An 8 lb weight is attached to a coil spring suspendedfrom a fixed support. The w

Question

An 8 lb weight is attached to a coil spring suspendedfrom a fixed support. The weight comes to rest in its equilibrium position, thereby stretching the spring 6 inches. The weight is then pulled down 9 inches beyond its equilibrium position and released at t=0. There is a resistive damping force that is proportional to velocity with a constant of 4 where velocity is measured in feet/seconds.

a.) Determine the displacement of the weight as a function of the time and graph this displacement. Neglect the force due to gravity.

Explanation / Answer

Spring stiffness k = Force / displacement = 8/(6/12) = 16 lb/ft

Damping coefficient c = Damping force / velocity = 4 lb-s/ft

Critical damping Cc = 2sqrt(km) = 2*sqrt(16*32.2*8) = 128.4 lb/s = 128.4 / 32.2 lb-s/ft = 4 lb-s/ft

Zeta = C/Cc = 4/4 = 1

Natural freq. w = sqrt (k/m) = sqrt (16 *32.2 / 8) = 8.025 rad/s

Equation of motion becomes: mx'' + cx' + kx = 0

For zeta = 1 case, Its solution is of the form x = (k1 + k2*t)*e^(-wt)

Thus, x' = -w*(k1 + k2*t)*e^(-wt) + k2*e^(-wt)

At t = 0, we get x = 9/12 = 0.75 ft

Initial velocity x'(0) = 0

Thus, k1 = 0.75 and k2 - w*k1 = 0

This gives k2 = 0.75*8.025 = 6.02

Thus, x = (0.75 + 6.02*t)*e^(-6.02*t)

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