User:Yeem/clock

Question

Rob Gerds: A slow running clock is correct only ONCE a day. True or False?

Answer

(very rough first draft)

To simply things, since the hour/minute/second hand of an analog clock represents one time, we can model the situation as a single hand that makes two full rotations. For an equation to model the normal movement of an accurate watch hand, let [math]\displaystyle{ f(t) }[/math] be a period function such as

[math]\displaystyle{ f(t) = sin(t) }[/math]

where [math]\displaystyle{ t }[/math] is in units of days. For the broken watch, let [math]\displaystyle{ g(t) }[/math] be a similar function,

[math]\displaystyle{ g(t) = sin(\omega t) }[/math]

where [math]\displaystyle{ \omega }[/math] is a frequency term that represents the relative speed of the broken clock. We will assume that the clocks begin from the same position, such as 12:00. We want to find the times at which the hands of the broken and accurate clocks align, or the values of [math]\displaystyle{ t }[/math] for which

[math]\displaystyle{ f(t) = g(t) }[/math]

[math]\displaystyle{ sin(t) = sin(\omega t) }[/math]

Since we are dealing with periodic functions, we know that there are multiple solutions of the form

[math]\displaystyle{ t = \omega t + 2\pi k }[/math]

where [math]\displaystyle{ k }[/math] is an integer and the period of an accurately moving watch hand in this case is [math]\displaystyle{ 2\pi }[/math]. Rearranging, we arrive at

[math]\displaystyle{ (1-\omega)t = 2\pi k }[/math]

Here, we can constrain [math]\displaystyle{ t }[/math] by focusing with a single day of two full rotations, or two periods totaling [math]\displaystyle{ 4\pi }[/math]. Furthermore, as our broken watch is defined as a slow watch, [math]\displaystyle{ \omega }[/math] must be less than 1. Therefore, our [math]\displaystyle{ (1-\omega)t }[/math] term must be between 0 and [math]\displaystyle{ 4\pi }[/math]. How does the other term vary with [math]\displaystyle{ k }[/math]? We can discard the solution where [math]\displaystyle{ k=-1 }[/math] as we are ignoring negative time; values of [math]\displaystyle{ k }[/math] lower than -1 can also be discarded. For [math]\displaystyle{ k=0 }[/math], we find the trivial solution where [math]\displaystyle{ t=0 }[/math], or when the two hands are initially aligned. For [math]\displaystyle{ k=1 }[/math],

[math]\displaystyle{ (1-\omega)t = 2\pi }[/math]

which depends on the value of [math]\displaystyle{ \omega }[/math]. For [math]\displaystyle{ \omega\le0.5 }[/math],

[math]\displaystyle{ t\le4\pi }[/math]

indicating the presence of a second solution. For [math]\displaystyle{ \omega\gt 0.5 }[/math],

[math]\displaystyle{ t\gt 4\pi }[/math]

which does not provide a solution, as the upper bound of [math]\displaystyle{ t }[/math] is [math]\displaystyle{ 4\pi }[/math]. Simiarly, all values of [math]\displaystyle{ k }[/math] above 2 provide extraneous solutions and can thus be disregarded.

Summary

If the slow clock runs less than half as fast as the accurate clock, it will be correct twice during the day: once at the very beginning, and again sometime after noon. If it is more than half as fast but still slower than the accurate clock, the only time that the hands of the clocks are in alignment is at initial conditions.