3D Spring Dynamics

At any instant of time, the kinetic energy of the system is:

$\mathcal{T} = \frac{\dot{x}^2+\dot{y}^2+\dot{z}^2}{2}$

The potential energy is the sum of the gravitational and spring potential energy:

$\mathcal{V} = 10z + x^2+y^2+z^2$

Applying Lagrange's equations from this point:

$\frac{d}{dt}\left(\frac{\partial \mathcal{T}}{\partial \dot{x}}\right)- \frac{\partial \mathcal{T}}{\partial x} + \frac{\partial \mathcal{V}}{\partial x}=0$ $\frac{d}{dt}\left(\frac{\partial \mathcal{T}}{\partial \dot{y}}\right)- \frac{\partial \mathcal{T}}{\partial y} + \frac{\partial \mathcal{V}}{\partial y}=0$ $\frac{d}{dt}\left(\frac{\partial \mathcal{T}}{\partial \dot{z}}\right)- \frac{\partial \mathcal{T}}{\partial z} + \frac{\partial \mathcal{V}}{\partial z}=0$

Evaluating this results in a system of three linear de-coupled differential equations:

$\ddot{x} +2x=0$ $\ddot{y} +2y=0$ $\ddot{z} +2z=-10$

General solution:

$x(t) = A_1 \cos(t \sqrt{2}) + B_1 \sin(t \sqrt{2})$ $y(t) = A_2 \cos(t \sqrt{2}) + B_2 \sin(t \sqrt{2})$ $z(t) = A_3 \cos(t \sqrt{2}) + B_3 \sin(t \sqrt{2})-5$

The unknown constants can be solved by applying the initial conditions. The required answer is:

$D = \sqrt{(x(15))^2 + (y(15))^2 + (z(15))^2} \approx 7.793$

A bit late on this one, but I solved it. I have nothing new to add, but here's a graph showing the position versus time:

This kind of problem is a nightmare to do using Newtonian mechanics, but much much easier to do using Lagrangian mechanics. With 3 generalised coordinates, $x, y, z$ , we should solve the Euler-Lagrange equation for all three variables.

The good thing about this system is that it isn't coupled (thankfully). I solved by numerically integrating the equation, although you can do it using the traditional periodic solution to the second-order system.