Suppose a mass of 4 kilograms stretches a spring by 0.8 meters, the damping forc

Question

Suppose a mass of 4 kilograms stretches a spring by 0.8 meters, the damping force is times the instantaneous velocity and there is no external driving force. At t = 0, the 4 kg mass will be released from rest at a position 1.5 meters above the equilibrium position. (a) Determine the spring constant (units will be Newtons/meter). (b) Set up (but do not solve) the initial value problem that models this problem using the orientation that positive displacement is directed downward (as in the textbook).

Explanation / Answer

Let in the rest stretched positon , the amount of extension be L

=> mg = kL

=> 4(9.8) = k(0.8)

=> k = 49

Now making Free Body Diagram of the Mass Appratus

=> mx'' = mg - kL - x' - kx

=> mx'' + x' + kx = 0

=> 4x'' + x' + 49x = 0

Critical Damping will occur when 2 - 4(4)(49) = 0 => = 28

=> Solution of the differential equation becomes x(t) = c1e-3.5t + c2te-3.5t

x'(t) = -3.5c1e-3.5t -3.5c2te-3.5t + c2te-3.5t

Plugging the initial conditions , x(0) = -1.5 and x'(0) = 0 => c1 = -1.5 and c2 = -5.25

=> x(t) = -1.5e-3.5t - -5.25te-3.5t

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