Displacement of zigzaging dust particle

  • Thread starter Pseudopro
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  • #1
Pseudopro
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Homework Statement


A particle of dust is bombarded by air molecules and follows a zigzag path at constant speed v.
(a)Assuming each step has a length d, find the distance traveled by the dust particle in time t.
(b)What is the length of the displacement vector after N steps where N is large? Assume that each step is taken in a random direction on the plane. (This problem assumes you are familiar with the scalar product of two vectors.

Homework Equations


v2=(v12+v22+...+vN2)/N


The Attempt at a Solution


The answer to part (a) seemed quite obvious: vt.
However, part (b) isn't so easy. I've been able to get the answer d[tex]\sqrt{N}[/tex] by doing vt=t x sqrt[(v12+v22+...+vN2)/N] but this is getting the answer for the sake of it. My method doesn't actually make any sense because it can cancel out to nothing. Please help!
(I believe this is called Brownian motion?) (the thing I don't get is how do I use dot product in this instance?)
 
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Answers and Replies

  • #2
tiny-tim
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Hi Pseudopro! :smile:

Hint: if you add two vectors whose lengths are fixed as a and b, but whose directions are random,

what is the average magnitude of the sum of those vectors? :wink:
 
  • #3
Pseudopro
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(a+b)/2?
 
  • #4
tiny-tim
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uhh? :confused:

call the angle θ, and work out the magnitude! :rolleyes:

(and then average over θ)
 
  • #5
Pseudopro
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Sorry, I don't think I'm doing this properly (or I don't have the knowledge).

Are you looking for something like this? [tex]\sqrt{a^{2}+b^{2}-2abcos\theta}[/tex]
 
  • #6
tiny-tim
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That's it! :smile:

ok, since θ can be anything, the average value of that is (roughly) … ? :wink:

(and then try the same thing for n vectors)
 
  • #7
Pseudopro
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average value of cos would be 0.5 I think - so it would end up as d... (wait does the assumption from part (a) still hold?) I don't see how I can get sqrt N... if I put d back into the equation, I just get exactly the same thing back again...
 
  • #8
tiny-tim
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average value of cos would be 0.5 I think …

uhh? :confused: average value of cos is zero
 
  • #9
Pseudopro
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oh yeah. my bad...
 
  • #10
Pseudopro
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Thanks a lot for your guidance. :smile: I've got it now :approve:
 
  • #11
CoolColin
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I am also doing this problem and understand the hints above, but I don't understand how the answer is d[itex]\sqrt{N}[/itex]. Can someone help?? :uhh:
 
  • #12
CoolColin
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I am also doing this problem and understand the hints above, but I don't understand how the answer is d√N. Can someone help?? :uhh:
 

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