Infinite value of amplìtude of an undamped driven oscillator at resonance

[phymath7]

TL;DR Summary

The expression of amplitude(squared) for a damped oscillator is given by:
$A^2=\frac{(\frac{F_0}{m})^2}{(\omega_0^2 -\omega^2)^2 +4(\beta\omega)^2}$

So if we set the damping constant $\beta=0$ that is if we consider an undamped oscillator the amplitude becomes infinity! What is the physical meaning of this phenomena? As we know energy fed into the system is proportional to $A^2$. So does this mean that an infinite amount of energy is being suplied to the system?But this is obviously not the case. So what's going on here?
Can this be explained by the work-energy theorem?

 

[Baluncore]

So what's going on here?

You appear to be discussing a driven energy storage device, a resonator.

You do not have sufficient energy available to reach an infinite amplitude.

An undamped oscillator or resonator has a bandwidth of zero, and an infinite Q.
It will take an infinite time = 1/bandwidth, for the amplitude to rise to infinity.

In reality, a lossless oscillator would be a resonator, a perpetual motion machine.

 

[phymath7]

You appear to be discussing a driven energy storage device, a resonator.

You do not have sufficient energy available to reach an infinite amplitude.

An undamped oscillator or resonator has a bandwidth of zero, and an infinite Q.
It will take an infinite time = 1/bandwidth, for the amplitude to rise to infinity.

In reality, a lossless oscillator would be a resonator, a perpetual motion machine.

I wonder how this can be explained by the work-energy theorem.

 

[Nugatory]

So what's going on here?

The $\beta=0$ case is not physically realizable because in an any real system $\beta$ may be very small but is never exactly zero. This is somewhat analogous to a spinning gyroscope - if the bearings were truly frictionless the thing would spin forever, but we know that any real gyroscope must eventually spin down. Frictionless bearings and perfectly undamped oscillators are valuable approximations but we cannot trust the math when we push it past the limits of validity of the approximation, which is what you're doing when you supply unlimited energy to get unlimited amplitude.

In practice either the effects of the very small but non-zero $\beta$ become too important to ignore, or something breaks.

 

[Baluncore]

I wonder how this can be explained by the work-energy theorem.

How is your version of "the work-energy theorem" stated.

 

[phymath7]

How is your version of "the work-energy theorem" stated.

The simplest version I know: Energy supplied to the system is equal to the change in kinetic energy

 

[Baluncore]

Energy supplied to the system is equal to the change in kinetic energy

A resonator is an energy storage device. The amplitude of the oscillation is a measure of the energy content. That energy is circulating within the resonator. You must supply that energy.

 

[Vanadium 50]

You might want to look up the Q-factor of an oscillator in your text or even something like Wikipedia.

 

[snorkack]

In reality, a lossless oscillator would be a resonator, a perpetual motion machine.

No, it isn´t. A perpetuum mobile is a device which moves forever and provides an output of energy. A, object which mover forever but without output is not a perpetuum mobile. For example a gyroscope that is truly frictionless because it floats freely in space. Or an electron in a ground state orbital of atom. Or a ring current in a superconducting circuit.
If you are providing input at a resonance frequency to a lossless oscillator, at first the oscillator is accumulating a large amount of energy, at a limited rate so it takes time. Then in practice one or more of the three things happens. One is that the oscillator is not precisely lossless, so at high amplitudes the importance of these small losses increases. The second is that the frequency is not exactly at resonance, so eventually the oscillator starts returning the energy. And the third is that the oscillator is not exactly linear at large amplitudes.

 

[Nugatory]

A perpetuum mobile is a device which moves forever and provides an output of energy. A, object which mover forever but without output is not a perpetuum mobile.

That is not right; you have confused perpetual motion of the first kind with perpetual motion in general. @Baluncore is correct: lossless oscillation would indeed be perpetual motion of the third kind.

In any case, this is an unhelpful digression from the thread topic.

 

[vanhees71]

The work-energy theorem is of course always right. In your case
$$m \ddot{x}=-m\omega^2 x + m A \cos (\omega t).$$
Then you multiply with $\dot{x}$ and integrate over an arbitrary time interval. I just take $(0,t)$. This gives
$$T(t)-T(0)=-m \omega^2 [x^2(t)-x^2(0)] + m A \int_0^t \mathrm{d} t' \dot{x}(t') \cos(\omega t')=W(t).$$
To evaluate this you need the solution of the equation of motion.

Standard methods give you the general solution
$$x(t)=\frac{A}{2 \omega} t \sin(\omega t) + C_1 \cos(\omega t) + C_2 \sin(\omega t).$$
Evaluating the "work integral", you'll see that it contains terms growing with $t^2$. That's because the amplitude of $x$ grows with $t$ (in the long-time limit).

That's known as the resonance catastrophe. In reality, of course, you'll never observe such a system. If you take it realized as a mass at a spring, the linear force law a la Hook becomes invalid for large strains, and at some point the spring will break.

 

[tech99]

The second is that the frequency is not exactly at resonance, so eventually the oscillator starts returning the energy.

I am not sure about this point. The frequency of oscillation in the resonator will always be exactly that of the driver, not the resonant frequency of the resonator.

 

[Baluncore]

I am not sure about this point. The frequency of oscillation in the resonator will always be exactly that of the driver, not the resonant frequency of the resonator.

Resonator or oscillator?
Lossy, or lossless, at what frequency?

If the frequency of drive to a resonator, is derived from that resonator oscillation, then it is a free running oscillator. A small pulse of energy can be added synchronously to the resonator on each cycle, from an external infinite DC source. If the resonator is lossless, then the amplitude will rise forever, at the frequency of the resonator. If a frequency determining parameter of the resonator changes, it is possible for the resonator, and so the oscillator, to wander in (lossless) centre frequency, without loss of the circulating energy.

Assuming that a resonator is lossless only at the natural centre frequency, implies that it will be progressively more lossy at higher and lower frequencies. If the resonator is driven with a very slightly off-centre frequency, the amplitude of the resonance will not rise without limit, but will oscillate at the drive frequency. That amplitude limit will be when the energy loss per cycle, is equal to the energy injected into the resonator per cycle. That comes back to the original assumption, that the resonator may be lossless at the centre frequency, but it is lossy at the drive frequency.

 

[boneh3ad]

Also of note: the equation provided in the OP comes from solving the linear second-order ODE that approximately governs a prototypical mass-spring-damper system. But these systems are only approximately linear in the real world. As oscillation amplitudes increase, the coefficients are not likely reasonably treated as constants (e.g., spring gets stiffer, damping gets larger) and additional nonlinear terms likely become non-negligible.

In a completely idealized system, the amplitude goes to infinity. In the real world, no system is ideal. They are only approximately ideal under certain ranges of conditions.