Kevin S. Huang

2007: Problem 25

Topic: Oscillatory Motion
Concepts: Buoyant force, Simple harmonic motion

Solution:

Starting at equilibrium, suppose we push the pogo down by a distance $x$ . Then the buoyant force increases by the weight of the water displaced:

F_b=\rho V_{\text{dis}}g=\rho (Lwx)g

Applying Newton’s 2nd law to the pogo (taking downwards to be positive), we have

F_{\text{net}}=-F_b=-\rho Lwgx=ma

By definition of acceleration, $a=\ddot{x}$ so

m\ddot{x}=-\rho Lwgx

\ddot{x}=-\frac{\rho Lwg}{m}x

This is of the form of simple harmonic motion $\ddot{z}=-\omega^2z$ so we can identify the angular frequency $\omega$ as

\omega=\sqrt{\frac{\rho Lwg}{m}}

The period is given by

T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{m}{\rho Lwg}}

so the answer is D.