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The hands of a clock

  1. Jun 18, 2010 #1
    Here's a cool problem. I thought I came up with it, but it turns out it's been a Google/Microsoft interview question. The solution is pretty cool.

    Name all the times in a day when the minute hand is exactly opposite the hour hand of a clock.

    Or you could do the similar problem, name the times when the hands are on top of each other.

    And before anyone suggests moving this to the math forum, the reason I decided to put it here is that creative use of math is something that physics students need to practice and the guys in the math forum would tear through this with little benefit or amusement.
     
  2. jcsd
  3. Jun 18, 2010 #2
    60 x 24 right?

    Edit: Nevermind, 24. I was thinking of the second hand.

    12:30, 1:35, 2:40, 3:45, 4:50, 5:55, 6:00, 7:05, 8:10, 9:15, 10:20, 11:25
     
    Last edited: Jun 18, 2010
  4. Jun 18, 2010 #3
    it would be 12:32.6, 1:37.6 and so on. At 12:30 the hour hand is exactly between the 12 and 1 so the minute hand would have to be past the six by the same amount.
     
  5. Jun 18, 2010 #4

    K^2

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    That's incorrect. Hour hand doesn't stay at exactly the same spot. It's not exactly on 12 at 12:30. It's between 12 and 1 at 12:30, which means that angle between hands is less than 180°.

    Think harder.
    Still wrong.
     
  6. Jun 18, 2010 #5
    bah! I am never going to get a job at google now :(
     
  7. Jun 18, 2010 #6
    I'm too lazy to work it out but I guess it involves an infinite sequence, since if you start at 12:30, you have to move the minute hand a bit because then the hour hand is half way between 12 and 1, then you have to move the hour hand a little bit more since you just moved the minute hand, and so on.

    OK now I have to try and work it out properly....
     
  8. Jun 18, 2010 #7

    K^2

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    You could. It's an easy series to set up and sum over. 30/12 + 30/12² + 30/12³... But now you'd have to do this for each possible position.

    There is a MUCH easier way.
     
  9. Jun 18, 2010 #8

    tiny-tim

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    Like Hercules never catching the tortoise? :biggrin:

    Physics is equations

    write equations for the positions of the two hands, and then a third equation saying that they're opposite each other. :wink:
     
  10. Jun 18, 2010 #9
    Oh cool, I got it from your hint, so it's at 6/11ths of the hour? (plus 5 minutes for each hour) That didn't agree with my series answer though, which was 2.5+5/11 minutes past half past 12, or 12:32.57.27, vs 12:32.43.64. I can't figure out what I did wrong... :(
     
  11. Jun 19, 2010 #10
    Hang on I just realized... at 6 o'clock the minute hand is on 0, so you can't just add 5 minutes each time. It should be

    t/60 = [(n+6)mod12]/11

    Where t is the time after the hour in minutes, n is the hour.
     
  12. Jun 19, 2010 #11

    K^2

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    It's not 5 minutes each time, because while the minute hand is catching up on these 5 minutes, the hour hand is still moving.

    You're still over-complicating the problem.

    To check yourself int the future, where is the minute hand at 6 o'clock?
     
  13. Jun 19, 2010 #12

    tiny-tim

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    to enlarge on this …

    when the time is h hours and m minutes, where is the hour hand, and where is the minute hand? :wink:
     
  14. Jun 19, 2010 #13

    Vanadium 50

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    I think you're all working too hard. If this were a linear problem, would anyone have any trouble with it?

    How much faster is one hand moving than the other?
     
  15. Jun 20, 2010 #14
    It's much simpler than all of this.
    The hands are straight 11 times in a 12 hour period, so just divide 12 hours by 11 (=1.0909 hours) and the hands will be straight every 1.0909 hours. (=1hr 5min and 27.27seconds)

    Start at 6:00:00, the next one is 7:05:27 and so on.
     
  16. Jun 20, 2010 #15
    I think we should do the second problem first:


    1.The minute when this happens is linearly propotional to the hour number.
    2.Round the clock(from 0:00 to 11:60),this happens 11 times(0,1,2,3....10).


    [tex]\therefore t_{min}=\frac{60n}{11}\ \ \ n\in\mathbb{Z}|\left[0,11\right][/tex]





    0:00,1:0545,2:109,3:1636,4:218..........
    Then first problem:
    6:00,7:0545,8:109,9:1636,10:218........
     
  17. Sep 21, 2010 #16
    Doesn't it just go past once every hour? So 24 times in a day?
     
  18. Sep 21, 2010 #17
    Not quite. If you start with the hour of, say, 1am, you know that the hands will be opposite to each other somewhere in the ballpark of 01:35 to 01:40. Where exactly you can determine with some extra math, but you KNOW that somewhere between 01:35 and 01:40, the two hands will be directly opposing. Same thing with the hour of 2am-- the hands will be opposite somewhere between 02:40 and 02:45.

    You can keep this up for a while, until you reach the hour of 5am. Between 05:00:00 and 05:59:59, the hour hand and minute hand are NEVER in opposition. Turns out they line up at PRECISELY 06:00:00.

    So for every hour, the hour hand and minute hand are directly opposite once-- EXCEPT for the hour of 5 (5am and 5pm). Therefore, there are only 22 times of the day when this takes place!

    Hence, it's pretty easy to see that the hands line up at direct opposite positions every 3927 and 3/11 seconds (3927.2727272727...). And since we know they're directly opposite at exactly 6:00 (am or pm), you can work backwards or forwards from there to figure out the exact times of day that this happens.

    DaveE
     
    Last edited: Sep 21, 2010
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