Resonance, intutive explanation

Question

In the case of forced, undamped oscillations, why is the amplitiude of the steady state oscillations bigger, when the frequency of the driving force gets closer to the natuaral frequency of the oscillating system. I know, that we can get this result by solving appropriate equation, but I am looking for an intuituive explanation. I know the "kid on a swing" explanationa, but it relates to driving in a form of short impulses, rather than a sinsuoidally changing force.

Answer 1

In the situation you describe, you have a system that wants to osculate at one frequency, but is driven at another. Using just some mathematical rearrangement, you can instead state that it is being driven at the oscillating frequency but that the driving frequency is changing phase over time. That change in phase corresponds to the difference in the two frequencies (the beat frequency). Doing that is just math.

So if you do this, you can see that you're going to have periods where you have a lot of pushes that are constructively interfering with the osculations, putting more energy into the system with every push. Later, when the phases are more out of phase (such as 90 to 270 degrees out of phase), you'll find they are destructively interfering).

It should be easy to see that, if you push close to the resonant frequency, you're going to get to have more constructive pushes in a row before you start to do destructive pushes. This means the constructive pushes get to add up longer before you start destroying them, and you get a higher maximum amplitude. If you're pushing at a frequency that's quit far from the resonant frequency, you only may get 1 or 2 constructive pushes before the destructive phase starts, and you can't put much energy into the system that way.