Energy transfer to a driven oscillator

Question

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

Answer 1

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 $\pm \pi/2$.

Answer 2

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

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

Answer 3

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.

Answer 4

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.

Answer 5

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.