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SHM and probabilities - help!

  1. Oct 11, 2004 #1
    Hey there. :smile:

    As part of our quantum assignment we;ve to also look at a simple classical oscillator (it's part b to a question about the one dimensional harmonic oscillator). Problem is, I can hardly remember a thing that isn't to do wih quantum! So any help here would be appreciated. We've to find an expession for the velocioty of the particle as a function of positon, and I think I'm going okay with it, it's just the limits that I don't know about (maximum displacements are a and -a).

    I've got it down to v(dv/dx)= (-w^2)x which gives me vdv=(-w^2)xdx, but I don't know which way to integrate between the limits. Is the amplitude at the bottom or the top? The example of it I looked at online as x as the upper limit and the amplitude as the bottom one, but I don't get that.....

    Secondly, we've to (as a result fo the first bit), show that the probability of locating the particle in an interval dx between the maximum displacements a and -a is given by the ewt P(x)dx=dx/(pi(a^2 - x^2)^(1/2)). I'd like to take a stab at it but I don't know how to work out the probability for it - is it related to the probability that we use in quantum mechanics? Sorry, I sound so dumb here but I've no idea how to work out the probability. I've checked the net but to no avail. Please help (it would be much appreciated!)!

  2. jcsd
  3. Oct 11, 2004 #2

    Tom Mattson

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    Hi, and welcome to PF! :smile:

    If your initial displacement is the amplitude, then yes the lower limit of integration should be at the bottom. The reason you integrate to an upper limit of x is because you're trying to find a function of x. So, you let the upper limit take on a variable value. If you integrated to another constant, your answer would be a constant.

    Sorry, what's an "ewt"?

    Short answer: No. You don't calculate it using QM at all.

    Longer answer: Sorta. The probability density for the classical SHO is the large-n limit for the probability density of the QM SHO. This must be the case, as per the correspondence principle.
  4. Oct 11, 2004 #3
    LOL, eqt, not ewt. You've just got to love my typing. ;)

    Think I've sorted it out ok-ish now. Discoverd that P(x)=(B^2)/v(x) where B^2 is a constant.

    Thanks for your help btw! :)
  5. Oct 11, 2004 #4

    Tom Mattson

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    Ha Ha, I thought you were trying to say exp(ωt), and I was wondering how that got in there!

    That's right. P(x) is inversely proportional to speed, which makes sense because the faster you are going in an interval from x to x+dx, the less time you spend there, and the less likely it is that you'll be found there.

    All you have to do is find B by normalizing P(x) to 1.
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