Consider a pendulum that is released from rest from an initial displacement of u0 radians. Solving the linear model (7) subject to the initial conditions u(0) u0, u(0) 0 gives . The period of oscillations predicted by this model is given by the familiar formula . The interesting thing about this formula for T is that it does not depend on the magnitude of the initial displacement u0. In other words, the linear model predicts that the time it would take the pendulum to swing from an initial displacement of, say, u0 p2 ( 90) to p2 and back again would be exactly the same as the time it would take to cycle from, say, u0 p360 ( 0.5) to p360. This is intuitively unreasonable; the actual period must depend on u0. If we assume that g 32 ft/s2 and l 32 ft, then the period of oscillation of the linear model is T 2p s. Let us compare this last number with the period T 2 1g/l 2 1l/g (t) 0 cos 1g/lt predicted by the nonlinear model when u0 p4. Using a numerical solver that is capable of generating hard data, approximate the solution of on the interval 0 t 2. As in 25, if t1 denotes the first time the pendulum reaches the position OP in Figure 5.3.3, then the period of the nonlinear pendulum is 4t1. Here is another way of solving the equation u(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 u(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 2p. d2 dt2 sin

# Consider a pendulum that is released from rest from an

## Problem 5.1.122 Chapter 5.3

Differential Equations with Boundary-Value Problems, | 8th Edition

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