Discussion
The period of oscillation of the plane of Foucault's pendulum is :
(\(\phi\) is the latitude)
Answer: B
The period is inversely proportional to sin \(\phi\).
The plane of a Foucault pendulum turns at the angular rate \(\Omega\sin\phi\), where \(\Omega\) is the earth's rotation rate, because only the vertical component of the earth's rotation turns the plane. The time for one complete turn is:
\[ \begin{aligned} T &= \frac{2\pi}{\Omega\sin\phi} \\ &= \frac{24\ \text{h}}{\sin\phi} \end{aligned} \]
At the poles (\(\phi = 90\)°) the plane turns once a day, and at the equator it does not turn at all. The period is not proportional to the latitude itself, nor to \(\cos\phi\).
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