A simplified model of the suspension of a car consists of a mass m, a spring wit

Question

A simplified model of the suspension of a car consists of a mass m, a spring witlh stiffness k, and a damper with damping coefficient, c. A bumpy road can be modeled by a sinusoidal up-and-down motion of the wheel y-Ysin(o). From the solution of the equation of motion for this model, the steady-state up-and- down motion of the car is given by x Xsinar-q). The ratio between amplitude X and amplitude Y is given by: m co Assuming m-2000 kg, k 500,000 N/m and c 28x 10' Ns/m, we want to determine the frequency (omega) for which X/Y 0.2. a) Write Matlab code which will solve for the required o using fzero and produce b) Write the Matlab code which will produce the required using roots and then c) I am doing a bisection search and after terations my EMAx is 0.0004883. What nicely formatted results. output the one required root. was the original search width and how many more iterations would I need to do to get Eux0.000001?

Explanation / Answer

m=2000; % m=2000kg
k=500000; % k=500,000 N/m
c=28000; % c=28*10^3 Ns/m
% Now we can define the function as (X/Y)-0.2, like below where w as
% variable
fn=@(w) sqrt((m*c*w^2)/(k*(k-m*w^2)+(w*c)^2))-0.2;
z=fzero(fn,[0 30]); % using 'fzero' function to find the root;

%%
% by converting the polynomial in p1*x^n+p2*x^(n-1)+......+p(n+1)

p=roots([0.04*c^2-0.04*k*m-m*c 0 0.04*k^2])
p=p(p>0);

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