A small mass m is in a box of mass M that is attached to a vertical spring of stiffness constant k. When displaced from its equilibrium position y0 to y1 and released, it executes simple harmonic motion. Calculate the reaction between m and M as a function of time. Does the
mass m always stay in contact with the box? If not, what determines that it will lose contact with the box? Calculate the value of h as measured from the equilibrium position for which it loses contact.
ma+bv+kx = 0
x(t) = Acos([tex]\omega[/tex]t+[tex]\delta[/tex])
v(t) = dx/dt = -A[tex]\omega[/tex]sin([tex]\omega[/tex]t+[tex]\delta[/tex])
a(t) = d2x/dt2 = -A[tex]\omega[/tex]2cos([tex]\omegat+\delta/tex])
The Attempt at a Solution
Think this is a bit off, since I know the weight doesn't stay in contact with the box, or move like it at all, but I'm not sure how to describe it's movement, especially when the box changes directions and it continues until hitting the other side of the box and being forced to change direction.
ma+bv+dx = Ma+bv+dx