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 ω(t) and average amplitude A(t) of the system change with time t.…

Answer: 2. Graph pair: \(\omega(t)\) rises and saturates; \(A(t)\) decays

Key idea: the frequency of a spring-mass system is \(\omega=\sqrt{k/m}\), and sand leaking vertically carries no horizontal momentum.

• As mass \(m\) decreases, \(\omega\) increases, and it levels off as the sand runs out (graph rising and saturating).
• The sand leaves vertically, so at the mean position the speed of the box is unchanged. Since \(v_{max}=A\omega\), a larger \(\omega\) means a smaller amplitude \(A=v_{max}/\omega\), so \(A\) decreases.

Only option 2 shows \(\omega\) rising and \(A\) falling. Option 1 shows constant \(A\), option 3 shows rising \(A\), and option 4 shows falling \(\omega\), all of which are wrong.