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´ ´ ESCUELA POLITECNICA DEL EJERCITO ´ CARRERA DE INGENIER´ MECANICA IA

ANALISIS MATEMATICO III

EJERCICIOS CAPITULO 5

Jonatan Aizaga C

Mec´nica 4to A a

1

Ejercicios 5.1 En los problemas 1 y 2 describa una posible interpretaci´n del problema f´ o ısico dado. 4 x + 3x = 0 32 4 x + 3x = 0 32 x(0) = −3; x (0) = 0 1) x + 24x = 0 xn = 0 r + 24 = 0 √ 02 − 4 ∗ 24 r= 2 √ ±2 ∗ 2 6i r= 2√ r = ±2 6i √ r1 = 2 6i √ r2 = −2 6i √ √ xn = C1e0 cos(2 6x) + iC2e0 sen(2 6x) √ √ xn = C1cos(2 6x) + iC2sen(2 6x) √ √ √ √ xn = −2 6C1sen(2 6x) + i2 6C2cos(2 6x) 0± √ √ x(0) = −3 = C1cos(2 6 ∗ 0) + iC2sen(2 6 ∗ 0) x(0) = −3 = C1cos(0) + iC2sen(0) C1 = −3
2

√ √ √ √ x (0) = 0 = −2 6C1sen(2 6 ∗ 0) + i2 6C2cos(2 6 ∗ 0) √ √ x (0) = 0 = −2 6C1sen(0) + i2 6C2cos(0) C2 = 0

√ √ xn = (−3)cos(2 6x) +i(0)sen(2 6x) √ xn = −3cos(2 6x)// 2

2)

1 x + 4x = 0 16

x(0) = 0,7; x (0) = 0 x + 64x = 0 xn = 0 r + 64 = 0 no factorable en los Reales √ 02 − 4 ∗ 64 r= 2 √ ±2 64i r= 2 r = ±8i 0± r1 = 8i xn = C1e cos(8x) + iC2e0 sen(8x) xn = C1cos(8x) + iC2sen(8x) xn = −8C1sen(8x) + i8C2cos(8x) x(0) = 7 = C1cos(8 ∗ 0) + iC2sen(8 ∗ 0) 10 7 = C1cos(0) + iC2sen(0) 10 7 C1 = 10 7 sen(8 ∗ 0) + i8C2cos(8 ∗ 0)10 56 sen(0) + 8iC2cos(0) 10 C2 = 0 7 cos(8x)// 10
0 2

r2 = −8i

x(0) =

x (0) = 0 = −8

x (0) = 0 = −

xn =

3

3)x + 25x = 0 x(0) = −2; x (0) = 10 xn = 0 r2 + 25 = 0 no factorable en los Reales √ 02 − 4 ∗ 25 r= 2 √ ±2 25i r= 2 r = ±5i 0± r1 = 5i xn = C1e cos(5x) + iC2e0 sen(5x) xn = C1cos(5x) + iC2sen(5x) xn = −5C1sen(5x) + i5C2cos(5x) x (0) = 10 = −5C1sen(5 ∗ 0) + i5C2cos(5 ∗ 0)x (0) = 10 = −5C1sen(0) + 5iC2cos(0) 10 = 5iC2 C2 = 2i
0

r2 = −5i

x(0) = −2 = C1cos(5 ∗ 0) + iC2sen(5 ∗ 0) x(0) = −2 = C1cos(0) + i(2i)sen(0) C1 = −2 xn = −2cos(5x) + i(2i)sen(5x) xn = −2cos(5x) + 2(−1)sen(5x) xn = −2cos(5x) + 2sen(5x)// 4

Halle la soluci´n del problema de valor inicial dado en la forma x(t) = o Asen(W t + φ)
1 4. 2 x + 8x = 0

x(0) = 1

x(0) = −2

1 2 r +8=0 2r2 = −16 ⇒ xn = c1 cos4t + c2 sen4t x(0) = 1 c1 = 1 x = −4c1 sen4t + 4c2 cos4t −2 = −4c1 sen0 + 4c2 cos0 −2 = 4c2 1 c2 = − 2 1 x = cos4t − sen4t 2 x = Asen(4t + φ) A= A= c2 + c 2 1 2 x(0) = −2 1 = c1 cos0 + c2 sen0 r = ±4i

12 1+ 2 √ 5 A= 2 c1 tanφ = c2 1 tanφ = 1 = −2 −2 φ = −1,11rad + π φ = −1,11rad

φ = 2,0344rad √ 5 ⇒x= sen(4t + 2,0344)// 2 5

5.x + 2x = 0 x(0) = −1 √ x(0) = −2 2

r2+ 2 = 0 r2 = −2 √ r = ± 2i √ √ ⇒ xn = c1 cos 2t + c2 sen 2t x(0) = −1 −1 = c1 cos0 + c2 sen0 c1 = −1 √ √ √ √ x = − 2c1 sen 2t + 2c2 cos 2t √ x(0) = −2 2 √ √ √ −2 2 = − 2c1 sen0 + 2c2 cos0 √ √ −2 2 = 2c2 c2 = −2 √ √ x = −cos 2t − 2sen 2t √ x = Asen( 2t + φ) A= A= c2 + c 2 1 2

1 + −22 √ A= 5 c1 tanφ = c2 −1 tanφ = −2 φ = 0,4636rad

φ = 0,4636rad + π φ = 3,6052rad √ √ ⇒ x = 5sen( 2t + 3,6052)//6

6.

1 4x

+ 16x = 0

x + 64x = 0 r2 + 64 = 0 r = +8, −8i Xt = C1 cos(8t) + C2 sin(8t) X( 0) = 4 4 = C1 cos(0) + C2 sin(0) 4 = C1 X( 0) = 16 16 = −8C1 sin(0) + 8C2 cos(0) 16 = 8C2 2 = C2 Xt = 4 cos(8t) + 2 sin(8t)

7

7.

1 10 x

+ 10x = 0

x + 100x = 0 r2 + 100 = 0 r = +10, −10i Xt = C1 cos(10t) + C2 sin(10t) X( 0) = 1 1 = C1 cos(0) + C2 sin(0) 1 = C1 X( 0) = 1 1 = −10C1sin(0) + 10C2 cos(t) 1 = 10C2 1 = C2 10 1 Xt = cos(10t) + sin(10t) 10

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8. x + x = 0 r2 + 1 = 0 r = +i, −i Xt = C1 cos(t) + C2 sin(t) X( 0) = −4 −4 = C1 cos(0) + C2 sin(0) −4 = C1 X( 0) = 3 3 = −C1 sin(0) + C2 cos(t) 3 = C2 1 = C2 10 Xt = −4 cos(t) + 3 sin(t)

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9. El periodo de oscilaciones libres no amortiguadoas de una masa sujeta a un resorte es π/4 segundos. Si la constante delresorte es 16 lb/pie, ¿Cual es el valor numerico del peso? T =π 4 k = 16 lb/pie ω = 2πf 2π =8 T k ω2 = m k m= 2 ω 16 m = 2 = 0, 25 8 ω= m = (0, 25)(32) lbpie/s2 m = 8lb// 10. Un Resorte esta suspendido de un Techo. Cuando el resorte es fijo de un extremo libre un cuerpo que pesa 60 lb se estira 1/2 pie, si se le quita dicho cuerpo y una persona asciende del extremo del resorte empieza a oscilar...