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I am going through "Optics" by E. Hecht and I think I am misnderstanding the following paragraph:

$x(t) = \frac{(q_e/m_e)}{(w_0^2-w^2)}E(t) $

This is the relative displacement between the negative cloud and the positive nucleus. [...]

Withouth a driving force (no incident wave), the oscillator will vibrate at its resonance frequency $w_0$.

So, I understand how the frequency of the incident wave would influence the oscillation, but how a particle is oscillating without any driving force?

From the equation: if $E(t)=0$, then why there is an oscillation at $w_0$?

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  • $\begingroup$ -1 Not clear what you are asking. You are surely familiar with a mass on a spring which continues oscillating after an applied force has been removed? ... The equation does not make sense because the displacement can be out of phase with the applied field. Please can you post an image of the surrounding text? $\endgroup$ – sammy gerbil Dec 1 '17 at 16:13
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The equation describing a driven (undamped) oscillator is: $$\ddot x +\omega_0^2x=f(t)$$ where $f(t)$ is the driving force, and $\omega_0$ the natural frequency of the oscillator. If $f(t)=0$, the solution is given by: $$x(t)=A\cos(\omega_0 t)+B\sin(\omega_0 t)$$ so the system oscillates nonetheless. If you add the force, the exact solution depends on the force that you apply. However, if the driving force is "oscillating", then it will be of the form $f(t) = C\sin{\omega t}$, where $\omega$ is the frequency of the oscillations of this force. If $\omega=\omega_0$ things are a bit more complicated, because you get resonant behaviour. In the easier case $\omega \neq \omega_0$, a solution will be given by: $$x(t)=\frac{1}{\omega_0^2-\omega^2} f(t)$$ By substitution, it's easy to check that this is a solution. However, nothing prevents you from adding the previous, undriven solution. After all, it gives zero! In other words, the general solution is given by:

$$x(t)=A\cos(\omega_0 t)+B\sin(\omega_0 t)+\frac{1}{\omega_0^2-\omega^2} f(t)$$

So as expected if $f(t)=0$ you recover the undriven solution, which is oscillating.

A typical example of this is a spring attached to a motor: even without the motor, the system will oscillate if perturbed due to the restoring force of the spring. In the case you mention, it's likely that the natural oscillations of the system are negligible compared to those caused by the incident wave, so that the undriven solution can be ignored in the equation you show; however if there is no incident wave, then the system will still oscillate (at its natural frequency).

Now your question might be: what causes the natural oscillations without the incident wave? Unfortunately I can't answer this without knowing more about the system. With this general picture you should be able to figure out the rest.

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This is a case with no damping. So, no energy loss is there in a cycle. So, after a cycle has started, it will not stop until some external force is applied. Do note that it takes a lot of energy to get the motion started but once it is there, you don't need force to keep it in motion. In actual cases, there is always friction and this won't happen. For an excellent discussion, read Feynman lectures volume 1 chapter 23.

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