Taking the initial conditions into consideration the

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Unformatted text preview: ation will eventually become steady, with an amplitude depending on the values of and . . As increases, the interval of forcing also increases. Hence the amplitude of the transient will increase with . Eventually, the forcing function will be constant. In fact, for large values of , even odd Further, for , ________________________________________________________________________ page 304 —————————————————————————— CHAPTER 6. —— Hence the steady state solution will oscillate about amplitude of . or , depending on , with an In the limit, as , the forcing function will be a periodic function, with period From Prob. , in Section , . . As increases, the duration and magnitude of the transient will increase without bound. 22 . Taking the initial conditions into consideration, the transform of the ODE is . Solving for the transform, . Using partial fractions, Since the denominator in the second term is irreducible, write Let Applying Theorem , term-by-term, the solution of the IVP is . ________________________________________________________________________ page 305 —————————————————————————— CHAPTER 6. —— For odd values of For even values of , the solution approaches , the solution approaches . . . The solution is a sum of damped sinusoids, each of frequency . Each term has an 'initial' amplitude of approximately For any given , the solution contains such terms. Although the amplitude will increase with , the amplitude will also be bounded by . . Suppose that the forcing function is replaced by methods in Chapter , the general solution of the differential equation is Note that and . Based on the . Using the method of undetermined coefficients, Based on the initial conditions, the solution of the IVP is ________________________________________________________________________ page 306 —————————————————————————— CHAPTER 6. —— Observe that both solutions have the same frequency, 23 . . Taking the initial conditions into consideration, the transform of the ODE is . Solving for the transform, . Using partial fractions, . Let . Applying Theorem , term-by-term, the solution of the IVP is . That is, ________________________________________________________________________ page 307 —————————————————————————— CHAPTER 6. —— . . . Based on the plot, the 'slow period' appears to be . The 'fast period' appears to be about . These values correspond to a 'slow frequency' of and a 'fast frequency' . . The natural frequency of the system is The forcing function is initially periodic, with period . Hence the corresponding forcing frequency is . Using the results in Section , the 'slow frequency' is given by and the 'fast frequency' is given by . Based on theses values, the 'slow period' is predicted as given as . and the 'fast period' is ________________________________________________________________________ page 308 —————————————————————————— CHAPTER 6. —— Section 6.5 2. Taking the Laplace transform of both sides of the ODE, we obtain . Applying the initial conditions, . Solving for the transform, . Applying Theorem , the solution of the IVP is 4. Taking th...
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