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Maybe it seems a dumb question, but I can't understand what the cycle in a damped oscillation is?

Let's take an example: In harmonic motion, one cycle is the smallest distinguishable part of wave that makes the pattern.

But in a damped oscillator, there is no such pattern, because the amplitude changes with time.

Ok. but we can define it by wavelength - one cycle is when a part of the wave travels one wavelength. But then, what is wavelength? Let's define it once again. Wavelength is distance between two points on a wave, that have the same phase. But phase is a fraction of the cycle, ...

If someone would write a clear definition of how a full cycle is defined, I would be very thankful.

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  • $\begingroup$ Take the time interval between points where the amplitude is zero. $\endgroup$ – ja72 Jan 31 '15 at 17:59
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"Cycle" in lightly damped oscillation is the time between successive zero crossings of the signal with the same slope. When you look at the equation of motion of a damped oscillator you see a oscillatory component multiplied by a damping term, for example

$$A=A(0)\sin(\omega t +\phi)e^{-kt}$$

Where $\omega$ (or if you like $\omega/2\pi$) is the frequency and $k$ the damping term.

Put differently, the zero crossing in a wave has well defined phase regardless of amplitude.

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  • $\begingroup$ But if take derivative of it, slope won't be the same for two points of damped wave. $\endgroup$ – Arlic Jan 31 '15 at 17:00
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    $\begingroup$ @Arlic that is true. However this is the best definition I can give. Zero crossings are the only unambiguous points to choose and their occurrence corresponds to the $\omega$ in my equation. It really depends on how you would like to define "cycle" - but this is a definition I am comfortable with. $\endgroup$ – Floris Jan 31 '15 at 17:24
  • $\begingroup$ I think that i have a answer for my question, but it need a criticism, because it is own made definition. Cicle is the smallest repatable segment between two points, where: 1. These point lies on one line and the line is parallel to direction of wave propagation. 2. These two points have the same sign of slope. I think, even computer would check with this definition if two points are in the same phase, what is wavelenght etc. $\endgroup$ – Arlic Jan 31 '15 at 17:54
  • $\begingroup$ If you add the word "always" as in "always the same sign of slope" then I think your definition is valid- and not fundamentally different from mine. $\endgroup$ – Floris Jan 31 '15 at 17:56
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Cicle is the smallest repatable segment between two points, where: 1. These point lies on one line and the line is parallel to direction of wave propagation. 2. These two points have the always same sign of slope.

Thanks to Floris in help of derivation of the definition.

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  • $\begingroup$ Is it just me? I really dislike it when questioners accept their own attempt at an answer, especially when a better one is available. $\endgroup$ – Rob Jeffries Apr 12 '15 at 12:47
  • $\begingroup$ yeah, I dont see the genius in this definition either :) $\endgroup$ – Ilja Apr 5 '16 at 19:32
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This is mainly a matter of definition, a damped oscillation is not periodic, since it dies away, that's why you cannot really talk about "period" or "cycle" in the usual way. What I've seen is definitions of "pseudo-period" and "pseudo-cycle" when one talks about the analogous concepts in damped oscillations.

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Wow: this is a very good question - and the answers all miss the most interesting point! :)

That's a good observation, there is no precise definition of where a cycle ends if you have some change in the amplitude. But do not be sad about it, take it positive: this explains (well, it's at least a very good analogy) the uncertainty principle.

In QM particles are described by wave packets. The position of the particle is somewhere in the packet, and the momentum is connected to the period of the underlying wave.
And now you see, if this packet has a finite length, i.e. if the amplitude changes along the packet, it's the same effect as with the damped oscillation: you can not define the wavelength well. The narrower you try to make it, i.e. the faster the amplitude changes with $x$, the less clear is a notion of wavelenght.
And there you are: $\Delta x\Delta p$ has some lower limit, the uncertainty is visualized!
(You can show the analogy more rigorously, it is somewhere in the second volume of the Feynman Lectures. It is indeed a perfect analogy, one can explain mathematically what one means by the width and by the uncertainty in wavelength of a wave packet, and and show that this product is bounded)

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