1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Time lost or gained by clock

  1. Dec 9, 2016 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    3. The attempt at a solution

    I usually have trouble with these clock problems . In this problem I can certainly rule out option d) . I considered a simple situation in which initially the clock is at 12' o clock . After 65 minutes , the minute hand would be at 1'o clock mark wheras the hour hand would be somewhere between 1 and 2 . This means when the two hands meet the actual time elapsed should be more than 65 minutes .

    Now the problem is how much time the clock has gained . Consider the hour hand . It covers 5 minute units in 60 minutes time . So, the time gained should be less than (5/60) times 86400 seconds . I am sure I am thinking incorrectly .

    Please help with the problem .

    Thanks .

    Attached Files:

  2. jcsd
  3. Dec 9, 2016 #2
    Yes, it's easy to get lost in this one. I think there must be more than one way to think about it, too. My approach was to think about how many times the minute hand catches up to the hour hand in one "12 hour" period.
  4. Dec 9, 2016 #3
    Hello ,

    Did you get one of the answers ? Could you elaborate your approach.
  5. Dec 9, 2016 #4
    I think the first thing should be to write an equation for the angle of rotation for each of the clock hands as a function of time (seconds or minutes). Then it seems to make sense to me to start at 1:00. At 1:00, write an equation for how much the minute hand moves relative to how much the hour hand moves until they align.

    After I started writing, but before I posted this message, I read @Cutter Ketch approach. I would recommend using his approach. The math is much simpler. :)
  6. Dec 9, 2016 #5
    The minute hand catches up with hour hand 11 times i.e after every 43200/11 = 3927 seconds . 65 minutes = 3900 seconds . This means the clock gains 27 seconds per hour .Or in other words 648 seconds in a day .So the answer should be option b) .

    Do you agree ?
  7. Dec 9, 2016 #6
    43200 is the number of seconds in 12 hours. So dividing that number by 11 gives the number of seconds in 12/11 of an hour, or 3927.2727 seconds. And (65)(60) = 3900 is the measured number of seconds (measured by a perfect clock) in 12/11 of an hour. So when you subtract those two numbers (3927.27 - 3900), the result does not give the clock error per hour. It gives the clock error per 12/11 of an hour. I hope I said that right.

    Edit: Actually I think this problem is easier to work using minutes.
  8. Dec 9, 2016 #7
    Almost correct. (and in multiple choice questions almost can be good enough). However, still wrong in two ways. First, it gets there 27 seconds early, but a true 65 minutes hasn't passed yet. After a true 65 minutes the margin will be slightly larger. Also, it's every 65 minutes, not every hour, so you scaled badly. Fix those things and your approach is correct.

    I had a slightly different approach. The hands meet 11 times per 12 hours as you note. So if the hands meet in 65 minutes, then the minute hand covered 1 hour in 11 * 65 / 12 minutes. Divide that into 24 hours and that is how many hours the false clock shows in 24 true hours.
  9. Dec 9, 2016 #8
    I don't know why I felt better working in hours. Your approach can quickly produce the right answer. You need the ratio false clock per true clock. You have that the false clock showed 3927 when only 3900 had truly passed. Take the ratio and multiply by 24 hours. (But don't use the rounded value of 3927. Keep all the digits)
  10. Dec 9, 2016 #9
    Another approach -

    Hour hand meets minutes hand after every 65 minutes . This means actual time elapsed when the two hands meet for the 11th time is 65 × 22 × 60 = 85800 seconds . But time elapsed as per the clock is 86400 seconds .

    The difference between the two is 600 seconds i.e option C) :rolleyes: .

    Do you agree/disagree with this line of reasoning ?
  11. Dec 9, 2016 #10
    Everything except "option c". As I hinted, your last try although slightly flawed got the right multiple choice answer. Here as in the previous case you've lost track of which is true and which is false. 600 seconds is how far ahead the false clock is when IT says 24 hours have passed. However it is fast, so 24 hours haven't actually passed yet. By the time 24 hours have truly passed it will be even a little further ahead.

    To get exactly how far ahead, you need the ratio of false time / true time for any interval you know. Multiply that by 24 true hours and you get how far ahead false time is after 24 true hours.
  12. Dec 9, 2016 #11
    I guess this is perhaps the exact reasoning problem setter had in his mind :smile: . This also looks to be the fastest way to arrive at the answer .

    Impressive analysis Cutter Ketch !

    As I said in OP I mess up big time in clock problems .
  13. Dec 9, 2016 #12
    I have not been able to understand your line of reasoning . Sorry .

    Is it similar to Cutter Ketch's reasoning ?
  14. Dec 9, 2016 #13
    Well, maybe it's just me, but I thought this was tricky and easy to get turned around. I certainly screwed up a few times.
  15. Dec 9, 2016 #14
    As I read post #10 by @Cutter Ketch I realized I had made the same mistake - that I had not calculated how far off the clock was in one day; I had calculated how far off the clock was in what the errant clock reported to be one day - which actually was only 23 hours and 50 minutes. That is why I came up with the wrong answer of 600 seconds. The clock reported that 24 hours had elapsed, when in reality, only 23 hours and 50 minutes had elapsed. So as Cutter pointed out, in the additional 10 minutes, more error is added to the 600 seconds.
    The ratio of reported time to actual time is 1440/1430.
    So for an actual time of 1440 minutes, the reported time would be (1440)(1440/1430) = 1450.07 minutes.
  16. Dec 9, 2016 #15
    @haruspex , would you like to add/point out something in this discussion .
  17. Dec 9, 2016 #16


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The least head-hurting way might be to figure out how slow the fast clock thinks the ideal clock is, then invert.
    According to the fast clock, in 24 hours its hands meet up 22 times, so the ideal clock measures 65*22 minutes.
    According to the fast clock, in 24 * 24*60/(65*22) hours the ideal clock measures 65*22 * 24*60/(65*22) minutes = 24 hours.
    According to the fast clock, in 24 hours and 10.07 minutes the ideal clock measures 24 hours.
  18. Dec 10, 2016 #17
    From post#7
    Would you like to review this ? I think true 65 minutes have passed .

    ##\frac{27.2727}{(65 × 60)} × 86400 = 604.2 sec##
    Last edited: Dec 10, 2016
  19. Dec 10, 2016 #18
    That's it!
  20. Dec 10, 2016 #19

    But you said in post#7 that when the clock was ahead by 27 seconds , true 65 minutes haven't passed :rolleyes: .

    I think true 65 minutes had passed .The previous post showed how 65 minutes gives correct answer .
  21. Dec 10, 2016 #20
    Here's what I got:
    When the clock is ahead by 27 seconds, the true time is 64 minutes and 21 seconds.
    When the clock is ahead by 27.272727272 seconds, the true time is 65 minutes and 0 seconds.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted