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Probability problem from Griffiths QM

  1. May 15, 2008 #1
    Howdy everybody. I've browsed this forum many a time but this is my first time posting. Anyways. I'm taking QM in the fall for the second time (the sole reason i didn't graduate this semester), and so i'm going through the griffiths book right now so i can ace this class next time around. I'm sort stuck on this problem (1.13). I'm researching for the university right now and so i could go talk to the prof, but (a) i don't want to look like to much of a kiss-ass by letting him know that i am already studying for his class and (b) i'm semi-terrified of this professor. so voila, here i am. any help would great.

    "a needle of length l is dropped at random onto a sheet of paper ruled with parallel lines a distance l apart. what is the probability that the needle will cross a line?"

    in the previous problem. i solved for the expectation values of theta, and its <x> projection. i'm thinking the solution will be a combination of the probabilities for theta, and also for where the needle falls on the y-axis (i.e. right on a line or somewhere between two lines) but i can't quite get a handle on the problem. any ideas?
  2. jcsd
  3. May 15, 2008 #2


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    If the center of the needle lands on a line the probability it will hit it is 1. If it lands half way between the lines the probability is 0. So figure the probability of hitting the lines as a function of the position of it's center relative to the lines. Then integrate that from 0 to l and divide by l.
  4. May 15, 2008 #3
    This is a classic problem, I think you should get a value that is inversely proportional to [tex]\pi[/tex] if you do it right.

    I think once you integrate over the center-of-mass business that Dick mentioned you also integrate over it's orientation measured by an angle [tex]\theta[/tex] as you mentioned, and there are only two orientations where the needle still hits the lines (only looking at the case where the center-of-mass is midway between two lines), but dividing by the integral over all variables, including the angle one will get you that silly pi at the end.
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