Undamped driven oscillation — Is there a phase delay?

[PHYSICSKOP]

TL;DR Summary

Is there any phase delay between displacement and driving force in undamped driven oscillation when the driving frequency is below resonant frequency?

I know that there is phase delay in damped driven oscillation but I want to know is there any phase delay in undamped driven oscillation when we apply sinusoidal driving force. When driving force is maximum, displacement is also maximum as well right?

 

[berkeman]

When there is no damping in a driven oscillator, the amplitude increases without bound, no?

 

[hutchphd]

The OP asked for below resonance.

 

[berkeman]

Ah, good point. So it will go through cycles then. Got it, thanks.

 

[hutchphd]

Off- resonance it will settle to a steady-state response (with a phase shift).

 

[berkeman]

Hmm, I'm no expert on this, but this animation was interesting for me:

 

[hutchphd]

I (we) have foundered on the semantic shore. The illustration is not steady- state because the input frequency is continuously changing and the system never "settles".
Of course the perfect system with no friction will not "settle" in any finite time. And so we always allow a little real world friction (in complex analysis we give the frequency a little imaginary part to allow the integrals to converge). This is then the "steady state" solution. In our case it will be an oscillation at the driving frequency with phase shift
So nothing incorrect was said here. It just depends upon how long you have to wait for steady state (infinity is a very long time...)

 

[sophiecentaur]

Ah, good point. So it will go through cycles then. Got it, thanks.

Did you mean to say that? There is no 'beat', if that's what you mean. The system just reaches (after several Q's worth of time) a steady amplitude. There is no component of power at the natural resonance frequency so it can't turn up.

 

[tech99]

With the electrical case, as the circuit is undamped there will be no resistive loss and the generator will see a pure reactance. For a parallel LC circuit it will be inductive (current lagging 90 degs) and for the series circuit it will be capacitive (current leading 90 degs).
In plotting the results be careful that the resistance of the generator is taken as zero or infinity as appropriate.
In the real world the circuit may be damped by the resistance of the generator.

 

[hutchphd]

In the real world the circuit may be damped by the resistance of the generator.

And the fact that every real accelerated charge will be radiating energy to the universe!

 

[vanhees71]

In my opinion the undamped case for a driven oscillator is a bit more subtle, because for the damped one you can argue that after a sufficiently long time (long against the inverse of the friction coefficient) after switching on the harmonic force the oscillator is in a stationary state oscillating with the frequency of the imposed harmonic force.

In the undamped case in general you have a superposition of the undamped natural oscillations and the ones imposed by the external harmonic force. Of course, as already mentioned above, there is the case where the frequency of the harmonic force coincides with the eigenfrequency of the oscillator, and then the amplitude grows linearly with time ("resonance catastrophe").

That said, let's do the calculation for the undamped oscillator. It's a bit simpler using the complex representation, i.e., introduce a complex variable $z(t)$ with the meaning that the physical quantity under consideration is $x(t)=\text{Re} z(t)$. Then we want to solve the equation of motion
$$\ddot{x}+\omega_0^2 x=A \exp(\mathrm{i} \omega t).$$
The general solution is the general solution of the homogeneous equation (with the rhs of the equation set to 0):
$$z_{\text{hom}}(t)=C_1 \cos(\omega_0 t) + C_2 \sin(\omega_0 t),$$
and an arbitrary special solution of the inhomogeneous equation.

It's intuitive to make the ansatz
$$z(t)=B \exp(\mathrm{i} \omega t).$$
As we'll see in a moment that always works, except for the resonance case. So let's first assume $\omega \neq \omega_0$. Plugging the ansatz into the equation we get
$$-(\omega^2-\omega_0^2) B = A \; \Rightarrow \; B=\frac{A}{\omega_0^2-\omega^2}.$$
Since both $A$ and $B$ are real, there's no phase shift between the external force and and that special solution of the inhomogeneous equation. I.e., you get
$$x_{\text{inh}}=\mathrm{Re} [B \exp(-\mathrm{i} \omega t)]=B \cos(\omega t) = \frac{A}{\omega_0^2-\omega^2} \cos(\omega t).$$
For the general solution
$$x(t)=C_1 \cos(\omega_0 t) + C_2 \sin(\omega_0 t) +\frac{A}{\omega_0^2-\omega^2} \cos(\omega t),$$
it doesn't make sense to talk about a phase shift, because it's a superposition of two harmonic motions with different frequencies. Only if $C_1=C_2=0$ you get a harmonic motion with the driving frequency $\omega$ and then there's no phase shift between $x$ and the external force.

The resonance case needs extra treatment. Here the idea is to use the ansatz of the rhs. of the equation with variation of the constant, i.e.,
$$z(t)=B(t) \exp(\mathrm{i} \omega_0 t).$$
Using
$$\ddot{z}=(\ddot{B} + 2 \mathrm{i} \omega_0 \dot{B} -\omega_0^2 B) \exp(\mathrm{i} \omega t),$$
in the equation of motion we find
$$\ddot{B} + 2 \mathrm{i} \omega_0 \dot{B} =A.$$
Since we only need one particular solution, we can solve this by making the ansatz
$$B(t)= B_0 t$$
leading to
$$B_0=\frac{A}{2 \mathrm{i} \omega_0}.$$
Thus the physical particular solution is
$$x_{\text{inh}}(t)=\mathrm{Re} B_0 t \exp(\mathrm{i} \omega_0 t) = \frac{A}{2 \omega_0} t \sin(\omega_0 t).$$
So the amplitude grows linearly with $t$, and for large times you can neglect the solutions of the homogeneous equations. The phase shift of the oscillations is obviously $-\pi/2$, because $\sin(\omega_0 t)=\cos(\omega_0 t-\pi/2)$, i.e., the phase of the oscillation of $x$ is by 90 degrees behind the phase of the external force, but still there's the growing amplitude of this oscillation!