What exactly is periodic motion?

1. May 30, 2015

SkyrimKhajiit

1. According to physicsclassroom.com, periodic motion is defined as "a motion that is regular and repeating." But the example included does not factor in damping (it's assumed that there's no air resistance and the spring will keep vibrating for eternity.)

http://www.physicsclassroom.com/class/waves/Lesson-0/Properties-of-Periodic-Motion

3. I know periodic motion means that the time it takes to complete each cycle is the same, so does that mean that in periodic motion there is never damping (so periodic motion is only a concept?) or can there still be damping where the amplitude decreases, eventually reaching 0...?

2. May 30, 2015

Staff: Mentor

There can be damping, or no damping, or even negative damping. This doesn't change the underlying periodicity (much).

3. May 30, 2015

SammyS

Staff Emeritus
For strictly periodic motion, there is no damping.

4. May 30, 2015

haruspex

I agree largely with SammyS (who is also supported by http://www2.hawaii.edu/~plam/ph170_summer/L13/13_Lecture_Lam.pdf). Periodic motion, strictly speaking, implies precise repetition of state, so any damping must be matched by a forcing. Harmonic oscillation does not imply such exact repetition, but allows scaling with time.

5. May 30, 2015

SkyrimKhajiit

So basically we call this periodic motion because there are little to no real-life perfect examples in physics, but technically periodic motion involves no damping.

^Is that correct?

Also, I wanted to clarify this passage:

"The time it [mass on a spring] takes to complete one back and forth cycle is always the same amount of time. If it takes the mass 3.2 seconds for the mass to complete the first back and forth cycle, then it will take 3.2 seconds to complete the seventh back and forth cycle. It's like clockwork. It's so predictable that you could set your watch by it. In Physics, a motion that is regular and repeating is referred to as a periodic motion. Most objects that vibrate do so in a regular and repeated fashion; their vibrations are periodic."

When it says that the time for each cycle is constant, this is still true even considering damping, right? So even though the amplitude will decrease after each subsequent cycle, the time will stay constant, or is this incorrect?

6. May 30, 2015

haruspex

As I posted, I don't accept that periodic absolutely implies no damping. It can still be precisely periodic if there is a forcing which, each cycle, compensates exactly for the energy lost by damping. In the real world, any forced damped system may settle into a steady periodic motion.
A pendulum clock driven by a constant force (hanging weights, say) is periodic, even though the pendulum swing is not exactly SHM, and even though it is damped.
If we relax the meaning of periodic to allow scaling with time (i.e., each cycle looks like the previous cycle, but attenuated or amplified by some factor), then we would only require the period of the cycle to be constant. Whether that will hold for a damped oscillation depends on the nature of the damping.
We can check this easily by applying a scaling factor to the displacement. If the equation has linear damping:
$\frac{d^2x}{dt^2}+b\frac{dx}{dt}+cx=0$
then replacing x everywhere by $\alpha x$ does not change the equation. The scaling factor cancels out. But if the equation is $\frac{d^2x}{dt^2}+b\left(\frac{dx}{dt}\right)^2+cx=0$ it won't cancel, so the period will almost surely change with time.

7. May 31, 2015

Staff: Mentor

Mathematically, a waveform whose peak amplitude is constantly varying, e.g., decaying and eventually dying away, is not periodic, because periodic means it repeats exactly and at fixed intervals, and a decaying waveform never repeats. This means a plucked guitar string's motion is not periodic in that formal sense. This does not stop us choosing to regard such a waveform as having a periodic oscillation and determining its period when it's convenient to do so, such as when the damping is slight. You do need to be guided by your textbook and context.

I think you'll find that in the real world there is not a single example where a mechanical vibration maintains an exactly constant frequency as it decays. But it is generally convenient to regard the frequency (hence period) as fixed where the error is miniscule.

8. May 31, 2015

SkyrimKhajiit

Yeah, that's kind of what I was getting at but I wasn't sure if it was right..

But about the time, is it true that the period (actual period, i.e. seconds) stays constant even if the system does not involve a periodic motion?

For example, Physics Classroom gives this graph and accompanying table:
http://www.physicsclassroom.com/Class/waves/u10l0b1.gif
http://i.gyazo.com/3138b953e2f7513307a6b6b7220f9ab5.png

The period is constant and yet the amplitude is decreasing..what about when the vibration becomes very little and eventually stops?

9. May 31, 2015

haruspex

I discussed all that in post #6. Is there something I need to explain more?

10. May 31, 2015

Staff: Mentor

When data is measured to only 2 or 3 significant figures and to this accuracy seems to be unchanging, you can't conclude that it is "constant" in the mathematical meaning of the word. Certainly, the period of oscillation of many systems doesn't change much, but in most areas of science we'd want measurements to be much more accurate than to just 3 figures before the phenomenon could be formally labelled "constant".

Your attachment's table of data is perfectly reasonable, the period over that comparatively small number of oscillations is unlikely to show much change when measured to only 3 sig figs. (But how annoying would be an electronic watch that lost even one half of a second in every 1000 seconds?)

Don't forget that the common mathematical model for a decaying vibration has it never completely dying away, it just gets smaller and smaller and motion becomes zero only at time infinity.