Para o Problema 5.13, calcule x 1 ( t ) e x 2 ( t ) para as seguintes condições iniciais:(a) x 1 ( 0 ) = 0,2 , x ˙ 1 ( 0 ) = x 2 ( 0 ) = x ˙ 2 ( 0 ) = 0 (b) x 1 ( 0 ) = 0,2 , x ˙ 1 ( 0 ) = x 2 ( 0 ) = 0 , x ˙ 2 ( 0 ) = 5,0

Passo 1

E ai galera tudo bem? O enunciado pede que a gente calcule x 1 ( t ) e x 2 ( t ) para as seguintes condições iniciais. Começamos com a letra “A”.

(a) ω 1 2 = 917.2 , σ 2 2 = 13082.8 , r 1 = 1.1805 , r 2 = - 0.8471

x 1 ( 0 ) = 0.2 , x 2 ( 0 ) = x ˙ 1 ( 0 ) = x ˙ 2 ( 0 ) = 0

Eq (5.18) dá

x 1 ( 1 ) = 1 ( - 0.8471 - 1.1805 ) [ ( - 0.8471 ) ( 0.2 ) ] = 0.08356

x 1 ( 2 ) = 1 ( - 0.8471 - 1.1805 ) [ ( - 1.1805 ) ( 0.2 ) ] = 0.11644

ϕ 1 = ϕ 2 = t a n - 1 ⁡ ( 0 ) = 0

x 1 ( t ) = 0.08356 c o s ⁡ 30.2853 t + 0.11644 c o s ⁡ 114.3801 t

x 2 ( t ) = ( 1.1805 ) ( 0.08356 ) c o s ⁡ 30 ⋅ 2853 t + ( - 0.8471 ) ( 0.11644 ) c o s ⁡ 114 ⋅ 3801 t = 0.09864 c o s ⁡ 30.2853 t - 0.09864 c o s ⁡ 114.3801 t

Passo 2

(b) x 1 ( 0 ) = 0.2 , x 2 ( 0 ) = x ˙ 1 ( 0 ) = 0 , x ˙ 2 ( 0 ) = 5.0

Eq (5.18) dá

x 1 ( 1 ) = 1 - 2 ⋅ 0276 { - 0.8471 ( 0.2 ) } 2 + ( 5 ) 2 917.2 1 / 2 = - 0.1167

x 1 ( 2 ) = 1 - 2.0276 { - 1 ⋅ 1805 ( 0.2 ) } 2 + ( - 5 ) 2 13082.8 1 / 2 = - 0.1184

ϕ 1 = t a n - 1 ⁡ 5.0 30.2853 ( - 0.8471 ) ( 0.2 ) = t a n - 1 ⁡ ( - 0.9745 ) = 135.7399 ∘

ϕ 2 = t a n - 1 ⁡ - 5.0 114.3801 ( - 1.1805 ) ( 0.2 ) = t a n - 1 ⁡ ( 0.1851 ) = 10.4895 ∘

x 1 ( t ) = - 0.1167 c o s ⁡ 30.2853 t + 135.7399 ∘

- 0.1184 c o s ⁡ 114.3801 t + 10.4895 ∘

x 2 ( t ) = ( 1.1805 ) ( - 0.1167 ) c o s ⁡ 30.2853 t + 135.7399 ∘ - 0.8471 ( - 0.1184 ) c o s ⁡ 114.3801 t + 10.4895 ∘ = - 0.1378 c o s ⁡ 30.2853 t + 135.7399 ∘

+ 0.1003 c o s ⁡ 114.3801 ∘ t + 10.4895 ∘

Resposta

( a )   - 0.1184 c o s ⁡ 114.3801 t + 10.4895 ∘

(b)   + 0.1003 c o s ⁡ 114.3801 ∘ t + 10.4895 ∘

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