In an oscillating spring mass system, a spring is connected to a box filled with sand. As the box oscillates, sand leaks slowly out of the box vertically so that the average frequency $\omega(t)$ and average amplitude $A(t)$ of the system change with time $t$. Which one of the following options schematically depicts these changes correctly?
Answer: (B) see figure
Frequency: $\omega = \sqrt{k/m}$. As sand leaks out, $m$ decreases, so $\omega$ increases with time.
Amplitude: the sand leaves vertically, with the same horizontal velocity as the box, so it exerts no horizontal force on the box. The box obeys $m(t)\ddot{x} = -kx$, i.e. $\ddot{x} + \omega(t)^2x = 0$ with $\omega$ changing slowly.
For such a slowly changing frequency, the quantity $\omega A^2$ stays constant (adiabatic invariant), so $A \propto \omega^{-1/2}$. As $\omega$ increases, the amplitude decreases.
(Energy view: each grain of sand carries away its own kinetic energy, so the mechanical energy of the box, $\tfrac{1}{2}kA^2$, decreases.)
So $\omega(t)$ increases and $A(t)$ decreases, as in option (2).
Solution by Sreeraj P, M.Sc Physics