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Probability density function of simple Mass-Spring system.

  1. Sep 14, 2015 #1
    1. The problem statement, all variables and given/known data
    We know that after long run of simple mass-spring system, there should be a probability of finding the mass at certain points between -A and A.. Obviously in probability of finding the particle near A or -A is higher than finding the particle at 0, because the speed is the highest at equilibrium point. We need $p(x)$, so that $p(x)$$dx$ is probability of finding the point between $x$ and $x+dx$.

    Given values:
    E - energy of the system
    m - mass of the point
    k - spring constant obeying Hook's law

    2. Relevant equations
    Standard mass-spring equations. We basically know everything about the system
    3. The attempt at a solution
    [itex]\int_{-A}^{A} p(x)dx=1[/itex]
    Particle has to be somewhere!

    I thought that it doesn't matter if it's sinusoidal or cosinusoidal oscillation, we can choose any. Meaning that we know everything about system, but still I can't move physical knowledge of the system into mathematical language of statistics.
  2. jcsd
  3. Sep 14, 2015 #2


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    The probability density at location ##x## should be proportional to the reciprocal of the velocity at that location, since it is proportional to the time spent in the incremental spatial area around that location. To be able to integrate wrt ##x##, you'll need to express the velocity in terms of ##x## rather than in terms of time ##t##.

    The integrand blows up at the extreme points because of the zero velocity, but I think it is still integrable.
  4. Sep 14, 2015 #3
    adnrewkirk, if the integral of 1/cosx blows up that means it is not correct answer. I actually thought that it is directly proportional to 1/v or 1/(v^2) but it is not the case, the integral does not make sense or it blows up using this logic.
  5. Sep 14, 2015 #4


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    Why do you feel that?

    It's perfectly valid to have a prob density function that is unbounded and 'hits infinity' at isolated points. What matters is that the integral over intervals including such points exists, and that is indeed the case here.
  6. Sep 15, 2015 #5

    Ray Vickson

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    You need some type of underlying probability model. One "reasonable" model might be that we choose the time t at random and see what is x(t) at that time. Of course, we cannot have a uniform random variable on the whole real line, but we could look at a large, finite interval ##[0, M]## and suppose that the time is a random variable ##T## on that interval. Then, given a distribution of ##T## we can perform the usual type of change of variable in probability theory to obtain a probability distribution of ##X##. In fact, if we let ##M## be an integer multiple of the period, we can obtain nice results; in fact, by periodicity, we might as well just restrict ourselves to a single period. I am going to re-scale time so that the motion is particularly simple. So, we look at the periodic motion ##x(t) = A \sin(2 \pi t), 0 \leq t \leq 1##. Let ##T \sim U([0,1]) ## be a uniformly-distributed random variable (time) on the interval ##[0,1]## and let the random variable ##X## be the function ##X = A \sin( 2 \pi T)##. We can perform a standard change-of variable computation to obtain the probability density of ##X## as
    [tex] f_X(x) = \frac{1}{\pi} \frac{1}{\sqrt{A^2 - x^2}}, \; -A < x < A [/tex]
    This satisfies the condition ##\int_A^A f_X(x) \, dx = 1##, and it also satisfies your intuitive property of being maximal near the endpoints.
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