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Why is the amount of energy transferred to a driven mechanical oscillator largest when the drive frequency is the same as the natural frequency of the oscillator? Why are they exactly out of phase by π/2?

Clarifying example experimental setup: http://av.ph.tum.de/Experiment/1000/Grafik/b1605.gif

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Could you be more precise about the coupling between the oscillators? The question very much depends on this; e.g. for completely independent oscillators there will be no transfer at all. I suppose you are talking about Fourier decomposition of some problem (a crystal?). If so, please provide also some information about the underlying problem. –  Marek Feb 4 '11 at 15:04
    
Honestly I do not have much of a clue about all this. I'm currently learning the basics about oscillation. I was referring to a simple example such as a spring pendulum that is moved up and down with a certain frequency which is the same as the frequence of the pendulum itself. –  user1751 Feb 4 '11 at 15:14
    
The question is simply one of finding the normal modes for a system of two SHO of frequencies $\omega$ and $\omega'$ with some coupling between them and showing that one of the modes has a dependence on $1/(\omega-\omega')$. Thus when $\omega \rightarrow \omega'$ the amplitude of this mode diverges. Ref: Feynman Lectures Vol. I –  user346 Feb 4 '11 at 15:40
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The example in Your link shows one oscillator, not two! And for one oscillator driven by a variable speed motor there are two results possible. Either the response is flat (no resonance, of course not found in experiment) or there is some maximum at some frequency. This frequency is called the natural frequency. Im Grunde wäre es also eine Frage wert, warum es überhaupt eine Resonanz gibt. –  Georg Feb 4 '11 at 16:18
    
@space_cadet: even two oscillators of identical frequency, will "split" into a system with two different resonances. (Difference is a question of coupling) Classic examples are coupled circuits in band filters (superhets) or Coupling in proton NMR. In the picture of the coupled pendulums, the two frequencies are the symmetric and antisymmetric vibration. –  Georg Feb 4 '11 at 17:28

5 Answers 5

If you have two decoupled oscillators, they satisfy differential equations $$-\frac{d^2}{dt^2}x_i=\omega^2_{i} x_i$$ where $i=1,2$. The solutions are clearly multiples of $\cos(\omega_i t+\phi_i)$. Now, consider two interacting oscillators. Each oscillator must know about the phase of the other, so the simplest dependence is to add a multiple of $x_2$ (a linear multiple) to the equation for $x_1$ and vice versa: $$-\frac{d^2}{dt^2}\vec x=\Omega \cdot \vec x$$ Here, $\Omega$ (capital omega) is a $2\times 2$ matrix, and if you don't know how to multiply matrices, you should learn it. The matrix $\Omega$ has diagonal elements $\omega_1^2$ and $\omega_2^2$ but we have just added real nonzero off-diagonal elements $\Omega_{12}$ and $\Omega_{21}$, too.

Such a system of two linear differential equations may be easily solved by diagonalizing the matrix $\Omega$. In other words, there exist two linear combinations of the two equations such that the second time-derivative of the combination of $x_1,x_2$ will only depend on the multiple of the same combination of $x_1,x_2$.

Diagonalizing the matrix means to find the eigenvalues. The eigenvalues of the $2\times 2$ matrix $\Omega$ are solutions to the charcteristic equation $$0 = \det(\Omega-\lambda\cdot{\bf 1}) = (\omega_1^2-\lambda)(\omega_2^2-\lambda)-\Omega_{12}\Omega_{21}$$ That's a quadratic equation for the eigenvalue $\lambda$ that has two solutions $$\lambda_\pm = \frac{\omega_1^2+\omega_2^2\pm\sqrt{(\omega_1^2-\omega_2^2)^2+4\Omega_{12}\Omega_{21}}}{2} $$ Note that under the square root, there is a difference of the squared frequencies. It's because the $+2\omega_1^2\omega_2^2$ term got overcompensated by $-4\omega_1^2\omega_2^2$ and switched the sign. For a fixed coupling between the two degrees of freedom, i.e. for a fixed $\Omega_{12}\Omega_{21}$, the square root - representing the difference between the two eigenfrequencies - is minimized for $\omega_1^2$ close to $\omega_2^2$. This proximity is what you want for a resonance, an effective transfer of energy.

If the energy is being transferred from one oscillator to the other by the (small) off-diagonal $\Omega_{12}$ and $\Omega_{21}$ elements, the kinetic energy stored in $x_1$ (plus the corresponding potential energy) will be slowly moved to the kinetic energy stored in $x_2$ (plus the corresponding potential energy). Energy conservation implies that the energies of the two oscillators have to go like $$E_1 = E \cos^2(\omega_d t), \quad E_2 = E \sin^2(\omega_d t)$$ because they sum up to constant (coefficients omitted). That means that $x_1^{max}$ itself has to go like $\cos(\omega_d t)$ and similarly for $x_2^{max}$ and $\sin$. The phase difference between $x_1$ and $x_2$ is therefore $\pm \pi/2$.

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I really appreciate your effort in expain it but as I said I do not have that much of a clue and therefore hardly understand your answer. Maybe I'm better off asking the question at some other portal ;) –  user1751 Feb 4 '11 at 15:53
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I thought that you asked exactly because you wanted to have a clue, not in order to keep having no clue. ;-) –  Luboš Motl Feb 4 '11 at 16:39
    
Well, I'm trying to get the basics and sadly was not able to follow your mathematics hehe –  user1751 Feb 4 '11 at 16:41
    
I wonder, what's so funny with lack of math knowledge. –  Georg Feb 4 '11 at 17:01

I think what you are raising a question about is a special case of the double pendulum

http://en.wikipedia.org/wiki/Double_pendulum

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No, I was talking about something like this: av.ph.tum.de/Experiment/1000/Grafik/b1605.gif –  user1751 Feb 4 '11 at 15:40
    
At the end, it's the same maths. –  Luboš Motl Feb 4 '11 at 16:39

If the frequencies are different, the averaged influence of one oscillator on the other cancels out over many cycles. Only if the frequencies are close together do we get the possibility of a resonance with influences building up constructively cycle after cycle.

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I guess your question is related to the phenomenon called "resonance". I shall give you a descriptive answer since you couldn't follow Lubos's rather mathematical answer. Hannes, whenever an oscillator is given some forced vibration it is forced to vibrate with the frequency of that 2nd oscillator and not with its own natural frequency. What does this mean? It means the two oscillators are vibrating without synchronization. They are not vibrating with "in phase". Most of the time they are "out of phase". The result is loss of energy in the form of heat and sound. On the other hand if the frequencies of the two oscillators match then they vibrate in perfect synchronism and they are always in phase. The loss becomes minimum. Almost all the energy of the two oscillators are manifested in the kinetic energy of the two oscillator system. Their amplitude becomes maximum. This is an example of resonance between two mechanical oscillators but the same principle holds in more general cases e.g. electrical circuit with a register, inductor and capacitor. When you tune your radio to a particular channel what you actually do is change the frequency of the circuit to match the signal frequency of that channel.

Hope, you have understood the phenomenon of resonance.

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The test setup pertains to forced oscillations, and not coupling of two oscillators. See Driven Harmonic Motion for more information.

The response amplitude depends on the ratio of (driven frequency) to (natural frequency) and is infinite when the two are equal and there is no damping. Damping, dampens then amplitude a but, and also changes the phase of the response. The reason is that the coupling force acts in the direction of motion, adding power (and hence energy) to the system.

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