Equation of motion of a suspension system is given as: Mä(t) + Cx(t) + ax²(t) + bx(t) = F(t), where the spring force is given with a non-linear function as K(x) = ax²(t) + bx(t). a. Find the linearized equation of motion of the system for the motion that it makes around steady state equilibrium point x, under the effect of constant F, force. b. Find the natural frequency and damping ratio of the linearized system. - c. Find the step response of the system (* Numerical values: a=2, b=5, M=1kg, C=3Ns/m, Fo=1N, xo=0.05m

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
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Equation of motion of a suspension system is given as: Mä(t) + Cx(t) + ax² (t) + bx(t) = F(t), where
the spring force is given with a non-linear function as K(x) = ax²(t) + bx(t).
%3D
a. Find the linearized equation of motion of the system for the motion that it makes around steady state
equilibrium point x, under the effect of constant F, force.
b. Find the natural frequency and damping ratio of the linearized system. -
c. Find the step response of the system (
Numerical values: a=2, b=5, M=1kg, C=3Ns/m, Fo=1N, xo=0.05m
Transcribed Image Text:Equation of motion of a suspension system is given as: Mä(t) + Cx(t) + ax² (t) + bx(t) = F(t), where the spring force is given with a non-linear function as K(x) = ax²(t) + bx(t). %3D a. Find the linearized equation of motion of the system for the motion that it makes around steady state equilibrium point x, under the effect of constant F, force. b. Find the natural frequency and damping ratio of the linearized system. - c. Find the step response of the system ( Numerical values: a=2, b=5, M=1kg, C=3Ns/m, Fo=1N, xo=0.05m
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