A 4-foot spring measures 8 feet long after a mass weighing 8 pounds is attached to it. The medium through which the mass moves offers a damping force numerically equal to 2 times the instantaneous velocity. Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of 7 ft/s. (Use g = 32 ft/s2 for the acceleration due to gravity.)

x(t) = 7te^(-6.83t)

Explanation:

m = mass attached

k = spring's constant

β = positive damping constant = 2

From W=mg, we can find m

m = W/g = 8/32 = 0.25 slugs

From Hooke's law, we can find k.

W = kΔs; k = W/Δs

k = 8/(8-4) = 8/4 = 2 lb/ft

Applying Newton's second law to the system, we have a Differential eqaution (DE) as;

m(d²x/dt²) = - kx - β(dx/dt)

Divide through by m;

(d²x/dt²) = -kx/m - β(dx/dt)

Rearranging, we have;

(d²x/dt²) + kx/m + (β/m)(dx/dt) = 0

x(t) is the displacement from the equilibrium position

Plugging in the relevant values to obtain ;

(d²x/dt²) + 2x/0.25 + (2/0.25)(dx/dt) = 0

(d²x/dt²) + 8x+ 8(dx/dt) = 0

The auxiliary equation of this is;

m² + 8m + 8 = 0

The roots are m = -1.17 and m= -6.83

So, the general solution is;

x(t) = A•e^(-1.17t) + B•t•e^(-6.83t)

From the initial condition,

x(0) = 0 ft and x'(0) = 7 ft /s

Thus,

At x(0) = 0; 0 = Ae^(0) + 0

So,A = 0

Thus, x(t) = B•t•e^(-6.83t)

Now, x'(t) = Be^(-6.83t)

At x'(0) = 7 ft /s

7 = Be^(0)

B = 7

Thus, the equation of motion is;

x(t) = 7te^(-6.83t)

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