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I am trying to build a model for a certain type of oscillatory behaviour with a kind of exponential dilatation.

How can I modify the function of a simple cosine oscillation $\psi(x)=A_0 \cos(2\pi\; \omega x)$ to an oscillation which period would increase exponentially along the abscissa from left to right?

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A very similar question was asked and answered here. –  user26872 Jun 23 '13 at 19:07
    
@oen thanks! this is exactly what I wanted, I was just playing with $e^x$ and $e^{-x}$ and too blind to think that the inverse should be $\log (x)$ of course. –  al-Hwarizmi Jun 23 '13 at 19:37
    
Glad to help. $\,\,\,$ –  user26872 Jun 23 '13 at 19:39

2 Answers 2

up vote 1 down vote accepted

I've turned my comment into an answer so the question can be closed.

A very similar question was asked and answered here.

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I believe I'm addressing your question with the following.

Your function can be rewritten as

$$\psi(x) = A_0\cos(2\pi\omega x ) = A_0 \cos\left(\frac{2\pi}{X} x\right)$$ where I've defined $X$ to be the period of oscillation:

$$ \omega = \frac{1}{X}$$

If you want to make the period of oscillation an exponential function of $x$ itself, this can be expressed as follows:

$$ X(x) = X_0e^{a x}$$

In this terminology, $X_0$ and $a$ are both constants which characterize the exponential behavior. $X_0$ is the period at $x=0$ and $a$ characterized how rapidly the period of oscillation varies with $x$.

Hence, the final expression is

$$\psi(x) = A_0\cos\left(\frac{2\pi }{X(x)}x\right)=A_0 \cos \left( \frac{2\pi}{X_0 e^{ax} }x\right) = A_0 \cos\left(\frac{2\pi}{X_0}x e^{-ax}\right)$$

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unfortunately not; I had exactly this problem see comments above –  al-Hwarizmi Jun 23 '13 at 19:37

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