$\mathrm{A} C D$ has a diameter of $12.0 \mathrm{~cm}$. If the $\mathrm{CD}$ is rotating at a constant frequency of $3.18 \mathrm{~Hz}$, then the period of the rotational motion is
A) $0.441 \mathrm{~s}$.
B) $0.582 \mathrm{~s}$.
C) $0.698 \mathrm{~s}$.
D) $0.750 \mathrm{~s}$.
E) $0.314 \mathrm{~s}$.

Question

Quizplus · Accepted Answer

The period of the rotational motion is the reciprocal of the frequency, so $T = \frac{1}{f} = \frac{1}{3.18 \mathrm{~Hz}} = 0.314 \mathrm{~s}$.

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