8
$\begingroup$

What really is a normal mode? Maybe it's because of my teachers but I find it really abstract. I know that "numerically" corresponds to the eigenvectors of the equation $\ddot{X}= -M^{-1}KX$ with the solution of the form $Ae^{-iwt}$ already plugged in, but what is the physical meaning of those eigenvectors in the first place? Is it the amplitude of the system when in normal mode? And (there you go) what does it mean to be in a normal mode?

I appreciate your answer in advance.

$\endgroup$
3
7
$\begingroup$

In a normal mode, all parts of the system are oscillating with one single frequency. If we write the positions of the objects as a column vector $X$, the (linearized undamped undriven) coupled differential equations can, in general, be written as $$\ddot{X} = MX$$ where $M$ represents the coefficients. To solve for the normal modes, we need to assume that the frequency is $\omega$, so we substitute $X = Ae^{i\omega t}$ to get $$-\omega^2 A = MA$$ which is the eigenvalue problem that we were looking for. Therefore, we can see that the eigenvalues $-\omega^2$ represent the pure frequencies of the normal modes, while the eigenvectors $A$ represent the relative amplitudes of each object.

This is entirely analogous to the ordinary differential equation $\ddot{x} = -\omega^2 x$ for a single oscillator. The general solution is then a linear combination of the normal modes.

$\endgroup$
2
  • $\begingroup$ What makes these frequencies special? Are normal modes nothing but easier solutions to help to derive the general solution? $\endgroup$ – arpg Jun 17 at 17:26
  • $\begingroup$ @arpg No, the normal modes not "easier solutions". They are fundamental. You must get the same solution regardless of how you solve the differential equations. For small systems (involving just two or three objects), you can try solving them by finding linear combinations of the equations that give simple harmonic motion, instead of using the eigenvector method. $\endgroup$ – Vincent Thacker Jun 17 at 23:03
3
$\begingroup$

Imagine a chain made of a line of springs with some mass $m$ located between each spring and the next.

Now suppose you wobble this system in some complicated wobbly motion (not a normal mode). All the masses will move up and down by different amounts and in some complicated motion. If you could hear the sound waves produced by the motion then it would sound like noise.

Now suppose instead that you carefully arrange that one of the masses goes up and down in a strictly periodic way: just one frequency, just one 'note' if you could hear it. The adjacent masses will pick up this motion too, and then the ones adjacent to those, etc. The motion throughout the system will stay complicated for a while, but eventually it may settle into a pattern where all the masses are going up and down at the same frequency. They don't have to be in phase: some may be going up while others go down, but this phase difference will be constant in time. Also they don't have to have the same amplitude: some may go up and down more than others. If you looked at the system it would like as if it had a standing wave, with all the motions having exactly the same frequency, repeating over and over.

This sort of motion is called a 'mode of oscillation' or just 'mode' for short. If, in the absence of damping, you do not need to provide an external force because the system just carries on oscillating at a single frequency throughout, then it is called a 'normal mode'.

For any given set of springs and masses, the normal mode motion can only take place at one of a discrete set of frequencies (this is what the eigenvalues tell you). And each normal mode has its own characteristic shape of displacements along the system (this is what the eigenvectors tell you).

I talked about springs and masses, but the concept is more general, and has quite a widespread application in physics, pretty much whenever something is more complicated than a single particle and can undergo some sort of oscillatory motion.

$\endgroup$
3
  • 2
    $\begingroup$ "the normal mode motion can only take place at one of a discrete set of frequencies" — and if you instead make one of the masses oscillate at a frequency not in this set, you'll get an evanescent mode with the peak at this mass. This decaying exponent will be the envelope of the more frequent (in space) oscillation pattern. $\endgroup$ – Ruslan Jun 17 at 16:48
  • $\begingroup$ "This sort of motion(...)" could you please be a little more specific? What does it take, physically, not in the math world, to be a normal mode - if somehow we put every system's object vibrating with the same frequency, that would be a normal mode, is that what it's all about? $\endgroup$ – arpg Jun 17 at 17:36
  • 1
    $\begingroup$ @arpg try searching for "normal mode gif" in google or similar, and use the image tab, and browse as you like e.g. here is one I found this way: mathematica.stackexchange.com/questions/187054/… $\endgroup$ – Andrew Steane Jun 17 at 18:35

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.