Why doesn't resonance result in infinite energy transfer?

[ckuze]

I have a question about resonance.

According to this graph, http://en.wikipedia.org/wiki/File:Resonance.PNG

When no damping force (the red line), there should be no energy loss due to frictional force or other damping force. So the energy should be fully transferd to the system from the excternal agent.

But I cannot understand why the amplitude of the forces oscillation is not infinity no matter the difference between the natural frequency and the driving frequency exist or not.

 

[Studiot]

Hello ckuze and welcome to Physics Forums.

The reason that only part of the energy of any frequency other than resonant will be transferred to the system is that this is the only frequency where the input will be in phase with the resonance at all times.

To explain this in physical terms consider the following mechanical example.

A heavy pendulum is hanging still but free to move.

Take a small mass on the end of a string and strike the pendulum sideways.

Some momentum will be transferred to the pendulum causing it to displace slightly side ways.
It will then tend to swing back to its zero position.

If we now strike the pendulum repeatedly with our light impulse device we can observe what happens.

If we always arrange the frequency of striking to hit the pendulum, just as it reaches the max displacement it will be traveling at zero velocity and so the maximum momentum will be transferred.
This is only the case if our frequency of impluse (striking) is in exact phase with the resonant frequency.

If, on the other hand, our impulse occurs at any other it will either meet the pendulum going away from or towards the striker some momentum transfer will be inefficient.
Sometimes impulse reinforce the motion of the pendulum and thus will speed it up. Sometimes it will oppose the motion and slow it down.

go well

 

[Bill_K]

Not only does energy flow into the system, it can also flow back out again. As the amplitude of the oscillation increases the backflow increases. Eventually an equilibrium is reached in which the energy flows in both directions balance each other, and from that point on the amplitude will grow no more.

 

[ckuze]

Hello ckuze and welcome to Physics Forums.

The reason that only part of the energy of any frequency other than resonant will be transferred to the system is that this is the only frequency where the input will be in phase with the resonance at all times.

To explain this in physical terms consider the following mechanical example.

A heavy pendulum is hanging still but free to move.

Take a small mass on the end of a string and strike the pendulum sideways.

Some momentum will be transferred to the pendulum causing it to displace slightly side ways.
It will then tend to swing back to its zero position.

If we now strike the pendulum repeatedly with our light impulse device we can observe what happens.

If we always arrange the frequency of striking to hit the pendulum, just as it reaches the max displacement it will be traveling at zero velocity and so the maximum momentum will be transferred.
This is only the case if our frequency of impluse (striking) is in exact phase with the resonant frequency.

If, on the other hand, our impulse occurs at any other it will either meet the pendulum going away from or towards the striker some momentum transfer will be inefficient.
Sometimes impulse reinforce the motion of the pendulum and thus will speed it up. Sometimes it will oppose the motion and slow it down.

go well

Thank you very much for your detailed solution.

So, if the resonance does not occur but it is still no damping force present, is the energy transfers back to the external agent or transfers to other position in a different state(PE, heat etc.)?

 

[Studiot]

So, if the resonance does not occur but it is still no damping force present, is the energy transfers back to the external agent or transfers to other position in a different state(PE, heat etc.)?

Neither of us said that.

My impactor meeting the pendulum coming the other way is an example of energy transfer from the pendulum to the impactor.

 

[ckuze]

So, the situation is the vibrating frequengy is equal to the driving frequency, but it still has a phase difference so the energy cannot completely transfer to the object?

But i still cannot understand where the input energy has gone?(I am so stupid and bad at English)

 

[Studiot]

But i still cannot understand where the input energy has gone?(

Perhaps if you posted more detail about the circumstances of your question?

 

[chrisbaird]

The input energy goes nowhere. It stays with the external agent. When an electromagnetic wave goes through a material that is non-lossy (very little damping), the energy stays in the wave instead of being absorbed my the materials. Think of sunlight going through glass: the energy goes right through.

 

[Studiot]

The input energy goes nowhere. It stays with the external agent. When an electromagnetic wave goes through a material that is non-lossy (very little damping), the energy stays in the wave instead of being absorbed my the materials. Think of sunlight going through glass: the energy goes right through.

If it goes right through, where is the resonance?

 

[ckuze]

Actually, I am now studying the A-Level in Hong Kong. My teacher told me that when damping exists, amplitude keeps constant at steady state because the input energy will lose due to damping force.I can understand this.

But I cannot understand why the amplitude still be constant when no damping force exists but forced osciilation still goes on.(The situation is : no damping, the natural frequqncy and driving frequency are not very close.) Studiot said that since it is not exactly in phase so that only part of energy transfers to the system, and the pendulum example is very detailed. But I still have a question about the 'remain part of energy'---the energy which is not transferred to the system, where they gone?

 

[vanhees71]

Well, let's do the calculation. For the undamped case you have

\ddot{x}+\omega^2 x=a \exp(-\mathrm{i} \Omega t).

You need the full solution of the homogeneous equation (i.e., the one where the rhs is set to 0) and one particular solution of the inhomogeneous equations.

The former you get by the usual ansatz

x(t)=A \exp(\lambda t).

This leads to the characteristic equation of the homogeneous equation

\lambda^2+\omega^2=0 \; \Rightarrow \; \lambda=\pm \mathrm{i} \omega,

and the full solution reads

x(t)=A_1 \exp(\mathrm{i} \omega t)+A_2 \exp(-\mathrm{i} \omega t).

For the homogeneous equation, you have to distinguish two cases:

(1) \quad \Omega \neq \omega

Then you make the "ansatz of the right-hand side"

x(t)=A \exp(-\mathrm{i} \Omega t),

leading to

(-\Omega^2+\omega^2)A=a \; \Rightarrow \; A=\frac{a}{\omega^2-\Omega^2}.

In this case the full solution reads

x(t)=A_1 \exp(\mathrm{i} \omega t)+A_2 \exp(-\mathrm{i} \omega t) + \frac{a}{\omega^2-\Omega^2} \exp(-\mathrm{i} \Omega t).

From this you see, there's never a stationary state since the transient state is not damped away as in the case with friction.

(2) \Omega=\omega

Then the Ansatz is

x(t)=y(t) \exp(-\mathrm{i} \omega t).

Plugging this into the ODE, you get

\ddot{y}-2\mathrm{i} \omega \dot{y}=a.

From this equation, we only need one particular solution, and thus we can make the ansatz

y(t)=C t,

which gives

C=\frac{\mathrm{i} a}{2 \omega},

and the full solution of the original equation reads

x(t)=A_1 \exp(\mathrm{i} \omega t)+ \left (A_2+\frac{\mathrm{i} a}{2 \omega} t \right) \exp(-\mathrm{i} \omega t).

As you see, in this case the amplitude grows linearly in time.

 

[Studiot]

Actually, I am now studying the A-Level

This is useful information to help frame answers.
From your original post which refers to a pretty sophisticated treatment I assumed your were studying this from a mechanical viewpoint at degree level.

For your information the example I gave came from my A level physics teacher back in the 1960s when he did exactly what I have described to demonstate resonance with a knotted hankerchief and a hanging 15kg weight.
Pushing a person on a swing is a similar example.

OK so let me start by stating that resonance is not primarily about energy - energy transfer or other energy considerations.

Resonance is basically about timing - which is another way of talking about phase.

Oscillation is about energy. In an oscillating system energy is continuously exchanged between two forms repeatedly cycling in a regular manner in time.

Resonance occurs between two such systems when energy or other quantity such as the momentum in my example is passed from one system to the other.
The systems do not have to be identical and the wave shapes of their oscillations do not have to be identical (again as in my example).
All that is required is that the repetition rate or frequency of both systems is the same and that the phase or timing of the transferred quantity matches.

I say that energy is not the best quantity to use to model the transfer because energy is a scalar and the point about resonance is that the timing and direction of the systems must match up.
Timing and direction is not taken into account with scalars. I could add energy to the pendulum with a blowtorch, but that would not induce resonance.
This was why I said that momentum is transferred. Momentum is a vector and considered in the A level syllabus.

As regards to the energy itself:

In theory, with no damping, each transfer of energy from the driving system to the resonating system will add without limit as you wondered at the beginning. All real systems impose limits so for instance the length of my pendulum limits the amplitude of the oscillation.
Energy not transferred from the driving system is obviously left in the driving system. This would be the case if my pendulum was struck glancing blows rather than head on impacts.

Hope this helps

 

[chrisbaird]

Yes, if you are exactly at resonance, then with zero damping, the oscillation will climb towards infinity. In real life there are few infinities. Usually we reach some other unaccounted limiting factor before reaching infinite amplitude, like the http://www.youtube.com/watch?v=3mclp9QmCGs". But this only happens right at resonance.

More often, the external agent is driving close to resonance but not exactly on resonance. If there is no damping factor, then the energy passes through.

 

[ckuze]

Thank you very much.