Unformatted text preview: — CHAPTER 6. —— After an initial transient, the solution oscillates about . 14. The specified function is defined by which can conveniently be expressed as 15. The function is defined by which can also be written as 16 . From Part , the solution is
, where Due to the damping term, the solution will decay to zero. The maximum will occur
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page 296 —————————————————————————— CHAPTER 6. ——
shortly after the forcing ceases. By plotting the various solutions, it appears that the
solution will reach a value of
, as long as
. Based on the graph, and numerical calculation, for . 17. We consider the initial value problem
,
with . . The specified function is defined by Taking the Laplace transform of both sides of the ODE, we obtain ________________________________________________________________________
page 297 —————————————————————————— CHAPTER 6. —— .
Applying the initial conditions,
.
Solving for the transform,
.
Using partial fractions, It follows that
.
Using Theorem , the solution of the IVP is
, in which . . Note that for So for , the solution is given by , the solution oscillates about , with an amplitude of
. ________________________________________________________________________
page 298 —————————————————————————— CHAPTER 6. —— ________________________________________________________________________
page 299 —————————————————————————— CHAPTER 6. ——
18 . . The forcing function can be expressed as Taking the Laplace transform of both sides of the ODE, we obtain
.
Applying the initial conditions,
.
Solving for the transform,
.
Using partial fractions, Let It follows that ________________________________________________________________________
page 300 —————————————————————————— CHAPTER 6. —— Based on Theorem , the solution of the IVP is . As the parameter decreases, the solution remains null for a longer period of time. ________________________________________________________________________
page 301 —————————————————————————— CHAPTER 6. ——
Since the magnitude of the impulsive force increases, the initial overshoot of the
response also increases. The duration of the impulse decreases. All solutions eventually
decay to
19 . From Part , ________________________________________________________________________
page 302 —————————————————————————— CHAPTER 6. ——
21 . . Taking the Laplace transform of both sides of the ODE, we obtain
.
Applying the initial conditions,
.
Solving for the transform,
.
Using partial fractions,
.
Let ________________________________________________________________________
page 303 —————————————————————————— CHAPTER 6. —— .
Applying Theorem , termbyterm, the solution of the IVP is
. Note that Hence . The ODE has no damping term. Each interval of forcing adds to the energy of the
system.
Hence the amplitude will increase. For
,
when
. Therefore the
oscill...
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 Spring '08
 Staff
 Continuity, Laplace, initial conditions

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