Answer:
(-0.377 - 1.117i)texp(-0.5 + 1.55i)t
Step-by-step explanation:
When the steel ball of mass m = 4 pounds is first suspended, the spring extends a distance of y₀ = 12 ft. From ky₀ = mg , k = mg/x₀ where k is the spring constant, and g = 32 ft/s² So, k = mg/y₀ = 4 × 32/12 = 10.67
The force of air resistance F = -4v
The spring force after it has been extended y ft is F = -ky
The net force on the steel ball is thus F = ma = -4v - ky
-4v - ky = ma (v = dy/dt = y' and a =dy/dt = y")
-4v - ky - ma = 0 ⇒ 4v + ky + ma = 0 (v = dy/dt = y' and a =dy/dt = y")
4y' + ky + my" = 0 ⇒ my" + 4y' + ky = 0
y" + 4y'/m + ky/m = 0
y" + 4y'/4 + 32y/12 = 0
y" + y' + 8y/3 = 0
3y" + 3y' + 8y = 0. Let y" = D², y' = D, y = 1
3D² + 3D + 8 = 0
Using the quadratic formula,
D = [-3 ±√(3² - 4 × 3 × 8)]/(2 × 3) = [-3 ±√(9 - 96)]/(6) = [-3 ±√( - 87)]/(6) = [-3 ±√( - 9.33)]/(6) = -0.5 ±√-1.55 = -0.5 ± 1.55i
y(t) = C₁exp(-0.5 + 1.55i)t + C₂texp(-0.5 + 1.55i)t
y'(t) = (-0.5 + 1.55i)C₁exp(-0.5 + 1.55i)t + (-0.5 + 1.55i)C₂texp(-0.5 + 1.55i)t + (-0.5 + 1.55i)C₂exp(-0.5 + 1.55i)t
y(0) = 0, y'(0) = 2
y(0) = C₁exp(-0.5 + 1.55i) × 0 + C₂× 0 ×exp(-0.5 + 1.55i) × 0
y(0) = C₁ exp(0) + 0 = C₁
0 = C₁
C₁ = 0
y'(0) = (-0.5 + 1.55i)C₁exp(-0.5 + 1.55i) × 0 + (-0.5 + 1.55i)C₂ × 0 × exp(-0.5 + 1.55i) × 0 + (-0.5 + 1.55i)C₂exp(-0.5 + 1.55i) × 0
2 = (-0.5 + 1.55i) × 0 × exp(0) + 0 + (-0.5 + 1.55i)C₂exp(0)
2 = 0 + 0 + (-0.5 + 1.55i)C₂
2 = (-0.5 + 1.55i)C₂
C₂ = 2/(-0.5 + 1.55i) = 2(-0.5 - 1.55i)/2.6525 = (-1 - 3.1i)/2.6525 = -0.377 - 1.117i
So,
y(t) = C₁exp(-0.5 + 1.55i)t + C₂texp(-0.5 + 1.55i)t
= 0 × exp(-0.5 + 1.55i)t + (-0.377 - 1.117i)texp(-0.5 + 1.55i)t
= 0 + (-0.377 - 1.117i)texp(-0.5 + 1.55i)t
= (-0.377 - 1.117i)texp(-0.5 + 1.55i)t
