# What is the qualitative cause for a driven oscillator to have a max. amplitude during resonance?

The steady-state motion of a driven oscillator is given by;$$x =\underset{\text{amplitude}} {\dfrac{F_0}{m({\omega_0}^2 - {\omega}^2)}} \cos\omega t.$$ As we see, the amplitude becomes maximum when $$\underset{\mbox{driven frequency}}{\omega} = \underset{\mbox{natural frequency}} {\omega_0}.$$

Coming now to the forced oscillator with damping, in order to maintain constant amplitude, the average power-input is given by $$\bar{P}(\omega) = \frac{F_0 ^2 \omega_0}{2kQ} \dfrac{1}{\left(\dfrac{\omega_0}{\omega} - \dfrac{\omega}{\omega_0}\right)^2 + \dfrac{1}{Q^2}}.$$ This also passes through maximum when $\omega = \omega_0$ i.e. at resonance.

So, these formulae tell that during resonance i.e. when the driven frequency is equal to the natural frequency, the amplitude & the power consumed by the oscillator is maximum.

• What is actually happening physically that makes the amplitude & the power consumption maximum during resonance?

• What is the reason that the oscillator stores maximum energy only when driven at natural frequencies?

I want to know what is happening actually that maximses the power absorbed & amplitude during resonance as is evident from the equations above.

• Jun 9, 2015 at 17:59
• @John Rennie: Sir, please read this question & you'll see the difference; I've asked here for the steady-state but there it was about the state when the force was first imparted; read your answer:) Moreover, I've asked here one more question: why the input-power is maximum at the natural frequency.
– user36790
Jun 9, 2015 at 18:05
• Comment to the title (v2): The resonance frequency is defined as the frequency with max. amplitude, so the title question (v2) seems to be a tautology. Did you mean to say natural frequency rather than resonance frequency? Mar 15, 2017 at 16:04
• The amplitude is not a maximum at energy, power transfer and velocity resonance for a damped system. The frequency at which amplitude resonance occurs depends on the amount of damping. Apr 1, 2017 at 9:41

The power transfer is maximised at resonance because the driving force and the velocity of the oscillator are in phase.

If you multiply two sinusoidal terms together (the force and the velocity) with a phase difference between them, then the product has its maximum average value when the phase difference is zero and a minimum value when the phase difference is $\pm \pi/2$.

Your steady state solution could be correct, but it is more usual to say that if the driving force is $F_0 \sin \omega t$, then the displacement $x \propto \sin(\omega t + \phi)$, where the phase difference $\phi$ is given by $$\phi = \tan^{-1}\left(\dfrac{-\gamma \omega}{\omega_0^{2} - \omega^{2}}\right),$$ and $\gamma$ is the damping coefficient.

You can see that when $\omega = \omega_0$ the phase difference between displacement and force is $-\pi/2$. But if you differentiate the displacement to get the velocity $$v \propto \cos(\omega t + \phi) = \sin(\omega t + \phi +\pi/2)$$ and at resonance the phase difference between velocity and force is zero.

If the power transfer is maximised, then this is also why the amplitude is maximised, since the velocity amplitude also increases with the amplitude of the displacement.

• Can you tell how much energy is stored at resonance in steady-state?
– user36790
Jun 11, 2015 at 9:36
• @user36790 You appear confused between cases where there is damping and where there is no damping. If there is no damping, then there is no "steady state" at resonance. For a damped system, the energy stored is $0.5 ka_{max}^2$, where $k$ is the "spring constant" and $a_{max}$ is the maximum amplitude. Note that in a damped system, resonance occurs at frequencies slightly lower than $\omega_0$. Jun 11, 2015 at 9:50

You wish for a qualitative answer, so my answer will deal with the physical aspects of the oscillation.

Lets consider a simple scenario of a block attached to a spring horizontally on the ground. Lets say this spring allows oscillations of time period of 4 seconds. This would mean that the block would move from mean to extreme, from one extreme to the next, and back to the mean in 4 seconds. So this spring will naturally move to the extreme position from the mean in 1 second.

Now consider a time varying external force which oscillates at a time period of 2 seconds [resonance]. Suppose the force initially acted in the right direction, and was at its maximum. So lets see what happens in the scenario initially.

From t=0s-1s:

The block will be pushed to the right by this external force. The magnitude of the force would gradually increase and will finally become zero at t=1s. By this time the block would also have moved to the extreme position. From t=1s-3s:

The block will reverse its direction and will move it the other extreme location on the other side. Since the external force also has the same frequency, it will also change its direction to the left and will increase in magnitude till the block crosses the mean position, thereafter decreasing and finally becoming zero at the extreme position. You can observe the interesting phenomenon here. As the block oscillates, the external force also moves hand-in-hand. At the extreme positions, when the block must be at rest at the extreme position, the external force is zero [spring force still acts on the block so that the block won't actually stay there. When the block is at mean position and moving with maximum velocity, the force is also maximum. So what happens is that the force permits natural movement of the block. At all times, force and velocity are in the same direction. When that happens, the block will accelerate increasing its velocity and its amplitude with time. Ideally, the block would soon achieve infinite amplitude, but damping is always present, so amplitude remains finite, but still quite large.

So, now we can easily relate for a non resonating force. The non resonating force may have a frequency either greater than or less than the resonance frequency. Lets take a look at both of these cases separately.

Case 1: External force frequency is less than resonance frequency

Here, suppose the block is moving to the right from its initial mean position. After reaching the extreme position, the object will have a tendency to return back to the mean position. But the external force, due to its lower frequency, hasnt yet reversed in direction. Thus the block will be moving leftwards, but the force is still rightwards, causing a deceleration, reducing speed and amplitude.

Case 2: External force frequency is more than resonance frequency

Here, the force will reverse in direction before the block reaches the extreme position. Here, block moves rightwards, but force become leftwards already, causing deceleration, and similar decrease in speed and amplitude

I hope this analogy is quite beneficial for you.

Although the mathematical model is generic, the physical systems it can represent are different in respect to what the parameters and input/output variables may represent. So in a general sense your question is difficult to answer.

But to illustrate you can take the simple pendulum for example, and more specifically the swing that you commonly find in a playground. As a qualitative explanation, the parent pushing the child in the swing with a frequency that doesn't match the pendulum natural frequency will not be able to build the same steady state amplitude as if they do push at the resonant frequency. By pushing at the same frequency as the natural frequency the parent can transfer the maximum amount of energy per push for two reasons: (1) the swing moves to the right position to receive the push and (2) the parent pushes as the swing moves away (not as it approaches).

So perhaps to put it in more general terms you want to look in any system for what frequency maximizes energy transfer between the driving input and the natural frequency of the system.

• What is the work of the driving force? To input power & hence energy, isn't it? When its input-power becomes equal to the power loss by damping, the oscillation reaches the steady-state. Can you tell as the quality factor increases i.e. the damping decreases, why does the power-input increases? The driving-force doesn't need to work hard & bother about the damping since it is now less as the quality factor is larger now. So, why does the input-power increases? Why does the force work more & hence apply large power even if the damping decreases??
– user36790
Jun 12, 2015 at 2:43
• @user36790 The Q of the system (inversely proportional to damping) is a measure of the energy capacity of the system at resonance. The higher the Q, the more energy the system can 'bank' at resonance. At resonance energy is bouncing back and forth between energy states. For example in the spring mass system, between kinetic and potential energy. Think of the juggler as an analogy. He juggles balls at some fixed rate. If you throw him additional balls at either too slow or too fast a rate, he misses them because of the timing. Jun 12, 2015 at 14:19
• user36790 ...But if you throw balls to him at the rate he's juggling them he can increase and maximize the number of balls in the air. Jun 12, 2015 at 14:22
• @user36790 As long as their is a sufficient supply of energy from the driving force, the system will fill it's internal energy with whatever capacity the Q allows, and beyond that leak any remaining energy that the damping allows. If you plug the leak - Q goes to infinity and the system theoretically can absorb infinite energy. But in real systems this cannot happen. The 2nd order system is ideal. Real systems have nonlinearity like saturation. They blow up or fall apart if you try to cram too much energy in them. Jun 12, 2015 at 14:28
• Really a great analogy, sir. So, the driving force apart from compensating the damped or leaked energy, provides energy for storage in the system. I'll need some time to ponder over these concepts. Thanks for responding, sir. Hope you'll help in the future also:)
– user36790
Jun 12, 2015 at 14:44