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

Correct question is;

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)

x(t) = 7te^(-2t√2)

Explanation:

We are given;

Weight; W = 8 lbs

mass; m = W/g

g = 32 ft/s²

Thus;

m = 8/32

m = ¼ slugs

From Newton's second law we can write the equation as;

m(d²x/dt²) = -kx - β(dx/dt)

Rearranging this, we have;

(d²x/dt²) + (β/m)(dx/dt) + (k/m)x = 0

Where;

β is damping constant = √2

k is spring constant = W/s

Where s = 8ft - 4ft = 4ft

k = 8/4

k = 2

Thus,we now have;

(d²x/dt²) + (√2/(¼))(dx/dt) + (2/(¼))x = 0

>> (d²x/dt²) + (4√2)dx/dt + 8x = 0

The auxiliary equation of this is;

m² + (4√2)m + 8 = 0

Using quadratic formula, we have;

m1 = m2 = -2√2

The general solution will be gotten from;

x_t = c1•e^(mt) + c2•t•e^(mt)

Plugging in the relevant values gives;

x_t = c1•e^(mt) + c2•t•e^(mt)

At initial condition of t = 0, x_t = 0 and thus; c1 = 0

Also at initial condition of t = 0, x'(0) = 7 and thus;

Since c1 = 0, then c2 = 7

Thus,equation of motion is;

x(t) = 7te^(-2t√2)

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360seconds i 1 hr