= Solution
For the <spring> oscillator,
$$
\frac{d}{dt}\left(\frac12m\dot x^2+\frac12kx^2\right)
=\dot x(m\ddot x+kx)=0,
$$
so \b[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 $A\,dy$ against this height difference gives $dU=2\rho gAy\,dy$, hence
$$
\boxed{U=\rho gAy^2,\qquad K=\frac12\rho Al\dot y^2.}
$$
<Energy> 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:
$$
\boxed{\ddot y+\omega_0^2 y=0,\qquad
\omega_0=\sqrt{\frac{2g}{l}},\qquad
f_0=\frac1{2\pi}\sqrt{\frac{2g}{l}}.}
$$
With negligible added volume, $y(0)=0$, $\dot y(0)=0.015\,\mathrm{m\,s^{-1}}$. Thus
$$
\boxed{y(t)=(2.395\,\mathrm{mm})\sin[(6.264\,\mathrm{s^{-1}})t],\qquad
f_0=0.997\,\mathrm{Hz},\quad T=1.003\,\mathrm s.}
$$
The graph starts at zero with positive slope and first reaches its maximum after $T/4$.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1-ia2014-u-tube.png]
{title=Original U-tube displacement solutions: the requested frictionless oscillation and the viscously damped response with exponential envelopes}
For <viscously damped U-tube oscillation>, the total <damping> <force> is $-\beta l\dot y$, not $-\beta\dot y$. Consequently
$$
\ddot y+\Gamma\dot y+\omega_0^2y=0,\qquad
\Gamma=\frac{\beta}{\rho A}=0.6667\,\mathrm{s^{-1}},\qquad
\boxed{Q=\frac{\omega_0}{\Gamma}=9.40.}
$$
The underdamped initial-value solution is $y=(v_0/\omega_d)e^{-\Gamma t/2}\sin\omega_dt$, where $\omega_d=\sqrt{\omega_0^2-\Gamma^2/4}=6.255\,\mathrm{s^{-1}}$. Setting its <amplitude> envelope equal to $0.1\,\mathrm{mm}$ gives
$$
\boxed{t=\frac2\Gamma\ln\!\left(\frac{v_0/\omega_d}{0.1\,\mathrm{mm}}\right)=9.53\,\mathrm s.}
$$
This is the decay of the envelope, not the first time the instantaneous displacement passes through $0.1\,\mathrm{mm}$.
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