Spring Compressed on Parabola (Part 2)

Spatial coordinates and time derivatives:

$x = x \\ y = x^2 \\ \dot{x} = \dot{x} \\ \dot{y} = 2 x \dot{x}$

Kinetic energy (for two masses):

$E = 2 \frac{1}{2} m v^2 = m (\dot{x}^2 + 4 x^2 \dot{x}^2) = m \dot{x}^2 (1 + 4 x^2)$

Spring potential energy:

$U_S = \frac{1}{2} k (2x - \ell_0)^2$

Gravitational potential energy:

$U_G = 2 \,m g \, x^2$

Lagrangian:

$L = E - U_S - U_G = m \dot{x}^2 (1 + 4 x^2) - \frac{1}{2} k (2x - \ell_0)^2 - 2 \,m g \, x^2$

Equation of motion:

$\frac{d}{dt} \frac{\partial{L}}{\partial{\dot{x}}} = \frac{\partial{L}}{\partial{x}}$

Evaluation:

$\frac{\partial{L}}{\partial{\dot{x}}} = 2 m \dot{x} (1 + 4 x^2) \\ \frac{d}{dt} \frac{\partial{L}}{\partial{\dot{x}}} = 2m \Big [ 8 x \dot{x}^2 + \ddot{x} (1 + 4 x^2) \Big ] \\ \frac{\partial{L}}{\partial{x}} = 8 m x \dot{x}^2 - 2 k (2x - \ell_0) - 4 \,m g \, x$

Left / right equivalence:

$2m \Big [ 8 x \dot{x}^2 + \ddot{x} (1 + 4 x^2) \Big ] = 8 m x \dot{x}^2 - 2 k (2x - \ell_0) - 4 \,m g \, x \\ 8 x \dot{x}^2 + \ddot{x} (1 + 4 x^2) = 4 x \dot{x}^2 - \frac{k}{m} (2x - \ell_0) - 2 g \, x \\ \ddot{x} (1 + 4 x^2) = -4 x \dot{x}^2 - \frac{k}{m} (2x - \ell_0) - 2 g \, x$

Second derivative of x:

$\ddot{x} = \frac{-4 x \dot{x}^2 - \frac{k}{m} (2x - \ell_0) - 2 g \, x}{1 + 4 x^2}$

Numerical integration, starting from appropriate initial conditions, yields a return time of $t \approx 1.215$