Spring Orbit

At an arbitrary instant of time, let the coordinates of the particle be $(x,y)$ and its velocity components be $(\dot{x},\dot{y})$ .

The kinetic energy of the particle is:

$T = \frac{1}{2}m\left(\dot{x}^2 + \dot{y}^2\right)$

The potential energy stored in the spring is:

$V = \frac{1}{2}k\left(\sqrt{x^2 + y^2} - L_o\right)^2$

From here, either using Lagrangian mechanics or equating the total time derivative of the total energy to zero yields a set of coupled non-linear differential equations that are solved numerically (using MATLAB) with the given initial conditions.

The result $X \approx 1.77$ (refer code below) is the required answer. What is more interesting about this problem is if one chooses to solve the equation for say 500 seconds. The trajectory of the particle is as shown below. This is a very interesting kind of periodicity one can observe in these elegant patterns. What type of motion does this qualify as?

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clear all
clc

% System Parameters:
K  = 10;
Lo = 1;
m  = 1;

% Time initialisation:
tf = 4;
dt = 1e-4;
t  = 0:dt:tf;
N  = length(t);

% Initial conditions:
dxs(:,1) = [0;3];
xs(:,1)  = [1;0];

for k = 2:N

    x  = xs(1,k-1);
    y  = xs(2,k-1);
    dx = dxs(1,k-1);
    dy = dxs(2,k-1);

    L = sqrt(x^2 + y^2);
    % Elongation of the spring:
    Le   = L - Lo;

    % Acceleration vector:
    ddxs = -(K/m)*(Le/L)*[x;y];

    % Modified Euler Integration:
    dxs(:,k) = dxs(:,k-1) + dt*ddxs;
    xs(:,k)  = xs(:,k-1)  + dt*dxs(:,k);

    % Check for finding the first crossing through the positive X-axis:
    if t(k) >= 0.5 && abs(xs(2,k)) <= 1e-4 && xs(1,k) > 0
        X = xs(1,k);
    end
end