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. Which one of the following options schematically depicts these changes correctly?

The correct answer is the first option (Option 1), which depicts that the average frequency $\omega \left(t\right)$ increases with time $t$ and the average amplitude $A\left(t\right)$ decreases with time $t$.

Step-by-Step Explanation:

1. Behavior of the angular frequency $\omega \left(t\right)$:
For a spring-mass system with spring constant $k$ and instantaneous mass $m\left(t\right)$, the angular frequency is given by the relation:

$\omega \left(t\right)=\sqrt{\frac{k}{m\left(t\right)}}$

As the sand slowly leaks out of the box, the total mass $m\left(t\right)$ of the oscillating system decreases over time. Because the frequency is inversely proportional to the square root of the mass, a decrease in mass $m\left(t\right)$ results in an increase in the angular frequency $\omega \left(t\right)$ over time.

2. Behavior of the amplitude $A\left(t\right)$:
Since the sand leaks out vertically (with zero horizontal velocity relative to the box), it carries away its instantaneous horizontal velocity and kinetic energy, but does not exert any horizontal force on the remaining box-sand system.
The horizontal equation of motion is:

$m\frac{d^{2}x}{d{t}^2}+kx=0$

For a system whose parameters vary slowly over time (adiabatic approximation), the energy $E$ of the system changes at a rate:

$\frac{dE}{dt}=\frac{1}{2}\frac{dm}{dt}{v}^2$

Averaging this expression over one complete cycle of oscillation:

$⟨\frac{dE}{dt}⟩=\frac{1}{2}\frac{dm}{dt}⟨v^2⟩$

Since the average kinetic energy is half of the total energy, we have $\frac{1}{2}m⟨v^2⟩=\frac{E}{2}$, which gives $⟨v^2⟩=\frac{E}{m}$. Substituting this back into the averaged rate of change of energy yields:

$\frac{dE}{dt}=\frac{1}{2m}\frac{dm}{dt}E$

Integrating this differential equation gives:

$E\propto m^{1/2}$

Since the mechanical energy of a spring-mass oscillator is related to its amplitude by $E=\frac{1}{2}k{A}^2$, we have:

$A^2\propto m^{1/2}\Rightarrow A\propto m^{1/4}$

Since the mass $m\left(t\right)$ decreases with time $t$, the average amplitude $A\left(t\right)$ also decreases with time.

Conclusion:
As the sand leaks out, the frequency $\omega \left(t\right)$ increases, while the amplitude $A\left(t\right)$ decreases. This behavior is correctly depicted by the graphs in the first option (Option 1).