}_{0}}, \\ & \dot{x}(0)={{{\dot{x}}}_{0}} \\ \end{align}$$

The assumed solution of previously derived differential equation can be written in the following form

$$x(t)=A\cos {{\omega }_{n}}t+B\sin {{\omega }_{n}}t$$

By applying initial conditions to assumed solution we will determine the constants of integration A and B. After we determine that we can write the assumed solution in the following form

$$x(t)={{x}_{0}}\cos {{\omega }_{n}}t+\frac{{{{\dot{x}}}_{0}}}{{{\omega }_{n}}}\sin {{\omega }_{n}}t$$

As we all know the

$${{\omega }_{n}}=\sqrt{\frac{k}{m}}$$

is the natural frequency of the system.

An alternate assumed solution can be written in the following form:

$$x\left( t \right)=A\sin \left( {{\omega }_{n}}t+\phi \right)$$

By expanding the previous equation using trigonometric identity we will get:

$$\begin{align} & \sin \left( a+b \right)=\sin a\cos b+\cos a\sin b \\ & x\left( t \right)=A\cos \phi \sin {{\omega }_{n}}t+A\sin \phi \cos {{\omega }_{n}}t \\ \end{align}$$

A and ϕ can be determined from the following expressions:

$$\begin{align} & A=\sqrt{x_{0}^{2}+{{\left( \frac{{{{\dot{x}}}_{0}}}{{{\omega }_{n}}} \right)}^{2}}}, \\ & \phi ={{\tan }^{-1}}\left( \frac{{{x}_{0}}{{\omega }_{n}}}{{{{\dot{x}}}_{0}}} \right) \\ \end{align}$$

The free vibration response of a one-degree-of- freedom system is shown in next  figure:

FIGURE 2 - Free vibration response of an undaped one-degree- of-freedom system

The initial conditions determine the energy initially present in the system. Potential energy is continually converted to kinetic energy and vice versa. Since energy is conserved, the system eventually returns to its initial state with its original kinetic and potential energies, completing the first cycle of motion. The subsequent motion duplicates the previous motion. The system takes the same amount of time to execute its second cycle as it does its first. Since no energy is dissipated from the system, the system executes cycles of motion indefinitely. 

            A motion which exactly repeats after some time is said to be period. A period is the amount of time it takes the system to execute one cycle. The frequency is the number of cycles the system executes in a period of time and is the reciprocal of the period.

$$\begin{align} & T=\frac{2\pi }{{{\omega }_{n}}}, \\ & f=\frac{{{\omega }_{n}}}{2\pi } \\ \end{align}$$

Example 1.1.

An engine of mass 500 kg is mounted on an elastic foundation of stiffness 7*10^5N/m. Determine the natural frequency of the system.

Solution:

The system is modeled as a hanging mass-spring system. The natural frequency will be determined using the following expression:

$${{\omega }_{n}}=\sqrt{\frac{k}{m}}=\sqrt{\frac{7\cdot {{10}^{5}}}{500}}=37.4rad/s=5.96Hz$$

Example 2.2

A wheel is mounted on a steel shaft (G=83*10^9 N/m^2) of length 1.5 m and radius 0.80cm. the wheel is rotated 5 degrees and released. The period of oscillation is observed as 2.3.s Determine the mass moment of inertia of the wheel. 

Figure 3 – Wheel mounted on torsional spring (shaft)

Solution:

The oscillations of the wheel about its equilibrium position are modeled as the torsional oscillations of a disk on a massless shaft, as illustrated in previous figure. The differential equation is:

$$\begin{align} & \sum{{{M}_{0}}}=I\ddot{\theta } \\ & -\frac{JG}{L}\theta =I\ddot{\theta } \\ & I\ddot{\theta }+\frac{JG}{L}\theta =0 \\ \end{align}$$

If we divide the previous equation by I we will get:

$$\ddot{\theta }+\frac{JG}{LI}\theta =0$$

The natural frequency of the system can be determined from:

$${{\omega }_{n}}=\sqrt{\frac{JG}{IL}}$$

The natural frequency can be also determined from:

$${{\omega }_{n}}=\frac{2\pi }{T}=\frac{2\pi }{2.3}=2.73rad/s$$

Thus the moment of inertia of the wheel is calculated form:

$$I=\frac{JG}{L\omega _{n}^{2}}=\frac{\frac{\pi }{2}{{\left( 0.008 \right)}^{4}}\left( 83\cdot {{10}^{9}} \right)}{1.5\cdot 2.73}=47.8kg{{m}^{2}}$$

Undaped free vibration

A vibration is the periodic motion of a body or system of connected bodies displaced from an equilibrium position. In general, there are two types of vibration Free and Forced. Free vibrations are vibrations that occur when no external force acts on the system. These vibrations are maintained by gravitational or elastic restoring forces, such as the swinging motion of a pendulum or the vibration of an elastic rod.  Forced vibration on the other hand are caused by an external period or intermittent force applied to the system. Free and forced vibrations can be damped or undamped. From idealistic point of view undamped vibrating systems can continue indefinitely. The reason why they go indefinitely is that the frictional effects are neglected in the analysis. Since in reality both internal and external friction forces are present, the motion of all vibrating bodies is actually damped.

The simplest type of vibrating motion is undamped free vibration, is represented by a model which consists of block and a spring. The model is shown in next figure.

FIGURE 1 - Undamped free vibrational system

Vibration motion occurs when the block is released from a displaced position x so that the spring pulls on the block. We say that block is oscillating around its equilibrium position which means that the block will attain a velocity such that it will proceed to move out of equilibrium when x= 0. One thing is very important in this theoretical analysis and that is the assumption that the surface is smooth. Smooth surface means that there is no friction between block and the surface. So block will oscillate without any friction.

The time dependent path of motion of the block can be determined by applying the equation of motion to the block when it is in the displaced position x. But before you derive the equation that describes the motion we need to make the Free Body Diagram. This method is used to draw the external and effective forces which acts on the body and in this case it’s the block of mass m. when you applied the method of free body diagram you can then apply the Newton second law from which we will derive the equation that describes the motion or in this case the vibration of the analyzed system.

The elastic restoring force F=kx is always directed toward the equilibrium position, whereas the acceleration a is assumed to act in the direction of a positive displacement. Since

We have

$$\begin{align} & \xrightarrow{+}\sum{{{F}_{x}}}=m{{a}_{x}} \\ & -F=ma \\ & -kx=m\ddot{x} \\ & m\ddot{x}+kx=0 \\ \end{align}$$

The previous equation describes the simple harmonic motion. The acceleration is proportional to the block displacement. Now we need to derive the expression for natural frequency. If we divide the previous equation with m we will get.

$$\begin{align} & m\ddot{x}+kx=0/:m \\ & \ddot{x}+\frac{k}{m}x=0, \\ & \ddot{x}+\omega _{n}^{2}x=0, \\ \end{align}$$

The Constant in the previous equation is called the natural frequency of the system, and in this case:

$${{\omega }_{n}}=\sqrt{\frac{k}{m}}$$

Now the solution to the differential equation which describes the motion of the analyzed system can be written in the following form:

$$x=A\sin {{\omega }_{n}}t+B\cos {{\omega }_{n}}t$$

If we look at the equation which is the solution of the system we will see that A are B are the unknowns and they represent two constants of the integration.

To prove that the previous equation is the solution of the system we need to insert it into the differential equation but before we do that we need to derive the first and the second derivation of the solution.

$$\begin{align} & \dot{x}=A{{\omega }_{n}}\cos {{\omega }_{n}}t-B{{\omega }_{n}}\sin {{\omega }_{n}}t, \\ & \ddot{x}=-A\omega _{n}^{2}\sin {{\omega }_{n}}t-B\omega _{n}^{2}\cos {{\omega }_{n}}t. \\ \end{align}$$

By substituting the second derivation of solution and solution of the system into the differential equation of the system we get:

$$\begin{align} & \ddot{x}+\omega _{n}^{2}x=0, \\ & -\omega _{n}^{2}\left( A\sin {{\omega }_{n}}t+B\omega _{n}^{2}\cos {{\omega }_{n}}t \right)+\omega _{n}^{2}\left( A\sin {{\omega }_{n}}t+B\cos {{\omega }_{n}}t \right)=0 \\ & 0=0 \\ \end{align}$$

From the previous expression we can see that the differential equation is satisfied for the proposed solution.

All that is left is to determine the constants of the integration. To determine the constants of the integration we need to apply the boundary condition.

For t=0, the x=x1 which means that at the beginning the system is at rest. By applying this boundary condition to the solution of the differential equation, we get:

$$\begin{align} & x=A\sin {{\omega }_{n}}t+B\cos {{\omega }_{n}}t \\ & {{x}_{1}}=A\sin 0+B\cos 0 \\ & B={{x}_{1}} \\ \end{align}$$

For t=0, the v=v1 and by applying that to the first derivation of solution we get:

$$\begin{align} & \dot{x}=A{{\omega }_{n}}\cos {{\omega }_{n}}t-B{{\omega }_{n}}\sin {{\omega }_{n}}t, \\ & {{v}_{1}}=A{{\omega }_{n}}\cos 0-B{{\omega }_{n}}\sin 0, \\ & A=\frac{{{v}_{1}}}{{{\omega }_{n}}} \\ \end{align}$$

So now we have determined the constants of integration. We can write the complete solution of the system in the following form.

$$x=\frac{{{x}_{1}}}{{{\omega }_{n}}}\sin {{\omega }_{n}}t+{{x}_{1}}\cos {{\omega }_{n}}t$$

The solution can be written in shorter form if we express the constants of the integration in following form:

$$\begin{align} & A=C\cos \phi , \\ & B=C\sin \phi . \\ \end{align}$$

The C and ϕ are constants of integration and they also need to be determined from boundary conditions. If we substitute the A and B into the solution we will get:

$$x=C\cos \phi \sin {{\omega }_{n}}t+C\sin \phi \cos {{\omega }_{n}}t$$

By applying trigonometric transformation:

$$\sin \left( \theta +\phi \right)=\sin \theta \cos \phi +\cos \theta \sin \phi $$

We will get:

$$x=C\sin \left( {{\omega }_{n}}t+\phi \right)$$

If this equation is plotted on an x versus the axis, the graph is obtained. The maximum displacement of the block from its equilibrium position is defined as the amplitude of vibration. From the figure the amplitude is C. The angle is called phase angle since it represents the amount by which the curve is displaced from the origin when t=0.

$$C=\sqrt{{{A}^{2}}+{{B}^{2}}}$$

Phase angle can be determined from the following expression:

$$\phi ={{\tan }^{-1}}\left( \frac{B}{A} \right)$$

Note that the sine curve completes one cycle in time or:

$$\tau =\frac{2\pi }{{{\omega }_{n}}}$$

Finally the frequency f is defined as the number of cycles completed per unit of time, which is the reciprocal of the period; that is

$$f=\frac{1}{\tau }=\frac{{{\omega }_{n}}}{2\pi }$$

The frequency is expressed in cycles/s. This ratio of units is called a hertz where 1 Hz = 1 cycle/s or 2PI rad/s. When the body or system of connected bodies is given initial displacement from its equilibrium position and released, it will vibrate at natural frequency. Provided the system has a single degree of freedom, that is, it requires only one coordinate to specify completely the position of the system at any time, then the vibration motion will have the same characteristics as the simple harmonic motion of the block and the spring just presented.