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Help with Buffon's… - QuestionCove
OpenStudy (unklerhaukus):

Help with Buffon's Needle 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 the needle will cross the line?

6 years ago
OpenStudy (unklerhaukus):

ps, i know what the answer is i am just having trouble showing it mathematically

6 years ago
OpenStudy (anonymous):

wow that is a cool problem

6 years ago
OpenStudy (anonymous):

What is the answer and how did you get it? I’m not really sure how to start… is this calc? Statistics? Calcistics?

6 years ago
OpenStudy (unklerhaukus):

i know right answer is π/2, i measured it experimentally and got pretty close , it is statistics but i'm getting the question from a QM text book

6 years ago
OpenStudy (anonymous):

Well… I’m thinking like this: We brake this down into an infinite series of cases, okay, by “angle of the pin relative to the parallel lines.” If the angle is 0°, then there is essentially a 0% chance the needle crosses a line, right? Discounting the width of the lines and needle. At 90° or π/2 radians, the chance is essentially 100%. Now… bear with me… for each of the infinite number of angles *between* 0 and π/2, the chance it lies on a line is equal to the fractional distance between lines that the needle covers, yeah? ... to be continued ...

6 years ago
OpenStudy (anonymous):

which is another way of saying, the ratio a base to the hypotenuse of a right triangle constructed such that the angle between the base and hypotenuse is x, right, where x is the angle at which the pin fell as discussed above.

6 years ago
OpenStudy (anonymous):

SOH CAH TOA, right, we want adjacent and hypotenuse, so CAH. Cosine.

6 years ago
OpenStudy (anonymous):

Now since we have an infinite series… we start to think, calculus! I’m thinking we need to, what, um, take the integral of the value of cos(x) from 0 to π/2? heh? and this gives us… I don’t know, actually. Does this sound right, @UnkleRhaukus? Can you take it from here? (Would love to see the solution)

6 years ago
OpenStudy (anonymous):

Wait, right, we want to get the average, so we integrate cos(x) from 0 to π/2, and then divide by π/2. This would be our answer.

6 years ago
OpenStudy (anonymous):

I think, anyway. On WolframAlpha: http://www.wolframalpha.com/input/?i=%28integral+of+cos%28x%29+from+0+to+π%2F2%29+%2F+%28π%2F2%29

6 years ago
OpenStudy (anonymous):

But that makes the answer 2/π and you are saying the answer is π/2. Man, I wish I weren’t talking to myself.

6 years ago
OpenStudy (anonymous):

Also… QM = Quantum Mechanics? And WHAT THE %#$ my username just changed to bennyv34 for a bit… weird…

6 years ago
OpenStudy (anonymous):

Hey @UnkleRhaukus… I was thinking, π/2 > 1 ∴ it can't be the answer. Probabilities are, of course, 0 ≤ p ≤ 1. So I bet you remembered it wrong and it was 2 / π. Also note taking the integral from 0 to 2π works too, if you divide by 2π, BUT you need to make the integral of abs(cos(x)) instead of just (cos) or it goes negative for a while and your answer erroneously becomes 0.

6 years ago
OpenStudy (unklerhaukus):

Yeah sorry for leaving you like that journeyman55 Quantum Mechanics is certainly weird. And yeah i apologize the answer IS 2/π, (NOT π/2) the integral of the absolute value of cos(x) over 0-π is equal to 1 , just as the integral of the absolute value of cos(x) over 0-π equals 2 i am just not sure how you got the normalizing constant (the divide by 2π bit) PS thankyou so much for taking such interest in this problem

6 years ago
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