Respuesta :
1 slug =
32 lb
f = kx
32 = k(2)
k = 16
c = 8 ( 8 times the
instantaneous velocity)
mx'' + cx' + kx =
8sin4t
x'' + 8x' + 16x =
8sin4t
Find for the
complimentary solution xh:
r² + 8r + 16 = 0
r² + 4r + 4r + 16 =
0
(r + 4)(r + 4) =
0
r = -4, -4 (repeated
roots)
xh = c₁e^(-4t) + c₂te^(-4t)
Find for the
particular solution xp:
xp = Acos(4t) +
Bsin(4t)
xp' = -4Asin(4t) +
4Bcos(4t)
xp'' = -16Acos(4t) -
16Bsin(4t)
x'' + 8x' + 16x =
8sin(4t)
-16Acos(4t) -
16Bsin(4t) + 8[ -4Asin(4t) + 4Bcos(4t) ] + 16 [ Acos(4t) + Bsin(4t) ] =
8sin(4t)
-16Acos(4t) -
16Bsin(4t) - 32Asin(4t) + 32Bcos(4t) + 16Acos(4t) + 16Bsin(4t) ] =
8sin(4t)
-32Asin(4t) +
32Bcos(4t) = 8sin(4t)
-4Asin(4t) + 4Bcos(4t)
= sin(4t)
We group like terms
and then solve for A and B:
4Bcos(4t) = 0
B = 0
-4Asin(4t) + 4Bcos(4t)
= sin(4t)
-4Asin(4t) =
sin(4t)
A = -¼
xp = Acos(4t) +
Bsin(4t)
xp = -¼cos(4t) + (0)
sin(4t)
xp = -¼cos(4t)
The general solution
is therefore:
x(t) = xh + xp
x(t) = c₁e^(-4t) + c₂te^(-4t) - ¼ cos(4t)
at t = 0 it starts
from rest that is initial velocity = 0
x'(0) = 0
at t = 0 it starts
from equilibrium
x(0) = 0
x(t) = c₁e^(-4t) + c₂te^(-4t) - ¼cos(4t)
0 = c₁ + c₂(0) - ¼cos(0)
c₁ = ¼
x(t) = c₁e^(-4t) + c₂te^(-4t) - ¼cos(4t)
x(t) =¼e^(-4t) + c₂te^(-4t) - ¼cos(4t)
x '(t) = -e^(-4t) + [
-4c₂te^(-4t) + c₂e^(-4t) ] + sin(4t)
x '(t) = -e^(-4t) - 4c₂te^(-4t) + c₂e^(-4t) + sin(4t)
x'(0) = 0
0 = -e^(0) - 4c₂(0) e^(0) + c₂e^(0) + sin(0)
0 = -1 + c₂ +
= -4c₁ - 4c₂(0) + c₂
0= -4(1/4) + c₂
c₂ = 1
x(t) =¼e^(-4t) + c₂te^(-4t) - ¼cos(4t)
x(t) =¼e^(-4t) +
te^(-4t) - ¼cos(4t)