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# Oscilações Forçadas com Força Externa Periódica

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npit,
que podemos reescrever como
\u221e
\u2211
n=1
[(2\u2212 n2pi2)Bn \u2212 3npiAn \u2212 bn] sen npit +
\u221e
\u2211
n=1
[(2\u2212 n2pi2)An + 3npiBn] cos npit = 0.
De onde obtemos o sistema{
(2\u2212 n2pi2)An + 3npiBn = 0
\u22123npiAn + (2\u2212 n2pi2)Bn = bn
que tem soluc¸a\u2dco
An = \u22123npibn
\u2206n
, Bn =
(2\u2212 n2pi2)bn
\u2206n
, para n = 1, 2, . . .
em que \u2206n = 9n
2pi2 + (2\u2212 n2pi2)2. Assim uma soluc¸a\u2dco particular da equac¸a\u2dco diferen-
cial e´
up(t) = \u2212
\u221e
\u2211
n=1
3npibn
\u2206n
cos npit +
\u221e
\u2211
n=1
(2\u2212 n2pi2)bn
\u2206n
sen npit
= 6
\u221e
\u2211
n=1
(\u22121)n \u2212 1
\u2206n
cos npit +
2
pi
\u221e
\u2211
n=1
(2\u2212 n2pi2)(1\u2212 (\u22121)n)
n\u2206n
sen npit
= \u221212
\u221e
\u2211
n=0
1
\u22062n+1
cos(2n + 1)pit +
4
pi
\u221e
\u2211
n=0
2\u2212 (2n + 1)2pi2
(2n + 1)\u22062n+1
sen(2n + 1)pit
que e´ a soluc¸a\u2dco estaciona´ria. Para encontrar up(t) fizemos a suposic¸a\u2dco de que as deri-
up(t) =
\u221e
\u2211
n=1
un(t).
11
Enta\u2dco
u\u2032p(t) =
\u221e
\u2211
n=1
u\u2032n(t),
u\u2032\u2032p(t) =
\u221e
\u2211
n=1
u\u2032\u2032n(t)
pois,
|u\u2032n(t)| \u2264 12
n
\u2206n
+
4
pi
(2\u2212 n2pi2)
\u2206n
,
|u\u2032\u2032n(t)| \u2264 12
n2
\u2206n
+
4
pi
n(2\u2212 n2pi2)
\u2206n
.
12
Exerc\u131´cios
-1
-0.5
0.5
1
1 2 3 4 5 6 7 8 t
y
Figura 6: Termo na\u2dco homoge\u2c6neo da equac¸a\u2dco do problema de valor inicial do Exerc\u131´cio
1.
1. Considere
f (t) =
{
t, se 0 \u2264 t < 1
2\u2212 t, se 1 \u2264 t < 2 e tal que f (t + 2) = f (t)
(a) Resolva o problema de valor inicial{
u\u2032\u2032 + \u3c920u = f (t),
u(0) = 0, u\u2032(0) = 0
(b) Encontre a soluc¸a\u2dco estaciona´ria do problema de valor inicial{
u\u2032\u2032 + 3u\u2032 + 2u = f (t),
u(0) = 0, u\u2032(0) = 0
13
Respostas dos Exerc\u131´cios
1. (a) A soluc¸a\u2dco geral da equac¸a\u2dco diferencial e´
u(t) = c1 cos \u3c90t + c2 sen\u3c90t + up(t),
em que up(t) e´ uma soluc¸a\u2dco particular. Como f e´ par, seccionalmente cont\u131´nua com derivada seccionalmente
cont\u131´nua, ela pode ser representada por sua se´rie de Fourier de cossenos:
f (t) =
a0
2
+
\u221e
\u2211
n=1
an cos npit
com
a0 = 2a0( f
(1)
0,1 ) = 1,
an = 2an( f
(1)
0,1 ) =
2
n2pi2
(cos npi \u2212 1) = 2
n2pi2
((\u22121)n \u2212 1),
f (t) =
1
2
\u2212 2
pi2
\u221e
\u2211
n=0
(\u22121)n \u2212 1
n2
cos npit
Vamos procurar uma soluc¸a\u2dco particular da forma
up(t) = A0 +
\u221e
\u2211
n=1
(An cos npit + Bn sen npit)
com coeficientes An, Bn a determinar. Vamos supor que a derivada da se´rie seja igual a se´rie das derivadas:
u\u2032p(t) =
\u221e
\u2211
n=1
(\u2212npiAn sen npit + npiBn cos npit).
u\u2032\u2032p(t) = \u2212
\u221e
\u2211
n=1
(n2pi2An cos npit + n
2
pi
2Bn sen npit)
Substituindo-se up(t) e u\u2032\u2032p(t) na equac¸a\u2dco diferencial obtemos
\u2212
\u221e
\u2211
n=1
n2pi2(An cos npit + Bn sen npit)
+ \u3c920(A0 +
\u221e
\u2211
n=1
(An cos npit + Bn sen npit)) =
a0
2
+
\u221e
\u2211
n=1
an cos npit
\u221e
\u2211
n=1
Bn(\u3c9
2
0 \u2212 n2pi2) sen npit + \u3c920A0 \u2212
a0
2
+
\u221e
\u2211
n=1
[An(\u3c9
2
0 \u2212 n2pi2)\u2212 an] cos npit = 0
De onde obtemos
A0 =
a0
2\u3c920
, An =
an
\u3c920 \u2212 n2pi2
, Bn = 0, para n = 1, 2, . . .
Assim uma soluc¸a\u2dco particular da equac¸a\u2dco diferencial e´
up(t) =
a0
2\u3c920
+
\u221e
\u2211
n=1
an
\u3c920 \u2212 n2pi2
cos npit
=
1
2\u3c920
+
2
pi2
\u221e
\u2211
n=1
(\u22121)n \u2212 1
n2(\u3c920 \u2212 n2pi2)
cos npit
14
A soluc¸a\u2dco geral e´ enta\u2dco
u(t) = c1 cos \u3c90t + c2 sen\u3c90t +
1
2\u3c920
+
2
pi2
\u221e
\u2211
n=1
(\u22121)n \u2212 1
n2(\u3c920 \u2212 n2pi2)
cos npit
Substituindo-se t = 0 e u = 0, obtemos
c1 =
2
pi2
\u221e
\u2211
n=1
1\u2212 (\u22121)n
\u3c920 \u2212 n2pi2
\u2212 1
2\u3c920
.
Substituindo-se t = 0 em
u\u2032(t) = \u2212\u3c90c1 sen\u3c90t + \u3c90c2 cos \u3c90t + 2
pi
\u221e
\u2211
n=1
1\u2212 (\u22121)n
n(\u3c920 \u2212 n2pi2)
sen npit
obtemos c2 = 0. Logo a soluc¸a\u2dco do PVI e´
u(t) =
(
2
pi2
\u221e
\u2211
n=1
1\u2212 (\u22121)n
\u3c920 \u2212 n2pi2
\u2212 1
2\u3c920
)
cos \u3c90t +
1
2\u3c920
+
2
pi2
\u221e
\u2211
n=1
(\u22121)n \u2212 1
n2(\u3c920 \u2212 n2pi2)
cos npit
=
(
4
pi2
\u221e
\u2211
n=0
1
\u3c920 \u2212 (2n + 1)2pi2
\u2212 1
2\u3c920
)
cos \u3c90t
+
1
2\u3c920
+
4
pi2
\u221e
\u2211
n=0
1
(2n + 1)2((2n + 1)2pi2 \u2212\u3c920)
cos(2n + 1)pit
-0.5
0.5
1 2 3 4 5 6 7 8 t
u
(b) A soluc¸a\u2dco geral da equac¸a\u2dco diferencial e´
u(t) = c1e
\u2212t + c2e\u22122t + up(t),
em que up(t) e´ uma soluc¸a\u2dco particular. Como f e´ par, seccionalmente cont\u131´nua com derivada seccionalmente
cont\u131´nua, ela pode ser representada por sua se´rie de Fourier de cossenos:
f (t) =
a0
2
+
\u221e
\u2211
n=1
an cos npit
com
a0 = 2a0( f
(1)
0,1 ) = 1,
an = 2an( f
(1)
0,1 ) =
2
n2pi2
(cos npi \u2212 1) = 2
n2pi2
((\u22121)n \u2212 1),
15
f (t) =
1
2
+
2
pi2
\u221e
\u2211
n=0
(\u22121)n \u2212 1
n2
cos npit
Vamos procurar uma soluc¸a\u2dco particular da forma
up(t) = A0 +
\u221e
\u2211
n=1
(An cos npit + Bn sen npit)
com coeficientes An, Bn a determinar. Vamos supor que a derivada da se´rie seja igual a se´rie das derivadas:
u\u2032p(t) =
\u221e
\u2211
n=1
(\u2212npiAn sen npit + npiBn cos npit),
u\u2032\u2032p(t) = \u2212
\u221e
\u2211
n=1
(n2pi2An cos npit + n
2
pi
2.Bn sen npit)
Substituindo-se up(t), u\u2032p(t) e u\u2032\u2032p(t) na equac¸a\u2dco diferencial obtemos
\u2212
\u221e
\u2211
n=1
n2pi2(An cos npit + Bn sen npit)
+ 3
\u221e
\u2211
n=1
(\u2212npiAn sen npit + npiBn cos npit)
+ 2(A0 +
\u221e
\u2211
n=1
(An cos npit + Bn sen npit)) =
a0
2
+
\u221e
\u2211
n=1
an cos npit
que podemos reescrever como
\u221e
\u2211
n=1
[(2\u2212 n2pi2)Bn \u2212 3npiAn] sen npit
+ 2A0 \u2212 a0
2
+
\u221e
\u2211
n=1
[(2\u2212 n2pi2)An + 3npiBn \u2212 an ] cos npit = 0.
De onde obtemos A0 =
a0
4
e o sistema de equac¸o\u2dces
{
(2\u2212 n2pi2)An + 3npiBn = an
\u22123npiAn + (2\u2212 n2pi2)Bn = 0
que tem soluc¸a\u2dco
An =
(2\u2212 n2pi2)an
\u2206n
, Bn =
3npian
\u2206n
, para n = 1, 2, . . .
em que \u2206n = 9n2pi2 + (2\u2212 n2pi2)2. Assim uma soluc¸a\u2dco particular da equac¸a\u2dco diferencial e´
up(t) =
a0
4
+
\u221e
\u2211
n=1
(2\u2212 n2pi2)an
\u2206n
cos npit +
\u221e
\u2211
n=1
3npian
\u2206n
sen npit
=
1
4
+
2
pi2
\u221e
\u2211
n=1
(2\u2212 n2pi2)((\u22121)n \u2212 1)
n2\u2206n
cos npit +
6
pi
\u221e
\u2211
n=1
(\u22121)n \u2212 1
n\u2206n
sen npit
=
1
4
+
4
pi2
\u221e
\u2211
n=0
(2n + 1)2pi2 \u2212 2
(2n + 1)2\u22062n+1
cos(2n + 1)pit\u2212 12
pi
\u221e
\u2211
n=0
1
(2n + 1)\u22062n+1
sen(2n + 1)pit
16
que e´ a soluc¸a\u2dco estaciona´ria.
-0.5
0.5
1 2 3 4 5 6 7 8 t
u
Exercícios