find solution of two variable wave equation and plot

I need a code to find the solution o a wave equation f(t)=Acos(wt-φ) when

I want to know what are the amplitude and frequency for a wave to have a rise of 1m in 100 years(876000h), but I have an additional condition. I want this to seem like a linear increase. Please take a look at the attached image for a better understanding. the cosine needs to have zero other condition is f(0)=0

I tried to use “syms” and “solve” to find the answers, but I was only getting w=0 as a solution, which doesn't make sense to me. I tried to do it, but I couldn't succeed

I would also like to see the plot of the resulting equation.

I hope this is clear, but if you have any other questions, please let me know.

Thanks in advance, I really appreciate it.

[Dyuman Joshi]

Can you show the code you have attempted yet?

[Contoso Donoso]

I realized my little code didn't have enough conditions to get the results I wanted, I asked ChatGPT help and came up with this code instead ( I changed hours to seconds):

% Given data points
t_data = [0, 31563000];
f_data = [0, 1];
% Define the function to fit
fun = @(A, w, t) A * cos(t * w - 90) + 0.5;
% Error function to minimize
error_func = @(x) sum((fun(x(1), x(2), t_data) - f_data).^2);
% Constraints
lb = [0.5, 0];             % Lower bounds for A and w
ub = [inf, pi/64];         % Upper bounds for A and w
% Initial guess for A and w
x0 = [1, 1e-7];
% Optimization using fmincon with 'sqp' algorithm
options = optimoptions('fmincon', 'Display', 'iter', 'Algorithm', 'sqp');
x_opt = fmincon(@(x) error_func(x), x0, [], [], [], [], lb, ub, [], options);
% Extract the optimized values for A and w
A_opt = x_opt(1);
w_opt = x_opt(2);
% Print the optimized values
disp(['Optimized A: ', num2str(A_opt)]);
disp(['Optimized w: ', num2str(w_opt)]);
% Plot the fitted curve
t_plot = linspace(0, 31563000, 1000);
f_plot = fun(A_opt, w_opt, t_plot);
plot(t_plot, f_plot);
hold on;
scatter(t_data, f_data, 'r', 'filled');
hold off;
xlabel('t');
ylabel('f(t)');
title('Fitted Harmonic Curve');
grid on;

It is not perfect because it doesn't start at zero as a wish (even though it is set as a condition), but I think it is a pretty good solution, and I think I can work with it.

I'll leave this solution as a reference for others or anyone who wishes to add any comments.

Thanks :)