5.3 Free vibration of a damped, single degree of freedom, linear spring mass system.
We analyzed vibration of several conservative systems in the preceding section. In each case, we found that if the system was set in motion, it continued to move indefinitely. This is counter to our everyday experience. Usually, if you start something vibrating, it will vibrate with a progressively decreasing amplitude and eventually stop moving.
The reason our simple models predict the wrong behavior is that we neglected energy dissipation. In this section, we explore the influence of energy dissipation on free vibration of a spring-mass system. As before, although we model a very simple system, the behavior we predict turns out to be representative of a wide range of real engineering systems.
5.3.1 Vibration of a damped spring-mass system
The spring mass dashpot system shown is released with velocity u0 from position s0 at time t = 0 . Find s(t).
Once again, we follow the standard approach to solving problems like this
(i) Get a differential equation for s using F=ma
(ii) Solve the differential equation.
You may have forgotten what a dashpot (or damper) does. Suppose we apply a force F to a dashpot, as shown in the figure. We would observe that the dashpot stretched at a rate proportional to the force
One can buy dampers (the shock absorbers in your car contain dampers): a damper generally consists of a plunger inside an oil filled cylinder, which dissipates energy by churning the oil. Thus, it is possible to make a spring-mass-damper system that looks very much like the one in the picture. More generally, however, the spring mass system is used to represent a complex mechanical system. In this case, the damper represents the combined effects of all the various mechanisms for dissipating energy in the system, including friction, air resistance, deformation losses, and so on.
To proceed, we draw a free body diagram, showing the forces exerted by the spring and damper on the mass.
Newton’s law then states that
This is our equation of motion for s.
Now, we check our list of solutions to differential equations, and see that we have a solution to:
We can get our equation into this form by setting
As before, ωn is known as the natural frequency of the system. We have discovered a new which is called the damping coefficient. It plays a very important role, as we shall see below.
Now, we can write down the solution for x:
Critically Damped System
where is known as the damped natural frequency of the system.
In all the preceding equations.
are the values of x and its time derivative at time t=0.
These expressions are rather too complicated to visualize what the system is doing for any given set of parameters. Instead,
contains a Java Applet that can be used to show animations along with graphs of the displacement. You can use the sliders to set the values of either m, k, and λ (in this case the program will calculate the values of ζ and ωn for you, and display the results), or alternatively, you can set the values of ζ and ωn directly. You can also choose values for the initial conditions x0 and v0 . When you press ‘start’ the applet will animate the behavior of the system, and will draw a graph of the position of the mass as a function of time. You can also choose to display the phase plane, which shows the velocity of the mass as a function of its position, if you wish. You can stop the animation at any time, change the parameters, and plot a new graph on top of the first to see what has changed. If you press `reset, all your graphs will be cleared, and you can start again.
Try the following tests to familiarize yourself with the behavior of the system
We now know the effects of energy dissipation on a vibrating system. One important conclusion is that if the energy dissipation is low, the system will vibrate. Furthermore, the frequency of vibration is very close to that of an undamped system. Consequently, if you want to predict the frequency of vibration of a system, you can simplify the calculation by neglecting damping.
5.3.2 Using Free Vibrations to Measure Properties of a System
We will describe one very important application of the results developed in the preceding section.
It often happens that we need to measure the dynamical properties of an engineering system. For example, we might want to measure the natural frequency and damping coefficient for a structure after it has been built, to make sure that design predictions were correct, and to use in future models of the system.
You can use the free vibration response to do this, as follows. First, you instrument your design by attaching accelerometers to appropriate points. You then use an impulse hammer to excite a particular mode of vibration, as discussed in Section 5.1.3. You use your accelerometer readings to determine the displacement at the point where the structure was excited: the results will be a graph similar to the one shown below.
We then identify a nice looking peak, and call the time there t0 , as shown.
The following quantities are then measured from the graph:
1. The period of oscillation. The period of oscillation was defined in Section 5.1.2: it is the time between two peaks, as shown. Since the signal is (supposedly) periodic, it is often best to estimate T as follows
where tn is the time at which the nth peak occurs, as shown in the picture
2. The Logarithmic Decrement. This is a new quantity, defined as follows
where x(tn) is the displacement at the nth peak, as shown. In principle, you should be able to pick any two neighboring peaks, and calculate δ . You should get the same answer, whichever peaks you choose. It is often more accurate to estimate δ using the following formula
This expression should give the same answer as the earlier definition.
Now, it turns out that we can deduce ωn and ζ from T and δ , as follows.
where Hence, the period of oscillation is
where we have noted that tn+1 = tn + T
Fortunately, this horrendous equation can be simplified greatly: substitute for T in terms of ωn and ζ , then cancel everything you possibly can to see that
Finally, we can solve for ωn and ζ to see that
Note that this procedure can never give us values for k, m or λ. However, if we wanted to find these, we could perform a static test on the structure. If we measure the deflection d under a static load F, then we know that
Once k had been found, m and λ are easily deduced from the relations