Solving Spring-Mass System: Force of Reaction as a Function of Time

[thenewbosco]

A spring of stiffness k supports a box of mass M in which is placed a block of mass m. If the system is pulled down a distancde d from the equilibrium position and released, find the force of reaction between the block and the bottom of the box as a function of time. Neglect any air resistance. For what value of d does the block just begin to leave the bottom of the box?

What i considered for this is as follows:
using Newtons second law:
for the overall system,
$$(m+M)a = kd - (m+M)g$$
then for the block of mass m:
$$ma=F_{normal}-mg$$
then i subbed in the second equation for ma in the first, and tried to rearrange for $$F_{normal}$$
however i am not sure how to get the force as a function of time, writing a as dv/dt doesn't help since i cannot rearrange it to integrate i think.

thanks for the help

 

[berkeman]

What would be the equation of motion without the extra block inside the first block?

 

[thenewbosco]

without the extra block it would just be
$$Ma=kd - Mg$$
is this correct? but then how do i proceed

 

[berkeman]

That is just a starting equation describing the forces. What physical motion results from those forces? What equation describes the SHM of the mass once the spring with the d distance preload is released?

 

[thenewbosco]

$$x(t)=d cos (\omega t)$$ is what comes to mind. so i can differentiate this twice to obtain acceleration as a function of time, and plug it into one of the equations i had earlier? is this a correct approach? if so what would do i solve for, thanks much appreciated

 

[thenewbosco]

no more help i can get on this one?