Please show work. A mass-spring system is shown in the figure on the right. Usin

Question

Please show work.

A mass-spring system is shown in the figure on the right. Using Newton's laws, the equations of motion are given by + Ffriction + Fspring = r(t). Common models of the spring force include Fspring = ky(linear spring), Fspring = k(1-a2y2)y (softening spring, common in micromachined mechanical resonators), and Fspring = k(l +a2y2)y (hardening spring)1. Assuming a hardening spring, and that the (wall) friction is of viscous type, i.e., Ffriction = b dy/dt, the equations of motion then are M d2y/dt2 + b dy/dt + k(1+a2 y2)y = r(t) For the numerical values M = 1,b = 0.2, k = 0.5,a = 0.1,r(t) 5, verify that the system has an equilibrium point with y = 6.8. (rough calculations are sufficient) Linearize the system around this equilibrium point, defining all variables of interest clearly.

Explanation / Answer

M*d^y/dt^2 + b*dy/dt + k(1+a^2y^2)y = r(t) -------------(1)

let y1 = y

y2 = y1dot = ydot ------------(2)

substituting in (1)

=> M*y2dot + b*y2 + k*(1+a^2*y1^2)*y1 = r(t)

writing in state space model

y1dot = y2

y2dot = 1/M*(r(t)-k*(1+a^2*y1^2)*y1-b*y2)

for equilibrium point

y1dot = y2dot = 0

=> y2 = 0 and

r(t)-k*(1+a^2*y1^2)*y1 = 0

substituting given values

5 -0.5* y1 -0.5* 0.01y1^3 = 0

y = 6.823 is a root

(can also be verified by substituting y = 6.8)

b)

let the system be

y1dot = f1(y1,y2)

y2dot = f2(y1,y2)

Jacobian matrix J = {df1/dy1 df1/dy2;df2/dy1 df2/dy2} = {0 1;(-k/M*(1+2*a^2*y1) -b/M}

J at y1 = 6.823 = { 0 1;-0.568 -0.2}

Linearized system

Ydot = J*Y where Y = [y1;y2]

y1dot = y2

y2dot = -0.568*y1 - 0.2*y2

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