Consider a pendulum that is released from rest from an initial displacement of 80

Chapter 3, Problem 25

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Consider a pendulum that is released from rest from an initial displacement of 80 radians. Solving the linear model (7) subject to the initial conditions 8(0) = 80, 8'(0) = 0 gives 8(t) = 80 cos Viflt. The period of oscillations predicted by this modelisgivenbythefamiliarformulaT= 27T/Vifl = 27TWg. The interesting thing about this formula for T is that it does not depend on the magnitude of the initial displacement 00 In other words, the linear model predicts that the time that it would take the pendulum to swing from an initial displacement of, say, 00 = 7r/2 (= 90") to -7r/2 and back again would be exactly the same time to cycle from, say, 00 = Tr/360 ( = 0.5") to -Tr/360. This is intuitively unreasonable; the actual period must depend on 00 for 0 :::;; t s; 2. As in 24, if t1 denotes the first time the pendulum reaches the position OP in Figure 3.11.3, then the period of the nonlinear pendulum is 4t1 Here is another way of solving the equation O(t) = 0. Experiment with small step sizes and advance the time starting at t = 0 and ending at t = 2. From your hard data, observe the time t1 when O(t) changes, for the first time, from positive to negative. Use the value t1 to determine the true value of the period of the nonlinear pendulum. Compute the percentage relative error in the period estimated by T = 27r. If we assume that g = 32 ft/s2 and l = 32 ft, then the period of oscillation of the linear model is T = 27r s. Let us compare this last number with the period predicted by the nonlinear model when 80 = 7r/4. Using a numerical solver that is capable of generating hard data, approximate the solution of d 2 () '1T -2 + sin()= 0, 0(0) = - 4 , O'(O) = 0

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