Ronald J. Anderson - Introduction to Mechanical Vibrations

A simple differentiation with respect to yields

(1.41)

Lagrange's Equation can then be written as

(1.42)

where now represents the generalized force corresponding to all externally applied forces that are neither conservative nor linear viscous in nature. Finally, we can transfer the Rayleigh Dissipation term to the left‐hand side and write Lagrange's Equation with dissipation as

(1.43)

Equilibrium solutions of the equations of motion are those where the degrees of freedom assume values which cause their first and second derivatives to go to zero. Under these conditions, there will be no tendency for the values of the degrees of freedom to change and the system will be in an equilibrium state .

Some equilibrium states are stable and some are unstable and, inevitably, systems are either in a stable equilibrium state or trying to get to one . The study of small motions around a stable equilibrium state is called Vibrations .

We start by considering the simple pendulum 5 shown in Figure 1.5. Using the angle as the single degree of freedom, the equation of motion is

(1.44)

Figure 1.5 A simple pendulum.

Once started in motion the pendulum will swing about the point of connection to the ground. In the case of the simple pendulum there is no mechanism for removing energy from the system as it swings (i.e. no friction or other forces that do work) so the motion, once started, will persist.

The motion will depend on the way in which it is started. That is, if the pendulum is rotated to some arbitrary starting angle, , and released from rest, it will swing through the position where and will eventually return to where it started before reversing and starting the cyclic motion over again. If the pendulum is stopped and returned to and then released, not from rest but with an initial velocity, the resulting motion will be different and the pendulum will pass through when it returns. The motion will, however, still be cyclic.

The question we ask now is Are there initial values of where the pendulum can be released from rest and remain stationary? These are the equilibrium states.

Consider Equation 1.44under the conditions that there is an initial angle and there is no angular velocity (i.e. so that does not change with time) and, further, that there is no angular acceleration (i.e. so that does not change with time and thus there will never be a change in ). This is an equilibrium position and Equation 1.44becomes the equilibrium condition .

(1.45)

Since , , and are never zero, this can only be satisfied by:

The total range of is . In this range, only (the pendulum hangs vertically downward) and (the pendulum stands upright) satisfy the requirements. These are the two equilibrium states for the pendulum.

There are formal methods for testing the stability of the equilibrium states but that we leave to courses on control systems. It is sufficient for us to be able to see that the state where the pendulum stands upright is unstable and the pendulum will try to get to the stable equilibrium position where .

The vibrations question is What will be the response of the system for small motions away from the stable equilibrium condition where ?