Does the Amplitude of an Undamped Driven Oscillator Stay Constant Over Time?

[adamjts]

The equations I'm getting when I solve the differential equations seem to imply that the amplitude of oscillation does not vary in time.

For example, if I have

x'' + ω₀²x = cos(ωt)

If we suppose that ω≠ω₀,

then the general solution should look something like:

x(t) = c₁cos(ω₀t) + c₂sin(ω₀t) + (1/(ω₀²-ω²))cos(ωt)

This is okay with me mostly. But then thinking about what happens when ω→ω₀ AND ω≠ω₀, then obviously the amplitude of the oscillator should be huge. However, It would seem that the amplitude does not depend on time. Which is to say, that the exact moment that we introduce this driving force, the amplitude of the oscillator instantaneously becomes enormous. Which is hard to believe, because I would expect the object to start deviating from its simple oscillations more slowly and grow in time.

I know that when ω=ω₀ that there is a factor of t in the amplitude, but that is not the case here.

The only explanation that I can think of so far is that the superposition of the two sinusoids makes it seem like the initial amplitudes are small. So when the driving force is introduced, the waves align such that the oscillating body does not seem to have a huge amplitude. But over some time, the waves will align such that the body does have evidently huge oscillations. This would imply, though, that the oscillations would become small again. In other words, we would expect long beats. Is this correct?

Or, maybe after the driver begins, the motion converges onto that of the driven oscillation?

 

[Hesch]

If we suppose that ω≠ω0,

I'm in doubt of the terms. Do you mean that

• ω₀ is the undamped frequency
• ω is the damped frequency

If the system is undamped , ω = ω₀. If you "drive" the system ( add/subtract energy ), the system is not undamped any longer.

And, yes, if you add energy to an undamped system, the accumulated energy will result in an enormous amplitude, and ω ≠ ω₀.

 

[nasu]

Which is to say, that the exact moment that we introduce this driving force, the amplitude of the oscillator instantaneously becomes enormous. Which is hard to believe, because I would expect the object to start deviating from its simple oscillations more slowly and grow in time.

The amplitude is not an instantaneous quantity. If the initial conditions are x=0 and v=0, then displacement increases gradually until it reaches a maximum value. That maximum value is the amplitude.
If you have damping, the oscillation stabilizes after a while to a steady state with the frequency of the driving force.
If there is no damping the transients (oscillations with natural frequency) should stay forever, theoretically.

 

[sophiecentaur]

It would seem that the amplitude does not depend on time. Which is to say, that the exact moment that we introduce this driving force, the amplitude of the oscillator instantaneously becomes enormous.

No. If there is no damping and the exciting frequency is exactly the same as the natural frequency (it would have to be), the energy in the resonator will be growing at the rate of the Power of signal that's being supplied to it.
In a real situation, there will be finite damping and the final level achieved for the energy in the resonator will be related to the Q (Quality factor or 'sharpness' of the response of the resonator) of the circuit and, when this level is finally reached, the power lost in the damping will be the same as the power being supplied.