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Temporal average

  1. May 2, 2010 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    The temporal average of a function f(t) over an interval T is given by [tex]\langle f(t) \rangle=\frac{1}{T} \int _{t}^{t+T} f(t') dt'[/tex].
    Let [tex]\tau =\frac{2 \pi}{\omega}[/tex]. Calulate the temporal average of [tex]\langle \sin ^2 (\vec k \cdot \vec r - \omega t) \rangle =\frac{1}{2}[/tex], when [tex]T=\tau[/tex] and [tex]T >> \tau[/tex].


    2. Relevant equations
    Already given in the problem description.


    3. The attempt at a solution
    I don't understand the problem. Aren't they asking me to prove that [tex]\langle \sin ^2 (\vec k \cdot \vec r - \omega t) \rangle =\frac{1}{2}[/tex] for [tex]T=\tau[/tex] and to calculate [tex]\langle \sin ^2 (\vec k \cdot \vec r - \omega t) \rangle[/tex] for [tex]T>> \tau[/tex]? If so, it's really not clear. If not, could you please explain with a bit more details? I can't start the problem since I don't understand it.
     
  2. jcsd
  3. May 2, 2010 #2
    Looks to me like the question is asking you to show that

    [tex]\left\langle \sin^2\left[\vec{k}\cdot\vec{r}-\omega t\right]\right\rangle=\frac{1}{2}[/tex]

    for both [itex]T=\tau[/itex] and [itex]T\gg\tau[/itex].
     
  4. May 2, 2010 #3

    fluidistic

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    Ok thanks a lot. I'll give some tries. :smile:
     
  5. May 3, 2010 #4

    gabbagabbahey

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    The first case, [itex]T=\tau[/itex] can be easily done with a useful trick....if you graph sin^2(u) and cos^2(u) over a full period, you should see that the area under them is equal and so

    [tex]\int_{u_0}^{u_0+2\pi}\sin^2(u)du=\int_{u_0}^{u_0+2\pi}\cos^2(u)du=\frac{1}{2}\left(\int_{u_0}^{u_0+2\pi}\sin^2(u)du+\int_{u_0}^{u_0+2\pi}\cos^2(u)du\right)=\frac{1}{2}\int_{u_0}^{u_0+2\pi}du=\pi[/tex]

    For the second case, I think you will want to use the double angle formula, and then expand your result in a Taylor series.
     
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