Physics Department LAB A - 120 Experiment 12 Damped Harmonic Motion References: Daniel Kleppner and Robert Kolenkow, An Introduction to Mechanics, McGraw -Hill 1973 pp. 414-418. Equipment: Air track, glider, blower, springs, pan, liquids (water, oil, glycerol),photogate timers, motion sensor, Science workshop interface, computer. Theory: In Expt. 11, it was assumed that only negligible frictional forces acted upon the body during its motion. This is often a very good approximation over a limited time, although eventually the initial energy is dissipated and the body comes to rest. In other cases, however, frictional effects cannot be neglected even over a short time; indeed, their presence may be desirable, as in the spring-shock absorber system. It often happens, or is arranged, that the dissipative force is proportional in magnitude to the body s speed (or the torque to the angular velocity, in rotational harmonic motion). Then we speak of damped harmonic motion, because the vibrations are quickly damped to negligible amplitude. An example is studied in this experiment. A glider of mass m, on an air track is fastened by two identical springs to fixed end blocks. Rigidly attached to the glider are two rods or arms (Fig. 1) that can dip into a tray of viscous liquid. Coil spring Glider Coil spring Block Viscous liquid Fig. 1 140618 1
The resulting drag force f can be represented fairly well as F = -β v (1) with v the instantaneous speed of the rods moving through the liquid; β is a constant, dependent upon the specific liquid and the geometry of the rods. (If the liquid is not very viscous e.g., water f is more nearly proportional to v 2 this is also an important case, but mathematically more difficult.) As was seen in Expt. 11, the springs exert a net force -kx on the glider when it is displaced a distance x from its equilibrium position x = 0; k is the net spring constant. Newton s second law becomes mx = kx βx (2) where the dots indicate time derivatives. This linear second-order differential equation can be solved exactly. The solutions take three different forms, depending on the relative sizes of k/m and β 2 /4m 2. 1. If k/m exceeds β 2 /4m 2, the motion is said to be underdamped, because vibrations still occur, although with steadily decreasing amplitude. The solution of Eq. 2 is then x = e!!" x! cosωt +!!!!!!! sinωt (3) γ = β / 2m (4) The angular frequency ω is given by ω = [(k/m) - γ 2 ] 1/2 (5) Here x o and v o are the values of x and dx/dt, respectively, at t = 0. Equation 3 is easily converted to the form x = A e - γ t cos(ω t - φ ) (6) in which A is related to the amplitude of the motion and -φ is the initial phase [write A cos φ = x 0, A sin φ = (v 0 +γ x 0 ) /ω]. Evidently A and φ are functions of x 0, v 0, m, k, and 140618 2
β. Figure 2 shows the behavior of x vs. t, starting arbitrarily at some maximum positive displacement (not necessarily at t = 0). Since ω is constant, recurring maxima are nearly equally spaced along the time axis, at intervals Δ t=2π/ω. From Eq. 6 we see that any one maximum value of x - when cos (ω t - φ ) = 1 is lower than its immediate predecessor by the fraction e γδt = e 2π(γ ω ) (7) Displacement x Envelope curve Envelope curve Fig. 2 Hence by timing the motion and using either Eq. 5 or Eq. 7 we can find the value of β; this is important because β strongly affects the whole motion and yet is often hard to estimate theoretically. 2 If, in contrast to the preceding analysis, γ 2 = β 2 / 4m 2 exceeds k/m, the motion is said to be overdamped. When released with any initial x 0 and v 0 at t = 0, the body goes through no oscillations; it returns, ever more slowly, toward the equilibrium point. The solution of Eq. 2 becomes x = (!!!!) ω! γ x! v! e!(!!!! )! + ω! + γ x! + v! e!!!!!! (8) 140618 3
γ is still given by Eq.. 4; ωʹ is ωʹ = [γ 2 - k/m] 1/2 (9) Since there is no longer a definite oscillation to be timed, the value of β is in general more difficult to determine than in underdamped motion 3 The dividing line between underdamped and overdamped motion is set by k/m = γ 2 = β 2 /4m 2 (10) This condition is called critical damping. In principle, Eq. 10 provides the easiest way of determining β, when k and m are known; in fact, adjusting the system exactly to critical damping (e.g., by altering the value of m) is hard because of the difficulty in recognizing just when oscillation disappears. Nevertheless, achieving critical damping is important in some practical cases. The solution of Eq. 2 for this case differs from both Eq. 3 and Eq. 8. It can be found in texts on mechanics; we omit it because in this experiment it would be of little use. Procedure: Weigh the glider and then connect it, on the air track, as shown in Fig. 1. Use a flexible pair of identical springs, set the rods so that they do not dip into the viscous liquid. Turn on the air blower, and displace the glider from the equilibrium point, NOT MORE THAN 20 CM. Release it; determine ω as in Expt. 11, by timing and counting. β is essentially zero in this case. Repeat, using stiffer springs. Observe that this procedure would allow calculation of the net spring constant k in each case, but in fact only k/m enters the equations henceforth. Retaining the stiffer springs, adjust the rods to dip into the viscous liquid. Displace the glider NOT MORE THAN 10 CM; release it, and observe the motion in repeated trials. By timing and counting, determine the new value of w. Using the scale attached to the air track, determine x at two or three successive maximum displacements of the glider, on the same side of x = 0. Do not include the original displacement from which the glider was released. (Why? Is A in Eq. 6 equal to x 0 even when v 0 is zero?) Calculate β from both Eqs. 5 and 7. Which calculation do you consider more reliable? 140618 4
Without altering the setting of the rods (so that β remains unchanged), return to using the more flexible springs. WITH AN INITIAL DISPLACEMENT OF NOT OVER 10 CM, observe the motion in repeated trials. Does it appear to be overdamped? (It may or may not be possible to achieve overdamping.) Test your opinion with previously determined values of β and m, and of k/m for these springs. Observe and record the qualitative effects of adding mass to the glider, changing nothing else. Should this process approach or recede from overdamping? If by trial and by calculation it appears that overdamping was achieved, try by changing the extra mass on the glider in small steps to reach critical damping. Then calculate β from Eq. 9, remembering that k/m changes with total mass. Computer Data Acquisition with Science Workshop Interface: An alternative method to do this experiment is to use a motion sensor, Pasco Capstone software, and the Pasco ScienceWorkshop 750 interface. The software will allow you collect position vs time data. You will essentially be able to produce the data in figure 2 in real time. From your position and time data (which you can save in Excel files, or in your favorite spreadsheet program), you should be able to easily determine the period of oscillation. (Advanced: You may also be able to curve fit the data and determine ω, γ, and β, your TA may be able to help you with this.) Questions: 1. Derive values of A and tan φ, for Eq. 6. 2. Between Eqs. 6 and 7 it was stated that successive maxima of x occur nearly at equal time intervals. Why nearly? (What else besides the cosine term in Eq. 6 helps fix such a maximum?) What condition on γ makes nearly approach exactly? 3. When very slight overdamping exists ωʹ is still positive but very small show that Eq. 8 is not a satisfactory solution of Eq. 2 as, at any fixed t, ωʹ 0. (Factor e iγt ; for e ±ω t use the first three terms of its power series expansion, (1+ 140618 5
±ωʹ t + (ωʹ t) 2 /2). Show that one of the essential initial conditions, v 0, disappears from Eq. 8 as ωʹ 0.) This result indicates why still another solution of Eq. 2 must be found, in the special case of critical damping a solution that will satisfy Eq. 2 and will retain v 0. 140618 6