Time — Question 7

Question 7

A repeated image can hide complete revolutions

A disk rotates uniformly counterclockwise. A marked radius starts at angle zero at $t=0$. A camera samples at $0.100\ \mathrm{s}$ intervals and records successive angles $0^\circ$, $72^\circ$, $144^\circ$, $216^\circ$, and $288^\circ$, as shown. Each image records orientation only, so adding any whole number of revolutions produces the same appearance.

An independent bound establishes $0<f<15\ \mathrm{rev/s}$, where $f$ is the constant rotation frequency. Take the sample times and angles as exact in this model.

1. Find every rotation frequency consistent with the observed increments and the independent bound. Explain why dividing $72^\circ$ by $0.100\ \mathrm{s}$ does not establish a unique answer.

2. Find the period associated with each allowed frequency. Distinguish the interval between camera samples from the disk’s physical period.

3. An extra instantaneous image is taken at $t=0.050\ \mathrm{s}$. Predict its orientation for each candidate and explain whether this distinguishes them.

4. Would the extra observation necessarily distinguish every possible frequency if the independent upper bound were removed? Give a counterexample using a higher frequency.

Question 7 - Solution

A repeated image can hide complete revolutions

(a) Retain the unseen whole turns.

The displayed increment is $72/360=0.2$ revolution. In each $0.100\ \mathrm{s}$ interval the disk could complete $k+0.2$ revolutions, where $k$ is any nonnegative integer. Hence $$f=\frac{k+0.2}{0.100}\ \mathrm{rev/s}=(2+10k)\ \mathrm{rev/s}.$$ With $0<f<15\ \mathrm{rev/s}$, the candidates are $\boxed{2\ \mathrm{rev/s}}$ and $\boxed{12\ \mathrm{rev/s}}$. The naive division chooses $k=0$ without evidence excluding one full hidden turn per frame interval.

(b) Compute periods from the candidate physical rates.

$$P_1=\frac12\ \mathrm s=0.500\ \mathrm s,\qquad P_2=\frac1{12}\ \mathrm s\approx0.08333\ \mathrm s.$$

The periods are $\boxed{0.500\ \mathrm{s}}$ and $\boxed{1/12\ \mathrm{s}\approx0.08333\ \mathrm{s}}$. The camera’s $0.100\ \mathrm{s}$ sampling interval is neither of these physical periods. It controls when orientation is observed, not how fast the disk rotates.

(c) Sample halfway between the original observations.

$$\theta(0.050\ \mathrm s)=360^\circ f(0.050\ \mathrm s)\pmod {360^\circ}=\begin{cases}36^\circ&f=2\ \mathrm{s^{-1}},\\216^\circ&f=12\ \mathrm{s^{-1}}.\end{cases}$$

At 0.050 s, the 2 rev/s disk completes 0.1 revolution and appears at $\boxed{36^\circ}$. The 12 rev/s disk completes 0.6 revolution and appears at $\boxed{216^\circ}$. These orientations differ, so the additional image resolves the two candidates under the stated bound.

(d) An observation can resolve one ambiguity without all ambiguities.

A 22 rev/s disk fits the original samples because it makes 2.2 turns per 0.100 s. At 0.050 s it makes 1.1 turns and appears at $36^\circ$, just like the 2 rev/s disk. Removing the frequency bound therefore restores ambiguity. A sampling design needs relevant prior bounds or further observations, not only a visually smooth sequence.