## Single Degree of Freedom Systems: Undamped Single Degree of Freedom System

A single degree of freedom (SDOF) system is one for which only a single coordinate is required to completely specify the configuration of the system. (This is a suitable working definition for now.) There are typically many possible choices for the coordinate to be used, although some are more natural than others. With a single degree of freedom system, we get one governing differential equation of motion. The specifics of the equation depend on the exact nature of the problem.

We begin our study of vibrations by considering free vibrations of a system. These are the easiest to deal with and understanding these systems is fundamental to understanding more advanced topics in vibration. To keep things as simple as possible, we begin studying undamped free vibrations. The effects of damping will be added in subsequent sections.

Typically in a free vibration problem we have a mass which is displaced from its equilibrium position and then released either from rest or with some initial velocity. Due to the restoring forces in the system, the mass will tend to move back to its equilibrium position but due to its own inertia it will overshoot the equilibrium postion. At some instant the mass will stop momentarily and start to return again to the equilibrium position, and the process will repeat itself (indefinitely if there is no damping). Our goal here is to determine and understand the mathematics used to describe this motion.

## Undamped Single Degree of Freedom Systems

Consider the system shown in Figure 2.1 which is probably the simplest possible vibratory system. Here a body of mass is attached to a rigid support by a spring with constant stiffness . The spring has an unstretched length . This is a SDOF system since we need only one coordinate (location of the mass) to completely specify the configuration of the system. There are many possible coordinates we could choose.

Figure 2.2 shows three possible coordinates, however obviously many others could be chosen as well. In this figure is measured from the *equilibrium position* while and are measured from other fixed locations. While all of these are valid choices for the coordinate, we will see from our study of vibrations that there are advantages to measuring displacements from the equilibrium position, and so we will chose to do so here.

To obtain the equation of motion we will use Newton’s Laws. To do so we consider the system when it is displaced in the **positive coordinate direction** and draw a FBD/MAD of the system in this configuration. We additionally **assume that the velocity and acceleration of the system are also positive (i.e. in the positive coordinate direction)** at this instant. By following this procedure, the signs of all of the terms in our governing equations will be correct and consistent. Note that the positive coordinate direction is determined by the choice of coordinate. In Figure 2.2, the positive direction for and is to the right while for it would be to the left. Any one of these is fine and will work, but once you have decided on your coordinate (and therefore the positive coordinate direction), you must stick with that choice in all that follows.

Figure 2.3 shows the FBD/MAD for our system using as the coordinate. Applying Newton’s Laws we get:

or

(2.1)

Dividing by results in:

and defining

(2.2)

allows equation 2.1 to be written as

(2.3)

Equation 2.3 is the *standard form* of the equation of motion for the undamped free vibrations of SDOF system. If a different coordinate had been used it would simply replace in equation 2.3. For example if we had chosen as the coordinate, the equation of motion would have the form

**EXAMPLE**

Determine the equation of motion for a simple pendulum of length and mass as shown below. Use as the coordinate and assume that only small motions occur.

**EXAMPLE**

The compound pendulum shown below has a mass and centroidal moment of inertia . The distance from the support to the center of mass is . Determine the equation of motion for this pendulum. Use as the coordinate and assume that only small motions occur.

### Solution of Equation 2.3

Equation 2.3 is a second order linear ordinary differential equation (ODE) with constant coefficients. Accordingly, we assume a soltuion of the form:

(2.4)

which when substituted into Equation 2.3 results in

or

which has two roots given by

As a result the solution to equation 2.3 becomes

(2.5)

(Note that and must be allowed to be complex numbers here.) To proceed we make use of Euler’s identity

to write

Defining*

the solution in equation 2.5 can be written as

(2.6)

* Note: Since the response of the system in equation 2.6 must be real, and must be real which implies that and are complex conjugates.

Equation 2.6 is the general solution to equation 2.3 and is the free undamped vibration response of a SDOF system. The response is simple harmonic motion which occurs at a frequency which is called the *natural frequency* of the system. For the simple spring mass system we are considering, we see from equation 2.2 that

For a general SDOF system, recall that the standard form of the equation of motion is

The natural frequency, which is a fundamental property of a vibrating system, can be determined *by inspection* if the equation of motion is expressed in this standard form. Note that in equation 2.3 is *always* in units of rad/s.

There are two unknown constants and in equation 2.6. We determine their values typically by considering the initial conditions for the system. Since we have a two unknown constants, we need to specify two initial conditions. These are typically the starting position and velocity of the mass. For example, at a start time we could have

(where or could be negative). Equation 2.6 then gives

so that

Differentiating equation 2.6

and using the second initial condition leads to

or

The response of the system is therefore given by

(2.7)

which

*completely*specifies the response of the system for all time .

### Other forms of the General response

While equation 2.7 represents the response of the SDOF undamped system, the solution may be expressed in other equivalent forms:

where

where

These relationships are shown in Figure 2.4. Note that

### Period and Frequency

The period is the time required for the harmonic motion to repeat itself (i.e. to complete one cycle). is related to the natural frequency by

(2.8)

so that

Another common unit for frequency measurement is Hertz (Hz) or cycles per second. Since there are radians per complete cycle, we note that

(2.9)

where is the natural frequency expressed in Hertz. Similar to Equation 2.8 we note that

### Rotating Vector Representation

We can also think of the response of the undamped SDOF system in terms of rotating vectors (another way of thinking about a subject is often helpful). Consider two vectors, with magnitudes and , fixed apart rotating CCW with an angular velocity as shown in Figure 2.5.

The (vector) sum of and is another vector with a magnitude

which also rotates at a rate . The fixed angles and can be seen to be

Now, guided by our previous results

we see that if we let

then the position can be interpreted as the *vertical component of the rotating vector* . As can be seen from Figure 2.5, this vertical component can be expressed as

which agrees with our previous results (noting that ).

### Energy Methods

For conservative systems we can also use an energy approach to obtain the equations of motion. Since the total mechanical energy is conserved in these systems, the sum of the kinetic energy and the potential energy will be a constant

(2.10)

where the actual value of the constant can be evaluated, if necessary, from the initial conditions. Differentiating Equation 2.10 with respect to time gives

which leads directly to the equation of motion for the system.

Consider the simple spring-mass system in Figure 2.1 with configuration described by the coordinate measured from the equilibrium position. The potential energy stored in the spring when the mass is at some location is given by

Similarly, since represents the speed of the mass, the kinetic energy in the system is

Since this system is conservative, the total mechanical energy can be expressed as

and differentiating with respect to time yields

(2.11)

Equation 2.11 must hold for all times and for all values of . The only way this can be satisfied (since for all times ) is to require that

at all times which is the equation of motion for the system obtained previously.

Note that if we have a conservative system and we assume harmonic motion at the start, we can obtain the natural frequency somewhat more directly. For such a system, the mechanical energy is continually converted between potential and kinetic energy. At the extremes of the motion, when the mass momentarily stops and changes direction, there is no kinetic energy in the system so that all of the energy is potential energy. Hence potential energy is at a maximum value . Similarly, when the mass is moving past the equilibrium position, the potential energy is a minimum while the kinetic energy has a maximum value . If we make the assumption that the potential energy of the system is zero at the equilibrium position, then these two maximum values for the energy will have the same value, i.e.

(2.12)

Again, consider the simple spring-mass system (with coordinate measured from the equilibrium position). If we assume the system undergoes simple harmonic motion given by

then the maximum displacement is so that

Similarly, the speed is given by

which has a maximum value of so that

Substituting these expressions into Equation 2.12 gives

which can be solved for as

which is the natural frequency of the system found previously.

This approach is known as *Rayleigh’s Method* and we will encounter variations of this approach later on in the course. Note that this approach is limited to conservative systems which have zero potential energy at the equilibrium position.

**EXAMPLE**

Using an energy approach, determine the equation of motion for a simple pendulum of length and mass as shown below. Use as the coordinate and assume that only small motions occur.

I have a task. The given equation is same as equation 2.1, initial velocity is 1m/s and displacement is 1m. How to find out the displacement and velocity during the first 5 seconds?