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This two properties are given in my module as formulas for clock related problems:

$(1)$ If both the hands start moving together from the same position, both the hands will coincide after $ 65\frac5{11} $ minutes.

$(2)$ Interchangeable positions of minute hand and hour hand occur when the original interval between the two hands is $\frac{60}{13}$ minute spaces or a multiple of this.

Could somebody help me to derive this two formulas?

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up vote 5 down vote accepted

We assume that the clock hands rotate at constant speed. That mathematical model does not describe all clocks well. In some clocks, the minute hand stays at say $17$ for almost $1$ minute, then moves very rapidly to $18$, with an irritating click.

1. For the first problem, it is clear that it will take a little more than an hour, say an hour plus $x$ minutes, where $x$ is well under $60$.

In $12$ hours the hour hand travels $360$ degrees. So it travels $30$ degrees per hour, and therefore $1/2$ degree per minute. In an hour and $x$ minutes the hour hand will have advanced by $30+x/2$ degrees.

In an hour the minute hand travels $360$ degrees, so it travels at $6$ degrees per minute. In an hour and $x$ minutes it will have travelled $360+6x$ degrees. So the minute hand will have advanced by $6x$ minutes. We therefore obtain the equation $$30+\frac{x}{2}=6x.$$ Solve for $x$.

2. We sketch the interchangeability argument. Take some clock time $x$, where $x$ is measured in minutes from $12$:$00$. So for example $1$:$00$ o'clock is called $60$.

At time $x$, the hour hand is $x/2$ degrees clockwise from straight up. The minute hand is at $6x-360m$ degrees clockwise from straight up, for some integer $m$ chosen to make $6x-360m$ less than $360$ degrees.

Take another time $y$ minutes after straight up. Then the hour hand is at $y/2$ degrees from straight up, and the minute hand is at $6y-360n$ for some integer $n$.

Suppose that the hour and minute hands are identical in appearance. For us to be confused between $x$ and $y$, we must have $x\ne y$ and $$\frac{x}{2}=6y-360 n \qquad \text{and} \qquad \frac{y}{2}=6x-360m.$$ We can use these equations to find all times $x\ne y$ such that times $x$ and $y$ are confused when the hands are identical, plus, of course, all times when the hands coincide.

But that's not what we want, since we were asked about $x-y$. Use the two equations to solve for this. We get $13(x-y)=720(m-n)$. Note that we have measured the "times" $x$ and $y$ in minutes after straight up, since that's what we used in part $1$. Convert to minute spaces.

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Start by finding a formula for the position of the hour hand at time $t$, and a formula for the position of the minute hand at time $t$. Do this by noting that the hour hand moves $2\pi$ radians in 12 hours, the minute hand, in 1 hour.

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for 1) the hour hand has an angular velocity of $360$ deg$/60$ min$=6$ deg/min, while the hour hand has an angular velocity of $(360/12)$deg$/60$ min$=1/2$ deg/min. so if they star off equal and go, to meet up we need $6t=1/2t$ to be a multiple of 360 deg, or $5.5t$ divisible by 360. this happens at $t=0,65\text{ and } 5/11$, etc.

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