4th oder runge kutta with 2 2nd order ode

Question

Noel Lou 2016-3-3

0
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https://ww2.mathworks.cn/matlabcentral/answers/271317-4th-oder-runge-kutta-with-2-2nd-order-ode

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Hi There!

MX''=Fn(sin θ - uCos θ )

MZ''=Fn(cos θ + uSin θ ) - Mg

Fn,M,θ,u is constant

fn/M = 0.866

M = 6000

θ = 30

u = 0.5774

i split my motion equation into 2 individual 1st ode,

X' = Vx

Z' = Vz

Vx'=[fn*(sin θ - uCos θ )]/M

Vz'={[fn(cos θ + uSin θ )]/M} - g

After trying many tutorials i still don't understand how to implement my motion equation into matlab, i tried to do the following for Vz' motion but it seems wrong, my right hand side of the equation does not have t or y. please help !

clc; % Clears the screen 
clear all; 
h=0.1; % step size 
thete=30; 
g=9.81; 
x = 0:h:3; % Calculates upto y(3) 
y = zeros(1,length(x)); 
y(1) = 5; % initial condition 
F_z = @(t,y) 0.866*(cos(thete)+0.5774*(sin(thete)))-g; % function 
for i=1:(length(x)-1) % calculation loop 
k_1 = F_z(x(i),y(i)); 
k_2 = F_z(x(i)+0.5*h,y(i)+0.5*h*k_1); 
k_3 = F_z((x(i)+0.5*h),(y(i)+0.5*h*k_2)); 
k_4 = F_z((x(i)+h),(y(i)+k_3*h)); 
y(i+1) = y(i) + (1/6)*(k_1+2*k_2+2*k_3+k_4)*h; % main equation 
end 
hold on %Plot graph 
plot(x,y,'+-', 'Linewidth', 1.5, 'color', 'blue')
xlabel('x') 
ylabel('y') 
legend('RK4')