## (1 point) A hollow steel ball weighing 4 pounds is suspended from a spring. This stretches the spring 12 feet. The ball is started in motion

Question

(1 point) A hollow steel ball weighing 4 pounds is suspended from a spring. This stretches the spring 12 feet. The ball is started in motion from the equilibrium position with a downward velocity of 2 feet per second. The air resistance (in pounds) of the moving ball numerically equals 4 times its velocity (in feet per second) . Suppose that after t seconds the ball is y feet below its rest position. Find y in terms of t. (Note that the positive direction is down.) Take as the gravitational acceleration 32 feet per second per second.

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2 weeks 2021-10-01T18:01:32+00:00 1 Answer 0

(-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

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