A mass of 1 slug, when attached to a spring, stretches it 2 feet and then comes to rest in the equilibrium position. starting at t = 0, an external force equal to f(t) = 8 sin 4t is applied to the system. find the equation of motion if the surrounding medium offers a damping force that is numerically equal to 8 times the instantaneous velocity.

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brittbb1357 brittbb1357

01-08-2017
Physics

contestada

A mass of 1 slug, when attached to a spring, stretches it 2 feet and then comes to rest in the equilibrium position. starting at t = 0, an external force equal to f(t) = 8 sin 4t is applied to the system. find the equation of motion if the surrounding medium offers a damping force that is numerically equal to 8 times the instantaneous velocity.

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barnuts barnuts · Answer 1 · 2017-08-14T14:17:32+02:00

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)