A particle of mass $m = 1$ is connected to one end of a spring of force constant $k = 2$ and natural length $\ell_0 = 0$ . The other end of the spring is fixed at the origin. The ambient gravitational acceleration $g = 10$ is in the negative $z$ direction.

At time $t = 0$ , the particle is at position $(x,y,z) = (1,1,1)$ with velocity $(\dot{x}, \dot{y}, \dot{z}) = (1,2,3)$ .

How far from the origin is the particle at time $t = 15$ ?

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Note:
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This is an ideal problem for practicing Lagrangian mechanics and seeing how it results in Newtonian equations

The answer is 7.793.

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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:

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