Suppose that a spring with mass 4 and spring constant k = 512 is immersed in a f

Question

Suppose that a spring with mass 4 and spring constant k = 512 is immersed in a fluid with damping constant c = 96. Find the position of the mass at time t if it starts from the equilibrium position and is given a push to start it with an initial velocity of 0.8 m/s.

SOLUTION The mass is m = 4 and the spring constant is k = 512, so the differential equation becomes

4*d^(2)*x/d*t^(2) +96*dx/dt+512*x=0

d^(2)*x/d*t^(2)+ 24*dx/dt=128*x=0

The auxiliary equation is r2 + 24r + 128 = (r + 8)( [ANSWER BOX HERE]) = 0 with roots -8 and -16, so the motion is overdamped and the solution is

x(t) = c1e-8t + c2e-16t

We are given that x(0) = 0, so c1 + c2 = 0. Differentiating, we get

x'(t) = -8c1e-8t - 16c2e-16t
x'(0) = -8c1 - 16c2 = 0.8

Since c2 = -c1, this gives 8c1 = 0.8 or c1 = [ANSWER BOX]

. Therefore

x(t) =  

Explanation / Answer

equation is r2+24r+128=0

r2+16r+8r+128=0

r(r+16)+8(r+16)=0

(r+8)(r+16)=0

so answer fo fill in the box is r+16

8c1 = 0.8

c1 = 0.8/8 =0.1

so c1=0.1

x(t) = c1e-8t +c2 e-16t

x(t) = 0.1e-8t + (-0.1)e-16t  as c1= -c2

x(t)=0.1e-8t - 0.1e-16t

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