What Is the Equation of Motion for a Mass-Spring System in a Falling Box?

[naggy]

Homework Statement

A mass m is attached to a spring(massless) that is located inside a massless box. The box is falling under gravity. When the box starts to fall the spring is in it's equilibrium position and the box sticks to the ground when it hits it.

-The box is a distance H from the ground
-Spring has spring constant k
-The mass on the spring is m



Find the equation of motion (and initial conditions) when
a)the box is falling and
b)when the box has landed.

Variables
x is movement from equilibrium position of spring
y is distance from ground to mass

Homework Equations

L=KE - PE
or
F=m\ddot{x}

The Attempt at a Solution

I prefer using Lagrangian equations. When the box is falling:
KE= \frac{1}{2}m\dot{x^2}
PE= mgy +\frac{1}{2}kx^2

Now can I connect y(distance from the ground to m) and x(movement from equilibrium position of mass) with y=constant + x and use the Euler lagrange equations?

I'm also not sure on intial conditions, it would be x(0)=0 and x'(0)=0 for the first eq. of motion

when the box lands, maybe x(tH)=H and x'(tH)=sqrt(2gH) ??

 

[Imperitor]

Homework Statement

I'm also not sure on intial conditions, it would be x(0)=0 and x'(0)=0 for the first eq. of motion

Just think what happens when the box hits the ground. It will stop but the mass on the spring will still have same velocity because nothing is stopping it. The only contribution of the fall on the system is an initial velocity. Hope that helps.