# Clock hands!

1. Feb 23, 2007

### Mk

How many times per day (00:00-24:00) do the hour and minute hand form a straight line together? Would this be the same for when they overlap? Say, the minute hand goes pi/1 hr and the hour hand goes (pi/12)/1 hr. Then what?

2. Feb 23, 2007

### Werg22

Construct a function and find it's solutions according to what you are looking for. As a hint, both hand start on 12 at 0:00. Now with the frequency of each hand, can you find how does the angle between them change with time?

I've put a "solution" here. I covered it though, it's only there if you really have no reason to try solving the problem by yourself (maybe a homework?). You can highlight it to see it, but I recommend to try to come up with the solution yourself.

Assume we are dealing with a unit circle. The velocity of a hand, actually equal to it’s angular velocity because we’re dealing with a unit circle, is equal to 2π/T, where T is the period of the hand. For the minute hand we have v = 2π and for the hour hand we have v2 = 2π/60. Now note that the distance along the circle covered by a hand is equal to v*t, where t is the time elapsed in minute. Moreover, the distance between the two hands is given by the difference of the respective covered distances. Since this is on a unit circle, this distance is equal to the angle that separates the two hands. We end up with something like this as a function θ(t) = 2π*t - 2π/60*t or θ(t) = 59*π*t/30. Now, it’s obvious that if θ is to be a multiple of π, t = k*30/59, where k is any integer number. However we have to be careful. If θ is a multiple of 2π, the hands are certainly not opposite, they are overlapping. This is equivalent to saying that the value k has to be odd. Let’s solve k*30/59 = 1440 (there are 1440 minutes in day). We have k = 2352. So there are 2352 instance of the day where is a multiple of π. Now, since of have the requirement of k being odd, there 2352/2 = 1176 solutions of interest.

Last edited: Feb 23, 2007