Question

Create a MATLAB script

An engineering company is designing and testing a car suspension system. The system has a conventional suspension design, consisting of a shock-absorber and spring at vertical speed (ie. up and down motion) of the wheel, and the spring's resistance is proportional to the vertical displacement of the wheel. The company would like to investigate how the actual performance of the suspension system compares to that of theory. Dynamic theory states that the equation of motion of the wheel in the vertical direction can be expressed as, dy dt2 + man + my = 0 where m is the mass of the wheel, k is the spring stiffness and c is the damping coefficient. This is a second order differential equation, and if for example, the car hits a hole at t-0, such that it is displaced from its equilibrium position with y yo, and dy/dt O, it will have a solution of the form, y(t) -e(Bo cos(pt) +vo (sin(pt) where p= -- provided - 4m2 To analyse the actual performance of the design, the suspension system was built and tested with the following displacements of the wheel recorded during the time period of 2.6 and 3.1 seconds (this dataset is known to contain experimental error): Time (s 2.6 2.65 2.72.75 2.8 2.85 2.92.953 3.05 3.1 Displacement 0.074 0.094 0.106 0.1130.1270.145 0.145 0.148 0.154 0.162 0.158 The design team are interested if the dataset can be characterised by expressions which are less complex that the above theory Plotting the theoretical displacement of the wheel which shows the first 6 roots when the suspension system has the following specifications: o mass acting on each wheel -3.6 x 103 kg o c 1.5 x103 Ns/m o k 1.5x 104 N/m o initial displacement 0.3 m Calculating the time for the first 3 occasions the wheel passes through the equilibrium position (i.e. the root); Plotting a graph of the experimental dataset of the wheel displacement; . Calculating the value of the coefficient of multiple determination (R2) and the standard error (O) when presuming the experimental dataset is Linear (i.e. y-a+ba) Characterised by a saturation growth equation (G.e.y

Answer 1

Code :=

%%%%%%%%%% Theroritical Ploting and Roots %%%%%%%%%%%%%
m = 3.6*10^3; c = 1.5*10^3;   k = 1.5*10^4; y0 = 0.3;
p = sqrt((k/m)-(c^2/(4*m^2)));
n = c/(2*m);
y = @(t) exp(-n*t).*(y0*cos(p*t) + y0*(n/p)*sin(p*t));

t = linspace(0,15);
plot(t,y(t))    % ploting theritical displacement
yline(0);       % Ploting zero line to identify root locations
xlabel("Time [s]"); ylabel("Displacement [m]"); title("Theoritical Plot")
% Form figure it's evident that first 3 roots are near 0.5, 2 and 4
% To calculate exact values we will use fzero function
root1 = fzero(y,0.5);   % 1st root using initial guess of 0.5
root2 = fzero(y,2);     % 2nd root using initial guess of 2
root3 = fzero(y,4);     % 3rd root using initial guess of 4
fprintf("\n First 3 roots : [%.4f %.4f %.4f] sec",[root1,root2,root3])
%%%%%%%%% Experimental Calculations %%%%%%%%%%%%%%%
warning off
Time = [2.6 2.65 2.7 2.75 2.8 2.85 2.9 2.95 3 3.05 3.1];
Disp = [0.074 0.094 0.106 0.113 0.127 0.145 0.145 0.148 0.154 0.162 0.158];
figure(2)
plot(Time,Disp,'*-')

[lin_Fit,gof_LinFit] = fit(Time',Disp','poly1');    % Linear Fit

% defininig fit type for saturation growth
myfittype = fittype('a*x/(b+x)','coefficients',{'a','b'});
[sat_growth_fit,gof_SatFit] = fit(Time',Disp',myfittype);
hold on
time_plot = linspace(2.6,3.1);
plot(time_plot,lin_Fit(time_plot),'LineWidth',1.2);     % Ploting Linear Fit
plot(time_plot,sat_growth_fit(time_plot),'LineWidth',1.2);   % ploting saturation growth fit
xlabel("Time"); ylabel("Displacement");
legend(["Experimental Data","Linear Fit","Saturation Growth fit"],'Location','NorthWest')
title("Experimental Plot");

% R2 and standard error for both
R2_LinFit = gof_LinFit.rsquare;     % Extracting Rsquare from goodness of fit (gof)
Std_err_LinFit = gof_LinFit.rmse;   % Extracting Standard error from gof

R2_SatFit = gof_SatFit.rsquare;
Std_err_SatFit = gof_SatFit.rmse;

fprintf("\n\n For Linear Fit : R2 = %.4f , Standard Error = %.4f", R2_LinFit,Std_err_LinFit);
fprintf("\n For Saturation Growth Fit : R2 = %.4f , Standard Error = %.4f\n", R2_SatFit,Std_err_SatFit);
warning on

Results:-

First 3 roots : [0.8239 2.3711 3.91821 sec F r Linear Fit : R2 = 0.9172 , Standard Error = o.ooee For Saturation Growth Fit : R2 = 0.4295 , Standard Error = 0.0232

Theoritical Plot 0.3 0.2 E 0.1 0 -0.1 -0.2 0.3 0 5 10 15 Time [s]

Experimental Plot 0.18 Experimental Data Linear Fit 0.16 Saturation Growth fit 0.14 0.12 0.1 0.08 0.06 2.6 2.65 2.7 2.75 2.8 2.85 2.9 2.95 3 3.05 3.1 Time