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amilapsn

  • one year ago

when does the minute hand and hour hand come one on one after 12 o' clock?

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  1. ParthKohli
    • one year ago
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    |dw:1433320516438:dw|It's around 1:05. You can use math to get a more precise answer.

  2. amilapsn
    • one year ago
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    thnk you

  3. ParthKohli
    • one year ago
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    Let's say it's 1 hr and \(x\) min. What does it mean when two hands coincide? How do we put it mathematically? It means that we can make an equation. What do we equate? The angles. We equate the angles. At 1 hour and \(x\) minutes, the angle created by the hour-hand wrt 12 is:\[30^{\circ}\left( 1+\dfrac{x}{60}\right)\]and that created by the minute hand wrt 12 is:\[360^{\circ }\cdot\dfrac{x}{60}\]Equate those two and you get\[x = \frac{60}{11} = 5.4545\cdots\]

  4. ParthKohli
    • one year ago
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    Therefore, the time is 1 hour and 5.4545... minutes. Or around 1:05:27.

  5. ganeshie8
    • one year ago
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    is it possible to model the situation wid vectors

  6. dan815
    • one year ago
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    ye

  7. dan815
    • one year ago
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    we know both angular velocity so

  8. dan815
    • one year ago
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    u can write position vectors for both and equate them together

  9. ParthKohli
    • one year ago
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    oh wow

  10. ganeshie8
    • one year ago
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    t is in hours : minute hand : <cos(2pi*60*t), sin(2pi*60*t)> hour hand : <cos(2pi*1*5m), sin(2pi*1*t)> how to solve

  11. dan815
    • one year ago
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    i think its better to write it out as arc length wrt to time

  12. dan815
    • one year ago
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    and equate the 2 sinusoids and u shud get 12 solutions which is easy to see from the clocks and the periods that are forming

  13. ParthKohli
    • one year ago
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    \[\rm \omega_{second}= 2\pi / T = 2\pi ~rad / 60 sec = \pi/30 ~rad ~sec^{-1}\]\[\rm \omega_{hour} = 2\pi / T = 2\pi ~rad / (60\times 60)sec = \pi /1800 ~ rad~sec^{-1}\]We equate the positions now? (The position of second hand can be a multiple of that of the hour hand.)

  14. dan815
    • one year ago
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    |dw:1433321503977:dw|

  15. ParthKohli
    • one year ago
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    Oh, whoops...

  16. ParthKohli
    • one year ago
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    \[\rm \omega _{hour} = \frac{2\pi}{60 \times 60 \times 24} ~rad ~ sec^{-1}\]

  17. dan815
    • one year ago
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    i guess we can just jump straight to arc angles since arclength/r is a constant

  18. ParthKohli
    • one year ago
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    I mean this isn't any different from what I did earlier. We're equating angles and using fancier terms. :P

  19. dan815
    • one year ago
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    ya its same thing lol, oky lets try to see vectors then

  20. dan815
    • one year ago
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    |dw:1433321910738:dw|

  21. dan815
    • one year ago
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    same simplifcation in polar form

  22. dan815
    • one year ago
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    its not really ideal to solve in cartesian form so

  23. dan815
    • one year ago
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    thats why wolfram giving up

  24. ganeshie8
    • one year ago
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    https://www.desmos.com/calculator/k3awg1pnpm

  25. dan815
    • one year ago
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    you solving sec and min there eh

  26. ganeshie8
    • one year ago
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    Oh right :/

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