Problem: Find the period of the motion $T$. Neglect the viscosity of the liquid.

Solution: Denote the density of the liquid by $\varrho$, so the density of the cylinder is $\gamma \varrho$. In equilibrium (i.e. when the net force acting on the cylinder is zero) the immersed part of the cylinder has height $\gamma h$.

Consider the system in a moment when the cylinder is displaced by distance $x_{1}$ downward and moves down with velocity $v_{1}$. As a result of the motion of cylinder the liquid level rises by some height $x_{2}$, and the liquid flows in the gap between the cylinder and beaker with some velocity $v_{2}$ upwards (see Fig. 1).

<img_4468>

Fig. 1

The relation between the aforementioned displacements and velocities are given by the continuity law:

$$
x_{1} s=x_{2}(S-s), \quad v_{1} s=v_{2}(S-s)
$$

In the following we express the potential and kinetic energy of the system. Compared to the equilibrium position the cylinder of mass $\gamma \varrho s h$ sunk by $x_{1}$, while the potential energy change caused by the redistribution of liquid can be imagined as the center of mass of liquid with mass $\varrho s x_{1}$ rises by distance $\gamma h+$ $x_{1} / 2+x_{2} / 2$. Taken the potential energy in the equilibrium state to be zero, the potential energy in the state indicated in the right figure can be written as

$$
E_{\mathrm{pot}}=-\gamma \varrho \operatorname{shg} x_{1}+\varrho s x_{1} g\left(\gamma h+\frac{x_{1}+x_{2}}{2}\right) .
$$

After opening the bracket the first two terms cancel each other:

$$
E_{\mathrm{pot}}=\frac{1}{2} \varrho \operatorname{sg} x_{1}\left(x_{1}+x_{2}\right)
$$

After expressing $x_{2}$ from continuity law and some simplification we get a quadratic expression for the potential energy:

$$
E_{\mathrm{pot}}=\frac{1}{2} \varrho s g x_{1}\left(x_{1}+\frac{s}{S-s} x_{1}\right)=\frac{1}{2} \varrho \frac{s S}{S-s} g x_{1}^{2} .
$$

Now let us calculate the kinetic energy of the system. The contribution from the cylinder is straightforward, $\gamma \varrho s h v_{1}^{2} / 2$, but the motion of the liquid is more complicated.

Note. We may notice that since $s /(S-s)=50$, the speed $v_{2}$ of the liquid in the narrow gap is 50 times larger than the typical speed of the liquid below the cylinder (which can be estimated to be in the range of $v_{1}$ ). And while the mass of the liquid below the cylinder is much larger than the mass of liquid inside the gap (the ratio is ca. 25 if the „few centimeters" in the problem text is taken to be $3.5 \mathrm{~cm}$ ), the kinetic energy is proportional to the square of the velocity, so the kinetic energy of the liquid inside the gap is roughly 100 times larger than the kinetic energy of the liquid below the cylinder.

Since the kinetic energy of the liquid below the cylinder is negligible, we can write the total kinetic energy of the system as:

$$
E_{\text {kin }}=\underbrace{\frac{1}{2} \gamma \varrho s h v_{1}^{2}}_{\text {cylinder }}+\underbrace{\frac{1}{2} \varrho(S-s)\left(\gamma h+x_{1}+x_{2}\right) v_{2}^{2}}_{\text {liquid }} .
$$

Here $x_{1}, x_{2} \ll \gamma h$, so we shall keep only the term containing $\gamma h$ in the second bracket:

$$
E_{\text {kin }}=\frac{1}{2} \gamma \varrho s h v_{1}^{2}+\frac{1}{2} \varrho(S-s) \gamma h v_{2}^{2}
$$

Expressing $v_{2}$ from continuity law gives the following:

$$
E_{\text {kin }}=\frac{1}{2} \gamma \varrho s h v_{1}^{2}+\frac{1}{2} \varrho \gamma h \frac{s^{2}}{S-s} v_{1}^{2}=\frac{1}{2} \varrho \gamma h \frac{s S}{S-s} v_{1}^{2} .
$$

The potential and kinetic energies can be written in the form

$$
E_{\mathrm{pot}}=\frac{1}{2} k_{\mathrm{eff}} x_{1}^{2}, \quad E_{\mathrm{kin}}=\frac{1}{2} m_{\mathrm{eff}} v_{1}^{2},
$$

where the effective spring constant and effective mass are given by

$$
k_{\mathrm{eff}}=\varrho \frac{s S}{S-s} g, \quad m_{\mathrm{eff}}=\varrho \gamma h \frac{s S}{S-s}
$$

So the oscillation is indeed harmonic, thus the angular frequency and the period are:

$$
\omega=\sqrt{\frac{k_{\mathrm{eff}}}{m_{\mathrm{eff}}}}=\sqrt{\frac{g}{\gamma h}}, \quad T=2 \pi \sqrt{\frac{\gamma h}{g}}=0.53 \mathrm{~s} .
$$

Note. The static restoring force, acting on the cylinder is due to the change (relative to the equilibrium position) of the hydrostatic pressure at its lower base:

$$
F=-\operatorname{seg}\left(x_{1}+x_{2}\right)=-\frac{s S}{S-s} \rho g x_{1}
$$

This immediately gives effective stiffness of the system $k_{\text {eff }}=\frac{s S}{S-s} \rho g$.

Alternatively, one may wish to integrate $\int F \mathrm{~d} x_{1}$ to get the potential energy

$$
E_{\mathrm{pot}}=\frac{s S}{S-s} \frac{\rho g}{2} x_{1}^{2}
$$

Answer: 0.53