Mass within a box hung from a spring

[Elvis 123456789]

Homework Statement

A block of mass m is placed inside a box of mass M , which is then hung from a spring with spring constant k. The system is pulled down some distance d and released at time t =0. Determine the reaction force between the block and the bottom of the box as a function of time. For what value of d will the block just lose contact with the bottom of the box when at the top of the vertical oscillation?

Homework Equations

F = ma
y = Acos(ωt + φ)
ω² = k/m

The Attempt at a Solution

A free-body diagram will show for mass "m" leads to
F_R - mg = ma -----> F_R = m(g + a)

the acceleration for the block "m" and the box "M" will be the same

so y = Acos(ωt + φ) ----> y = dcos(ωt) where ω = sqrt (k/(M + m) )

differentiating twice gives

a = -dω²cos(ωt)

plugging back into the equation for the reaction force gives

F_R = m(g -dω²cos(ωt))

for the second part set F_R = 0 which gives

d = g/[ω²cos(ωt)]

does this look okay to you guys?

 

[BvU]

No. PF isn't really in the business of stamp-approving homework.

But at t=0 I get something less than mg for F_R, which seems strange to me.

 

[haruspex]

y = dcos(ωt)

Are you defining up or down as positive?