1 Oscillations SHM review –Analogy with simple pendulum SHM using differential equations –Auxiliary Equation Complex solutions Forcing a real solution The damped harmonic oscillator –Equation of motion –Auxiliary equation –Three damping cases Under damping –General solution Over damping –General solution –Solution in terms of initial conditions Critical Damping –Break down of auxiliary equation method and how to fix it –General solution –Solution in terms of initial conditions –Over damping as ideal damping Phase diagrams –Un-damped phase diagram –Obtaining phase equations directly –The under-damped logarithmic spiral –Critical damping example Harmonic oscillations in two dimensions –Lissajous figures

2 What will we do in this chapter? This is the first of several lectures on the the harmonic oscillator. We begin by reviewing our previous solution for SHM and use similar techniques to solve for a simple pendulum. We next solve the SHM using the auxiliary equation technique from linear differential equation theory. This allows us to extend our treatment to the case of a damped harmonic oscillator with a damping force proportional to drag. Three damping cases are considered: under damped, over damped, and critically damped. The critically damped case -- besides being very practical -- brings a new wrinkle to the auxiliary equation technique. We discuss the phase diagram which is a plot of trajectories in a phase space consisting of p and x. Several methods for computing these trajectories are discussed and the under damped, un- damped, and critically damped examples are drawn. We conclude with a brief discussion of harmonic motion in two dimensions and Lissajou figures.

3 Simple harmonic motion We already are familiar with this problem applied to the mass spring problem. We write the force law and potential and the solution which we obtained using energy conservation. We also show that the simple pendulum put in small oscillation has an analogous potential and kinetic energy expression and hence will have the same formal solution as the spring-mass system.

4 SHM solution by DE methods Complex numbers snuck in our solution! Indeed B 1,2 should be taken as complex numbers as well. But x(t) must be real? One way around this problem is to insist that x is real by demanding x* = x This really means that rather than having arbitrary B 1 and B 2 we really only have one arbitrary number say B 1. Fortunately B 1 is complex and thus has an arbitrary amplitude and phase. We need two arbitrary real quantities in order to match the initial position and phase.

5 DE methods (continued) If we had chosen B 1 =-i  exp(-i  ) we would have obtained an equally valid solution x = 2  sin(  t -  ). As another example we could include a linear drag force acting on the mass along with the spring.

6 Damping cases Solutions will depend quite a bit on relationship between  and . The under damped case will have complex auxiliary roots and will have oscillatory behavior. The over damped case will have real roots and thus have a pure exponential time evolution. The critically damped case, with a single root, has some non-intuitive aspects to its solution. Under damping The under damped oscillator has two constants (phase and amplitude) to match initial position and velocity. The solution dies away while oscillating but with a frequency other than  = (k/m) 1/2.

7 An under-damped case Here is a plot of x(t) for the under-damped case for three different choices of phase. We have chosen the damping coefficient  to be 1/4 of the effective frequency    so you can see a few wiggles before the oscillation fades out.

8 Over damped case In this case both the B 1 and B 2 coefficients must be real so that x is real. The two terms are both there to allow us to match the initial position and velocity boundary conditions. Since    both 1 =   and 2 =   are both positive and thus both terms correspond to exponential decay. We can re-arrange the blue form to match initial conditions. The initial condition form is easy to confirm by expanding to 1st order in t.

9 Critical damping To get a real x, B 3 must be real. But this can’t possibly be right! With only 1 real coefficient we can’t match the initial position and velocity! Appendix C of M&T says in the case of degenerate roots, the full solution is of the following form which can be confirmed by direct substitution. In terms of initial conditions : It is interesting to note that the critically damped case actually dies away faster than the over damped case. Consider the case where v 0 =0. The critically damped case will fall off according to exp(-  t) The over damped case will have a exp[-(    t] piece which dies off faster than the critically damped case. But it will also contain a exp[-(    t] piece which dies off slower than the critically damped case. For many applications: vibration abatement, shocks, screen door dampers, one strives for critical damping.

10 Example of Critical Damping We use solutions with x 0 =1 and v 0 =0 and consider the case of critical damping and two cases of over damping. The critical damped case dies out much faster and the over damped case dies out more slowly as the over damping ratio increases.

11 Phase Diagram It is fashionable to view the motion of mechanical systems as trajectories in a phase space which is a plot of p (or v) versus x. Oscillations are a perfect example of such plots. A first “phase diagram” is for an un-damped oscillator. Since we have a purely position dependent one dimensional force we know that energy is conserved. This curve is an ellipse with an area monotonic in energy. Without damping the particle will execute endless elliptical orbits of fixed energy in this phase space x p In Newtonian mechanics, we specify the initial conditions as x(0) and p(0) (or v(0)). This corresponds to one point in the phase diagram which specifies the initial energy or the phase space ellipse. We note that with this choice of coordinates, we have a clockwise phase space orbit. This is because of the negative sign in the equation of motion: Essentially this means that p must decrease when x is positive and must increase when x is negative. The particular phase space orbit will depend on the nature of the forces but in general we know that no two phase space orbits can intersect. Otherwise several motions would be possible for the same set of initial conditions.

12 Phase Diagrams (continued) x p We show the phase diagram for an under damped oscillation. Again we have a clockwise motion but this time the energy continuously decreases with time and eventually disappears. This sort of curve is called a logarithmic spiral. We could in principle obtain this directly from out solution. The text shows a way of getting the spiral by a variable transformation. One can construct the phase diagram directly from x(t) and its derivative or (in the absence of dissipation) from the conservation of energy. Often the phase diagram can be obtained through a clever separation of variables:

13 Critically damped phase diagram What’s wrong with this picture? It is very straight-forward to draw the phase diagram once one has x(t) and its derivative. We plot plot x and v for 50 times for 3 sets of initial conditions. You can easily visualize the motion including the maximum displacement and velocity and retrograde motion Whoops I messed it up. How did I know?

14 Harmonic motion in two directions Lissajous figure is a plot of y versus x as t is varied. To the right is a Lissajous figure for the case A x = A y and  x =  y and  x =0 for 4 different  y phases. With these conventions, the figure maximally opens up into a circle when  y =0. Lissajous figure’s are often a great way to measure a phase difference on an oscilloscope.

15 Lissajou figure with unequal frequency

16 A cool trig identity

17 Lissajous figures with different frequencies To the right is are Lissajous figures for the case A x = A y and  y = 3  x and  x =0 for 3 different  y phases. The figure maximally opens up when  y =90.

18 Lissajous with irrational frequency ratios To the right are a Lissajous figure for the case A x = A y, and  x =  y =0 with the indicated frequency ratios When the frequency ratio is not a simple fraction (or irrational) the path does not close and finally fills up the whole screen. How would this figure look if ?