0
$\begingroup$

The principles of resonance are such that the greatest amplitude of a vibrational or electrical wave peak at a certain frequency, and they peak while consuming less power overall to maintain it.

I'm wondering, since an electron has a resonant frequency, and so does a system of water molecules, brick, person, and planet; how would an incoming vibrational wave of infinite 'Q' (so basically it has a very accurate and narrow bandwidth resonant frequency) tell the difference between two objects of the same resonant frequency?

So if you have an apple that resonates at 10.8Hz, and an apple that resonates at 10.3Hz, and your incoming wave is at 10.8Hz with infinite Q, obviously all of the power goes to the first apple, and it peaks.

But if you have an 11Hz apple and an 11Hz pear, (two different objects), what happens?

I would think that if you had two identical apples at 11Hz, it would split the power between them such that the new amplitude is some lesser exponential than it would have been if there was only one apple at 11Hz.

What if the wave is not vibrational, but electromagnetic, same thing right?

$\endgroup$
1
$\begingroup$

As a general rule, two "degenerate" oscillators (i.e. two oscillators with the same frequency) will no longer have the same frequency if they interact with each other, however weakly; the degeneracy is usually broken. So if you have two identical resonators with a fundamental of 11 Hz, and you bring them together, the combined system will have resonant frequencies of something like 11.001 Hz and 10.999 Hz. Exactly how big the shift is will depend on how strong the resonators are "coupled" with each other, and it may be imperceptibly small depending on the limits of your measurement.

This general phenomenon, of degeneracy breaking, is pretty independent of the exact details of the system or the precise form that the energy takes in its "kinetic" and "potential" forms (or how energy is transferred between the systems.)

$\endgroup$
2
$\begingroup$

For macroscopic systems some physical means must be present for energy to flow between potential and kinetic states, or magnetic and electrical fields. The flow must also not be subject to energy loss. Apples do not have the properties that allow such a flow of energy. Apples do not 'ring' when they are subjected to an impulse of energy. Almost all the energy is lost to heat, deformation or destruction of the apple.

$\endgroup$
  • $\begingroup$ Surely an elastic property relating to the positions of atoms in an Apple such that an incoming wave is dispersed along every atom until they reach the end, at which point they are reflected backwards, counts as the potential system. And any more incoming waves act as the kinetic, or "in action" force. $\endgroup$ – ARMATAV May 18 '15 at 23:31
  • 1
    $\begingroup$ It's a very different world at the atomic level. Resonance that we observe from an apple at the atomic level, if any might result in properties such as the color we see, but that's not a resonant property of the entire structure we call an apple. $\endgroup$ – docscience May 19 '15 at 0:40
  • 2
    $\begingroup$ @ARMATAV Also while you might be able to attribute an elastic nature to properties of atoms or molecules, the elasticity doesn't necessarily emerge at the larger structural levels. Crystalline structures including metals do, but allot of organic or biochemicals do not. Apples and pears don't ring. At least I've never heard one ring. $\endgroup$ – docscience May 19 '15 at 0:46
  • $\begingroup$ Not with vibrations, but with the movement of the electrons when an alternating electric field is applied to them. Then wouldn't the entire object 'apple' have a system that resonated at a certain frequency and resulted in stabilized waves if you were to look at each individual atom? Basically, I can make an apple a electric resonantor, just not a 'sonic' resonator. $\endgroup$ – ARMATAV May 19 '15 at 1:11
  • 1
    $\begingroup$ @ARMATAV DeBroglie interpreted Bohr's electron shells as standing (resonant) waves within the atom. But that's each atom, and not necessarily in phase with surrounding atoms. I'm gathering you believe that there is a possible phase synchronization of the DeBroglie waves. I don't believe that's possible. Without synchronization you get allot of cancellations and neutralization in any building of resonance across your apple. I think that's what Michael Seifert is also saying. If you could somehow do that though it would be a cool invention. $\endgroup$ – docscience May 19 '15 at 22:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.