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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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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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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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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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