? Resorte 1:Veamos la relacion entre γ y ω0 entonces γ = 3.006x10−3 1s

y ω0 = 3.82 rads donde γ < ω0 luego el movimiento es subcrıtico.

? Resorte 2: la relacion entre γ y ω0 entonces γ = 0.02212 1s y ω0 =

5.53 rads donde γ < ω0 luego el movimiento es subcrıtico

? Resorte 3: la relacion entre γ y ω0 entonces γ = 6.935x10−3 1s y

ω0 = 6.39 rads donde γ < ω0 luego el movimiento es subcrıtico

♠ Ecuacion diferencial y solucion

? Resorte 1:

md2x

dt2+ γ

dx

dt+ kx = 0

Remplazando

0.15d2x

dt2+ 3.006x10−3

dx

dt+ 2.19x = 0

Cuya ecuacion caracterıstica esta dada por:

0.15r2 + 3.006x10−3r + 2.19 = 0

Cuyas soluciones estan dadas por

r1 = −0.01002 + 3.821i

r2 = −0.01002− 3.821i

Luego la solucion general sera:

x(t) = C1e−0.01002tcos(3.821t) + C2e

−0.01002tsen(3.821t)

? Resorte 2:

0.15d2x

dt2+ 0.02212

dx

dt+ 4.59x = 0

Cuya ecuacion caracterıstica esta dada por:

0.15r2 + 0.02212r + 4.59 = 0

Cuyas soluciones estan dadas por

r1 = −0.0742 + 5.531i

r2 = −0.0742− 5.531i

Luego la solucion general sera:

x(t) = C1e−0.0742tcos(5.531t) + C2e

−0.0742tsen(5.531t)

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? Resorte 3:

0.15d2x

dt2+ 6.935x10−3

dx

dt+ 6.125x = 0

Cuya ecuacion caracterıstica esta dada por:

0.15r2 + 6.935x10−3r + 6.125 = 0

Cuyas soluciones estan dadas por

r1 = −0.02311 + 6.39i

r2 = −0.02311− 6.39i

Luego la solucion general sera:

x(t) = C1e−0.02311tcos(6.39t) + C2e

−0.02311tsen(6.39t)

♠ Velocidad y Aceleracion

? Resorte 1:Derivando x(t)

v(t) = e−0.01002tcos(3.821t)(C1(−0.01002)+3.821C2)+e−0.01002tsen(3.821t)(−0.01002C2−3.821C1)

Derivando v(t) y reemplazando nuestras condiciones iniciales x(0) =0.67m y v(0) = 0

a(t) = e−0.01002t((0.076C1−14.63C2)sen(3.821t)+(−14.63C1−0.076C2)cos(3.821t))

Valor de las constantes C1 y C2

C1 = 0.67

C2 = 1.75× 10−3

Remplazando

x(t) = 0.67e−0.01002tcos(3.821t) + 1.75× 10−3e−0.01002tsen(3.821t)

v(t) = (−2.665×10−5)e−0.01002tcos(3.821t)+(−2.56)e−0.01002tsen(3.821t)

a(t) = (0.02531)e−0.01002tsen(3.821t)− (9.80)e−0.01002tcos(3.821t)

? Resorte 2:Derivando x(t)

v(t) = e−0.0742tcos(5.531t)(C1(−0.0742)+5.531C2)+e−0.0742tsen(5.531t)(−0.0742C2−5.531C1)

Derivando v(t) y reemplazando nuestras condiciones iniciales x(0) =0.32m y v(0) = 0

a(t) = e−0.0742t((0.832C1−31.002C2)sen(5.531t)+(−31.002C1−0.832C2)cos(5.531t))

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Valor de las constantes C1 y C2

C1 = 0.32

C2 = 4.29× 10−3

Remplazando

x(t) = 0.32e−0.0742tcos(5.531t) + 4.29× 10−3e−0.0742tsen(5.531t)

v(t) = −1.601× 10−5e−0.0742tcos(5.531t)− 1.770e−0.0742tsen(5.531t)

a(t) = 0.13324e−0.0742tsen(5.531t)− 9.9242e−0.0742tcos(5.531t)

? Resorte 3: Derivando x(t)

v(t) = e−0.02311tcos(6.39t)(C1(−0.02311)+6.39C2)+e−0.02311tsen(6.39t)(−0.02311C2−6.39C1)

Derivando v(t) y reemplazando nuestras condiciones iniciales x(0) =0.24m y v(0) = 0

a(t) = e−0.02311t((0.2964C1−40.979C2)sen(6.39t)+(−40.979C1−0.2964C2)cos(6.39t))

Valor de las constantes C1 y C2

C1 = 0.24

C2 = 8.67× 10−4

Remplazando

x(t) = 0.24e−0.02311tcos(6.39t) + 8.67× 10−4e−0.02311tsen(6.39t)

v(t) = −6.27× 10−6e−0.02311tcos(6.39t)− 1.5336e−0.02311tsen(6.39t)

a(t) = 0.03460e−0.02311tsen(6.39t)− 9.8352e−0.02311tcos(6.39t)

♠ Energia

? Resorte 1:

E = Ec + Ep =1

2mv2 +

1

2kx2

=1

20.15(e−0.01002tcos(3.821t)(−2.66x10−5)+e−0.01002tsen(3.821t)(−2.5601))2

+1

22.19(0.67e−0.01002tcos(3.821t)+1.75×10−3e−0.01002tsen(3.821t))2

8