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202306021010 Pastime Exercise 000

To tell time on an analog clock, you look at where the hands are pointing.

The short/small hand tells you the hour, the long/big hand tells you the minute of the current hour, and the thinnest hand indicates the seconds of the current minute.

Extracted from Malcolm McKinsey. (2023). How To Tell Time; Read An Analog Clock

Roughwork.

Define two periodic functions by

\begin{aligned} \theta_\textrm{h}(t) &:[0,12)\in [T\textrm{ (in hours)}]\rightarrow [0,2\pi )\in [\Theta\textrm{ (in radians)}] \\ \theta_\textrm{m}(t) & :[0,1)\in [T\textrm{ (in hours)}]\rightarrow [0,2\pi )\in [\Theta\textrm{ (in radians)}] \\ \end{aligned}

for angular displacement \theta of the hour hand \textrm{h} and the minute hand \textrm{m} as of time t, with

\begin{aligned} \theta_\textrm{h}(t+T_\textrm{h}) & =\theta_\textrm{h}(t)\textrm{ where }T_\textrm{h}=12\,\mathrm{hr} \\ \theta_\textrm{m}(t+T_\textrm{m}) & =\theta_\textrm{m}(t)\textrm{ where }T_\textrm{m}=1\,\mathrm{hr} \\ \end{aligned}

such that

\begin{aligned} \dot{\theta}_\textrm{h} & = \frac{\mathrm{d}}{\mathrm{d}t}(\theta_\textrm{h}) = \frac{2\pi}{T_\textrm{h}} =\textrm{Const.}\\ \dot{\theta}_\textrm{m} & =\frac{\mathrm{d}}{\mathrm{d}t}(\theta_\textrm{m}) = \frac{2\pi}{T_\textrm{m}}=\textrm{Const.} \\ \end{aligned}

observing an isomorphism \varphi between \theta_\textrm{m}(t) and the restriction \theta_\textrm{h}\big|_{[0,1)}(t) of \theta_\textrm{h} to [0,1) at one-hour time intervals:

\begin{aligned} & \quad\enspace \theta_\textrm{m} :[0,1)\rightarrow [0,2\pi ) \textrm{ by }\theta_\textrm{m}(t)=\bigg(\frac{2\pi}{T_\textrm{m}}\bigg) t \\ & \cong \theta_\textrm{h}\big|_{[0,1)} :[0,1)\rightarrow\bigg[ 0,\frac{\pi}{6}\bigg) \textrm{ by }\theta_\textrm{h}(t)=\bigg(\frac{2\pi}{T_\textrm{h}}\bigg) t \\ \end{aligned}

Then, let

\begin{aligned} \textrm{HH12} & =\{00,01,02,\dots ,09,10,11\} \\ \textrm{MI} & = \{00,01,02,\dots ,57,58,59\} \\ \end{aligned}

so that digital format of clocks is given as the Cartesian product:

\textrm{hh:mm}=\textrm{HH12}\times \textrm{MI}.

Synchronised at the start time:

\theta_\textrm{h}(\textrm{00:00})=\theta_\textrm{m}(\textrm{00:00})=0.

(a) Find the time(s) exact to minutes when the hour hand and the minute hand are perpendicular to each other, i.e.,

\displaystyle{|\theta_\textrm{h}-\theta_\textrm{m}|=\frac{\pi}{2}\textrm{ \scriptsize{OR} }}\frac{3\pi}{2}

(b) Find also the time(s) exact to minutes when the hour hand and the minute hand are parallel to each other, i.e.,

|\theta_\textrm{h}-\theta_\textrm{m}|=0\textrm{ \scriptsize{OR} }\pi

(c) Is the aforementioned isomorphism \theta_\textrm{h}\stackrel{\varphi}{\cong}\theta_\textrm{m} perturbed when the clock is losing or gaining time \delta t?

This problem is not to be attempted.