Find the period of small oscillations of a water pogo, which is a stick of mass m in the shape of a box (a rectangular parallelepiped.) The stick has a length $L$, a width $w$ and a height $h$ and is bobbing up and down in water of density $\rho$ . Assume that the water pogo is oriented such that the length $L$ and width $w$ are horizontal at all times. Hint: The buoyant force on an object is given by $F_{buoy} = \rho Vg$, where $V$ is the volume of the medium displaced by the object and $\rho$ is the density of the medium. Assume that at equilibrium, the pogo is floating.

$ \textbf{(A)}\ 2\pi \sqrt{\frac{L}{g}} $

$ \textbf{(B)}\ \pi \sqrt{\frac{\rho w^2L^2 g}{mh^2}} $

$ \textbf{(C)}\ 2\pi \sqrt{\frac{mh^2}{\rho L^2w^2 g}} $

$\textbf{(D)}\ 2\pi \sqrt{\frac{m}{\rho wLg}}$

$\textbf{(E)}\ \pi \sqrt{\frac{m}{\rho wLg}}$
1. **Equilibrium Condition**: At equilibrium, the buoyant force equals the gravitational force acting on the stick. Therefore, we have:
   \[
   F_{\text{buoy}} = mg
   \]
   where \( F_{\text{buoy}} = \rho V g \) and \( V \) is the volume of the water displaced by the stick at equilibrium. Since the stick is floating, the volume of the water displaced is equal to the volume of the stick submerged in water.

2. **Displacement and Buoyant Force**: If the stick is displaced downwards by a small distance \(\Delta h\), the additional volume of water displaced is:
   \[
   \Delta V = w L \Delta h
   \]
   The increase in the buoyant force due to this additional displacement is:
   \[
   \Delta F_{\text{buoy}} = \rho \Delta V g = \rho w L \Delta h g
   \]

3. **Newton's Second Law**: According to Newton's second law, the net force acting on the stick is equal to the mass of the stick times its acceleration. The net force here is the restoring force due to the buoyant force:
   \[
   -\rho w L \Delta h g = m \frac{d^2 (\Delta h)}{dt^2}
   \]
   This can be rewritten as:
   \[
   \frac{d^2 (\Delta h)}{dt^2} + \left( \frac{\rho w L g}{m} \right) \Delta h = 0
   \]

4. **Simple Harmonic Motion**: The above equation is in the form of the differential equation for simple harmonic motion:
   \[
   \frac{d^2 x}{dt^2} + \omega^2 x = 0
   \]
   where \(\omega^2 = \frac{\rho w L g}{m}\). Therefore, the angular frequency \(\omega\) is:
   \[
   \omega = \sqrt{\frac{\rho w L g}{m}}
   \]

5. **Period of Oscillation**: The period \(T\) of the oscillation is given by:
   \[
   T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{\rho w L g}}
   \]

Conclusion:
\[
\boxed{2\pi \sqrt{\frac{m}{\rho w L g}}}
\]