By Hamilton W.R.

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Vectoria leaves it as a problem for you to obtain the analytic solution for the critically damped case. Unfortunately, Vectoria must leave us for now, since her cell phone has just rung. Mike's plane has just landed and she's o® to the Metropolis International Airport to pick him up after he clears immigration and customs. PROBLEMS: Problem 1-4: Critical damping Given ! = 1, what ¯ value corresponds to critical damping of the SHO? Make a phase-plane portrait for this case using the DEplot command and the initial condition x(0) = 1, x(0) _ = 0.

Clearly, stable focal points occur for p > 0, and unstable focal points for p < 0. The underdamped SHO is an example of a stable focal point. CHAPTER 2. PHASE-PLANE ANALYSIS 50 For q > 0, now consider the case p = 0. , ¸1;2 = § i q, and the general solution is of the undamped oscillatory form u = A cos(q t) + B sin(q t). The solution is characteristic of trajectories in the vicinity of a vortex point. Finally, we examine the situation in which q < 0. Independent of the value or sign of p, the roots ¸1 and ¸2 are real but of opposite signs.

9 for the second-order ODE de, given the _ = 0. The time range is from t = 0 to 50, and initial condition3 x(0) = 1, x(0) the time step size in the underlying numerical scheme is taken to be 0:05. 9 decreases in amplitude in an oscillatory manner, again characteristic of the underdamped SHO. Finally, since the SHO equation de is linear with constant coe±cients, an analytic solution can be easily obtained for x(t) using the dsolve command. > dsolve(fde,x(0)=1,D(x)(0)=0g,x(t)); ! Ã p Ã p t t 3 11 t 3 11 t 1 p (¡ (¡ ) ) + e 10 cos 11 e 10 sin x (t) = 33 10 10 3 Note that the derivative condition x(0) _ = 0 is entered as D(x)(0)=0, where D is the di®erential operator.