Maximum distance a mass falls on a spring before reversing? If a mass, $m$, is released vertically from a spring's un-stretched equilibrium position, what will be the distance the mass will fall before reversing?
I was thinking the point of maximum distance would be when $F_{net} = 0$, so $mg-kx = 0$, and you would solve for $x$.
Is this correct or is there some other solution?
 A: Assuming that the initial speed is $0\; \rm m/s$ then the maximum distance will be $\dfrac{2mg}{k}$. (so your result is almost right).
The reason is that the 'new' equilibrium position will be when the net force on the mass is zero, i.e. when $mg=kx\implies x_{\rm new \;equilibrium}=mg/k$.
However, that is the distance between the unstretched position and the new equilibrium position, which would correspond to the amplitude of your oscillations; therefore, the mass will continue past this new equilibrium.
Thus, to get the maximum distance, multiply the amplitude by $2$ and you obtain
$$x=\dfrac{2mg}{k}$$
A: Use conservation of energy. At the moment the mass stops falling and reverses direction, it has zero kinetic energy, and a potential energy $-mgh$, where $h$ is its original height. This is equal to the energy stored in the spring, $kx^2/2$, where $x = h$ (since the spring is stretched by the amount the mass falls).
Equating this yields the solution $x = 2mg/k$.
A: For a complete solution of the motion, solve the force relationship  $md^2x(t)/dt^2 = -kx(t)$ where $x(t)$ is the time dependent distance of the spring downward from its vertical equilibrium hanging position $a + mg/k$. $a$ is the unstretched length of the spring.  Then find $x$ where $dx(t)/dt = 0$.  See for example Fowles, Analytical Mechanics.
