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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 , termbyterm, 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 , termbyterm, 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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 Spring '08
 Staff
 Continuity

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