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Simple harmonic oscillator- the probability density function

  1. Mar 1, 2009 #1
    How to find the probability density function of a simple harmonic oscillator? I know that for one normal node is should be a parabola but what is the formula and how do we derive it?
     
  2. jcsd
  3. Mar 1, 2009 #2
    Probability is equal to the time spent in interval (x,x+dx) normed to total time it takes mass to run through entire available x space ( from -A to A, A is amplitude ). So you have

    dP = 2*dt/T .

    Take your dt from energy conservation law ( v = dx/dt ! ) and integrate from -A to A. IIRC your distribution should have two poles at amplitudes, maybe some pi somewhere ... don't know, do your math =) Let us know what have you found!

    edit:
    Your integration limits should not be from -A to A but over the range where you want to calculate probability of finding your mass! Upper integral will give you 1, of course.
     
    Last edited: Mar 1, 2009
  4. Mar 1, 2009 #3
    You will find that although the particle average position is zero, the most of the time it spends at the ends where its velocity is small.

    Bob.
     
  5. Mar 1, 2009 #4
    I know that, I was more curious about the function p(x)....
     
  6. Oct 12, 2009 #5
    I'm struggling with the same problem here. Trying to find the probability density function for a simple harmonic oscillator.

    I'm having trouble deriving P from the information suggested here.

    It was suggested that we should use:

    dP = 2*dt/T

    and the fact that

    v = dx/dt

    Now we have:

    x(t) = Acos(ωt)

    and

    v(t) = -Aωsin(ωt)

    Now if we integrate dP we get

    P = int( 2/(T*v) dx)

    But since I only have v(t), but not v(x), I'm not sure how to go about this.

    I know that a(x) = -ω2x

    but I'm not sure if that can be of any aid?

    Anyone have a clue?

    regards
    Frímann
     
  7. Oct 12, 2009 #6
    dx=vdt so P =2 int(dt)/T = 1.
     
  8. Oct 12, 2009 #7
    Which is what we started with ( dp = 2 dt/T ), with normalization added: P(-infinity < x < +infinity) = 1

    But I'm supposed to find P as a function of x, which probably looks something like a parabola. Now isn't it neccesarry for me to substitute that dt with dx/v ?

    Then I would need to define v as a function of x, right?

    Regards
    Frímann Kjerúlf
     
  9. Oct 12, 2009 #8
    Maybe Asin(ωt) = A[1-cos2(ωt)]1/2 = A[1-x2/A2]1/2, so you obtain v(x)?
     
  10. Oct 12, 2009 #9
    Thats seems to be about right, although I don't understand how you got there :)

    I got the same results doing:

    Isolate x:

    x(t) = Acos(ωt) => t(x) = arccos(x/A)/ω

    differentiate t(x):

    dt/dx = 1 / (A^2 - x^2)^1/2

    which is equal to 1/v :

    v(x)= dx/dt = (A^2 - x^2)^1/2

    which is the same result you got.

    Thanks :)
    Frímann
     
  11. Oct 12, 2009 #10
    I used the trigonometry, but your way is also right.
     
  12. Aug 24, 2011 #11
    What is about a Probability density to find particle with amplitude A or -A?

    Near point A and -A.
     
    Last edited: Aug 24, 2011
  13. Oct 24, 2012 #12
    I tried to solve for the functional form of the probability.
    I set A = w = 1 and ignored 2/T for simplicity.

    So, probability of finding oscillator at position x is proportional to the amount of time it spends at every dx---

    i.e
    Prob. prop to ...
    int(dt) = int(dx/v)

    v(x) = 1/(1-x^2)^1/2 (*There's two ways to get this--see post #8 and #9*)

    Integration of 1/(1-x^2)^1/2 dx

    yields Arcsin(x), which is negative when x is negative--->doesn't make sense as a probability distribution.

    Any thoughts?
     
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