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I Period of anharmonic oscillator

  1. Mar 5, 2016 #1

    ChrisVer

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    Well I numerically solved for the potential [itex]V(x)=x^4[/itex], the period:
    \begin{equation}
    T = \sqrt{8m} \int_0^a \frac{dx}{\sqrt{V(a) - V(x)}}
    \end{equation}

    where [itex]a[/itex] was the amplitude of the oscillation and [itex]m[/itex] the mass of the particle.
    Nevertheless, what I was asked to plot was the above period [itex]T(a)[/itex] for [itex]a\in [0.,2.][/itex].
    My problem however is that the period of small values of [itex]a[/itex], diverges. I was able to see how this is the case mathematically, by expanding the square root and obtaining an expression for [itex]T[/itex] that goes for [itex]V(x)=x^p[/itex] as:
    [itex]T \sim \frac{1}{a^{p/2-1}} [/itex]
    which reproduces also the [itex]V=x^2[/itex] result of constant period.

    However I cannot picture what is happening physically. Does it have to do with the flatness of the potential at so small a's and so x's?
    Any idea?
     
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  3. Mar 5, 2016 #2

    Simon Bridge

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    Imagine the potential curve as a literal valley, with the particle a small object sliding along the surface.
    You start the object stationary at x=a, let it go, and measure how long it takes to return... and call that number T.
    What happens to T as the start position gets closer to zero?
     
  4. Mar 6, 2016 #3

    ChrisVer

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    yup, the potential is pretty flat as you move to smaller x....so the body will "hardly" slide...
    in explicitly [itex]a\rightarrow0 \Rightarrow T \rightarrow \infty[/itex]

    But then again, I remember some quote saying that any potential around a (local) minimum can be described to a good approximation by an Harmonic Oscillator potential with the minimum at that point... (Taylor expanding the potential around the local minimum)...What about its application to this case?
     
  5. Mar 6, 2016 #4

    Simon Bridge

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    The other interpretation to starting at the minimum point is that T=0 because it take no time at all to return to it's start position.
    So there is a singularity there :)

    An arbitrary potential can be expanded in a Taylor series, and it is common to take low order's of that series as an approximation.
    So have you taken the taylor expansion for your potential to see what the first few terms are?
     
  6. Mar 6, 2016 #5

    ChrisVer

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    Oh I see... I tried to but I was carried away and didn't think that the 2nd order of the expansion that I [only] kept was also vanishing.
    so it's not a rule for all types of potentials, rather than those who move slower than [itex]V''(x_0) \ne 0[/itex]...
    so from power potentials [itex]V(x)= x^p[/itex] no [itex]p \ge 3[/itex] can be described by an harmonic oscillator...
     
  7. Mar 6, 2016 #6

    Simon Bridge

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    V(x) = x^4

    expanding as a taylor series about x=a:

    ##V(x) = 4a^3(x-a) + 6a^2(x-a)^2 + 4a(x-a)^3 + 1(x-a)^4 + 0 + 0 + \cdots##... put a=0.
    ##V(x) = 0 + 0 + 0 + x^4 + 0 + 0 + \cdots## ... surprise surprise.

    This is telling you that the HO approximation to the quartic potential is V=0.
    In this approximation: T=0 or infinity everywhere.
    The approximation is valid for x "sufficiently close to" 0 ... how close is "sufficient" depends on the sensitivity of your instruments.
    It's like how the approximation that the Earth is flat is really good inside a typical room but not if you are looking from orbit.
     
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