Solution: Pendulum Clocks

We publish the solution to the Pendulum Clocks amusement.

The divertimento:

A watchmaker has two misadjusted pendulum clocks. One is one minute ahead a day and the other one is one and a half minutes behind a day. If they are set correctly at 12 midnight today, when will they match again and what time will it really be?

The solution:

The angular speed of the minute hand of the first clock is $$\frac{2 \pi + \pi / (30 \cdot 24)}{60} \frac{rad}{min},$$ while that of the second will be $$\frac{2 \pi – \pi / (20\cdot 24)}{60}\frac{rad}{min}.$$ The two clocks will coincide in the same hour when the difference between the angles that describe one and other minute hand is of 12 hours. That is, that the time \(t\) elapsed until they coincide again fulfils $$ \Big(\frac{2 \pi + \pi / (30\cdot 24)}{60} – \frac{2 \pi – \pi /( 20\cdot 24)}{60}\Big) t = 12 \cdot 2 \pi,$$ so that $$t= \frac{24\cdot 24 \cdot 60 \cdot 60}{5} \, min= 288 \text{ days},$$ which means they will coincide when it is midnight on the 288th day.

The first clock will have travelled $$\frac{2 \pi + \pi / (30 \cdot 24)}{60} \cdot \frac{24 \cdot 24 \cdot 60 \cdot 60}{5} rad,$$ which are \(6916.8= (288 \cdot 24 + 4.8)\) turns, that is to say, it will mark the 4 hours 48 minutes of the day 288. The second clock will have travelled $$\frac{2 \pi – \pi / (20 \cdot 24)}{60} \cdot \frac{24 \cdot 24 \cdot 60 \cdot 60}{5} rad,$$ which are \(6904.8= 287 \cdot 24 + 16.8\) turns, which means it will mark the 16 hours 48 minutes of the day 287.