 Solution

For the spring oscillator,so the total mechanical energy is conserved.

For U-tube liquid oscillation, define positive $y$ as the fall of the left surface, equal to the rise of the right surface. The hydrostatic height difference is $2y$. Moving a volume $Ady$ against this height difference gives $dU=2\rho gAydy$, henceEnergy conservation yields $\rho Al\dot{y}(\ddot{y}+2gy/l)=0$. Equivalently the net pressure force is $-2\rho gAy$, so the equation holds also at turning points:With negligible added volume, $y(0)=0$, $\dot{y}(0)=0.015\mathrm{m}{\mathrm{s}}^{-1}$. ThusThe graph starts at zero with positive slope and first reaches its maximum after $T/4$.

For viscously damped U-tube oscillation, the total damping force is $-\beta l\dot{y}$, not $-\beta \dot{y}$. ConsequentlyThe underdamped initial-value solution is $y=(v_0/{\omega }_d)e^{-\Gamma t/2}\sin {\omega }_{d}t$, where ${\omega }_d=\sqrt{{\omega }_0^2-{\Gamma }^2/4}=6.255{\mathrm{s}}^{-1}$. Setting its amplitude envelope equal to $0.1\mathrm{mm}$ givesThis is the decay of the envelope, not the first time the instantaneous displacement passes through $0.1\mathrm{mm}$.