I have a spring of length $L$ when unstretched with one end fixed to the roof. At the lower end I place a mass $m$ and drop gently so it stretches by $x_1$ at equilibrium so that $mg=kx_1$. Now I place another mass $m$ on it and let it drop so that it oscillates. What will be the amplitude of this oscillation?
The equilibrium of the combined masses would be at $x_2$ below the initial equilibrium position and it can be shown that $x_1 = x_2= x$. By intution I feel it should oscillate with amplitude $x$ about the second equilibrium position. But when I try to work it out on energy conservation I get a different answer.
The elastic potential energy of the spring at the farthest position is $½k(x+y)^2$ where $y$ is the maximum position below initial equilibrium position. The changes in potential energy that are converted to elastic spring energy are $[mg(x_1+y)+ mgy]=mg(x_1+2y)$. Equating these and putting $mg= kx$ and solving the quadratic for $y$ gives $y=2.414x$ since $y$ cannot be negative.
I would be grateful if someone could point out where I am going wrong.