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Help with loop and tracking values.

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John Biscanti
John Biscanti 2020-9-19
关闭： John D'Errico 2020-9-19
Hello,
I have an array full of acceleration values called B. B is calculated using the function at the bottom of the page. The rows of B correspond to the time and the columns correspond to a variable called zeta. Acceleration is calculated using these values of zeta and t that increment each iteration.
I need to figure out how to find the values of zeta which make the acceleration values in B less than or equal to 0.05 when t is greater than or equal to 0.05
How would I do this? Note: each row the time increases by 0.001 seconds, and each column of zeta increases by 0.1. Thank you.
I am specifically talking about the second for loop with ii:n2 and jj being the length of zeta
Code:
clc
clear
T0=1;
tf = .01;
h= 0.001;
y0 = [0, 0];
n = int32(ceil(tf/h))+1; % determine the number of steps
t = linspace(0, tf, n); % Generate a time step vector
y = zeros(n,2); % Allocate the array for numerical solutions
y(1,:) = y0'; % The first row of y is the initial condition
A = zeros(n,21); % Allocate the array of acceleration values
%----------------------------------------------------------------------
% TODO: Implement the loop
%----------------------------------------------------------------------
zeta = 0:0.01:0.05;
for ii = 1:(n)
for jj = 1:length(zeta)
k1 = func2(t,zeta(jj),y(ii,:))';
k2 = func2(t+h/2,zeta(jj),(y(ii,:)+ h/2*k1))';
y(ii+1,:) = y(ii,:) + h*k2;
A(ii,jj) = -10000*y(ii,1)-200*zeta(jj)*y(ii,2)+1;
end
end
%Amax = max(A,[],'all') %Amax
%A(11,6) %sanity check
T02=1;
tf2 = .2;
h2= 0.001;
y02 = [0, 0];
n2 = int32(ceil(tf2/h2))+1; % determine the number of steps
t2 = linspace(0, tf2, n2); % Generate a time step vector
y2 = zeros(n2,2); % Allocate the array for numerical solutions
y2(1,:) = y0'; % The first row of y is the initial condition
B = zeros(n2,21); % Allocate the array of acceleration values
zeta2 = 0:0.1:2;
for ii = 1:(n2)
for jj = 1:length(zeta2)
k1 = func2(t2,zeta2(jj),y(ii,:))';
k2 = func2(t2+h2/2,zeta2(jj),(y(ii,:)+ h2/2*k1))';
y(ii+1,:) = y(ii,:) + h2*k2;
B(ii,jj) = -10000*y(ii,1)-200*zeta2(jj)*y(ii,2)+1;
end
end
end

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