What Determines the Loss of Contact in a Mass-Spring System?

[Kreyopresny]

Homework Statement

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.

Homework Equations

ma+bv+kx = 0
x(t) = Acos($$\omega$$t+$$\delta$$)
v(t) = dx/dt = -A$$\omega$$sin($$\omega$$t+$$\delta$$)
a(t) = d²x/dt² = -A$$\omega$$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

 

[Doc Al]

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.

They are not looking for anything quite so complicated, such as describing the trajectory of the small mass as it bounces around inside the box.

Start by assuming that the small mass remains in contact with the box. For a given acceleration, what's the contact force between m and M? What tells you when the small mass is about to lose contact?