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Consider a driven harmonic oscillator under a sinusoidal force $x''(t) + \gamma x'(t)+ \omega_0^2 x(t) = F(t)$. In the regime of light damping ($\omega_0/\gamma > 0.5$), we find resonance (maximum amplitude) in displacement, velocity and acceleration at driving frequencies of:

  • $\omega_{dis.} = \omega_0\sqrt{1 - \frac{\gamma^2}{2\omega_0^2}} < \omega_o$
  • $\omega_{vel.} = \omega_0$
  • $\omega_{acc.} = \frac{\omega_0}{\sqrt{1 - \frac{\gamma^2}{2\omega_0^2}}} > \omega_0$.

I understand the mathematical derivation of each but is there an intuitive reason why there should be different resonant driving frequencies for the different quantities? It seems quite symmetrical that $\omega_{dis.} < \omega_0, \,\omega_{vel.} = \omega_0, \, \omega_{acc.} > \omega_0$ and there is an inverse relationship between $\omega_{dis.} = \omega_0^2/\omega_{acc.}$. So is there a natural explanation to why this should be the case?

Edit 1:

By natural explanation, I mean an intuitive physical explanation (perhaps considering energies) rather than intuition behind the mathematical derivation. The answer here is along the lines of what I mean. However, I do not understand the reasoning:

The key here is that the maximum AMPLITUDE is not reached at the same frequency as the maximum POWER DISSIPATED. For the former, you would like the frequency to be slightly lower (because you dissipate a certain amount of power per cycle).

Edit 2:

I think that the confusion I have may stem from the simple question: why does the frequency that produces maximum power absorption ($\omega = \omega_0$ when the driving force is in phase with the velocity) not lead to a maximum in the potential energy (i.e. peak in the amplitude response) of the oscillator?

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  • $\begingroup$ Sorry, I'm not sure I follow what you mean. Would you be able to expand further? I understand that displacement and acceleration are half a cycle out of phase, and velocity is orthogonal to both, in the complex domain. But how does this relate to their resonance frequencies? $\endgroup$
    – user246795
    Jan 3 at 16:04
  • $\begingroup$ But can that be used to explain why $\omega_{dis.} < \omega_0$ and $\omega_{acc.} > \omega_0$? Is there an intuitive reason why they shouldn't all be $\omega_0$? $\endgroup$
    – user246795
    Jan 3 at 16:51
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    $\begingroup$ Related: physics.stackexchange.com/q/153197/2451 , physics.stackexchange.com/q/228279/2451 and links therein. $\endgroup$
    – Qmechanic
    Jan 4 at 12:42

5 Answers 5

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It may be easier to think about this in the frequency domain.

$x''(t) + \gamma x'(t)+ \omega_0^2 x(t) = F(t)$ transforms to $-\omega^2X(\omega) + \gamma j \omega X(\omega)+\omega_0^2 X(\omega) = F(\omega)$

which gives the familiar transfer function form $X(\omega) = F(\omega)/(\omega_0^2 + \gamma j \omega - \omega^2 )$

The resonance is usually defined as the point at which the first and last terms in the demoninator cancel, i.e. $\omega=\omega_0$. However, this is not necessarily the peak of the response. The peak of the response depends on what you are measuring. As you point out, amplitude $X(\omega)$, peaks at the lowest frequency and then successive derivatives (velocity $j\omega X(\omega)$, acceleration $-\omega^2 X(\omega)$, jerk $-j\omega^3 X(\omega)$, etc.) peak at increasingly higher frequencies because of the successive multiplications by the linear frequency factor $j\omega$.
The bottom line is that the resonance is at $\omega = \omega_0$, but where the response peaks depends on what particular property you are measuring.

[Edit] Asking "why does the potential energy peak at a lower frequency than the kinetic energy" is a good way of phrasing the question.
The energy loss mechanism in the system is a linear damping term proportional to velocity. So the kinetic energy reaches a peak when the energy loss (power dissipation) reaches a peak. However, this is not true for the potential energy. The potential energy reaches a peak at a slightly lower frequency where there is still a large 'magnification' from the resonance but where the velocity is slightly lower and thus slightly less of the energy is lost to dissipation.

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    $\begingroup$ But is there a intuitive explanation for why the peaks should become successively higher, rather than "it's how the math works out when you apply successive factors of $j\omega$"? $\endgroup$
    – user246795
    Jan 4 at 9:30
  • $\begingroup$ @user246795 I see your edit #2. It's a good question. I'll try to add to my answer. $\endgroup$
    – Roger Wood
    Jan 5 at 3:20
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I think you are using the term "resonance" in a totally unusual and very personal meaning.

What I understand from the expressions you write is your question is that you call "resonance" the value of the frequency for which a particular quantity is maximal for a given input.

If your real question is why is the peak of the acceleration at a higher frequency than the peak of the velocity which itself is at a higher frequency than the peak of the displacement, the answer is given by Roger Wood :

As you point out, amplitude 𝑋(πœ”), peaks at the lowest frequency and then successive derivatives (velocity π‘—πœ”π‘‹(πœ”), acceleration βˆ’πœ”2𝑋(πœ”), jerk βˆ’π‘—πœ”3𝑋(πœ”), etc.) peak at increasingly higher frequencies because of the successive multiplications by the linear frequency factor π‘—πœ”.

But the only "resonance" is at $\omega=\omega_0$. Changing the usual definition of a word is generally a bad idea.


EDIT

To turn the perfectly correct answer by Roger Wood into a more "intuitive" one :

The maximum amplitude of the displacement for a given driving force is at some frequency (which happens to be just below $\omega_0$, but this remark is irrelevant to what follows). Since it is a maximum, the slope is zero. If you increase the frequency ever so slightly, the response decreases very, very slowly because there is no downward slope, just a downwards curvature which means you have to build up the down slope by getting away before it starts to go down "seriously". The amplitude of the velocity is the amplitude of the displacement multiplied by $\omega$. If you increase $\omega$ ever so slightly, the product of (displacement times $\omega$) increases by almost the same "ever so slightly" because of the increase $\omega$, since the decrease of the amplitude of the displacement has not yet built up.

So you see the amplitude of the velocity still increases at the "top" of the amplitude of the displacement. Soon enough the decrease of the amplitude of the displacement will build up and dominate the increase due to multiplying by $\omega$, and you will reach the maximum of the velocity for a slightly higher $\omega$ (which happens to be just $\omega_0$ but this is not important.

Repeat exactly the same reasoning, and you'll see that the maximum of the acceleration is still for a slightly higher frequency.

EDIT 2 as an answer to your edit 2

What you call "maximum power absorption" and I would call "maximal dissipation" is obtained at maximal velocity, because it is velocity that causes dissipation, not potential energy.

For exactly the reason I gave in my first edit, maximal velocity is not attained at maximal amplitude of displacement (which is maximal potential energy) but at a slightly (for weak damping) higher frequency.

EDIT 3

In reaction to Farcher's comment : I had assumed you considered an electrical circuit. This is why I said the usual definition of resonance is $\omega=\omega_0$, maximal velocity.

For Mechanics, indeed, the usual the usual definition of resonance is maximal displacement. This is indeed an inconsistency, due to an old historical situation.

But my point is, in a given setting, whether Mechanics or electricity, there is one usual definition of resonance. So keep to the usual definition in that whatever setting your are dealing with, even if there is another usual definition for a different setting.

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    $\begingroup$ Yes, that is my real question. By resonance, I was meaning maximal of the particular quantity. More specifically however, I wanted to try understand why the frequencies of the various peaks are all different. The mathematical derivation is not the issue. $\endgroup$
    – user246795
    Jan 4 at 10:20
  • $\begingroup$ @user246795 Did you understand that Roger Wood answer's does indeed give an intuitive reason, or should I explain it more precisely ? $\endgroup$
    – Alfred
    Jan 4 at 11:15
  • $\begingroup$ I guess I didn't, if you could that would great thanks! $\endgroup$
    – user246795
    Jan 4 at 11:30
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    $\begingroup$ I had not considered the meaning of multiplying by $j\omega$ like that. So essentially are you describing the intuition of the chain-rule result $d[\omega X(\omega)]/d\omega = X(\omega) + \omega dX/d\omega$? Since at the peak of the displacement $dX/d\omega = 0$, it means for a zero in the slope $d[\omega X(\omega)]/d\omega$ we need a higher $\omega$ to make the term $\omega dX/d\omega$ more negative. This is insightful, but in my original question I meant "is there an intuitive (e.g. considering energy dissipation etc.) reason for the differences in frequencies". I'll add an edit. $\endgroup$
    – user246795
    Jan 4 at 12:29
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    $\begingroup$ @Alfred But the only "resonance" is at Ο‰=Ο‰0.Changing the usual definition of a word is generally a bad idea. You might well have studied resonance in Mechanics?When you did so you most likely looked for an amplitude peak and called it resonance?Condition for resonance and ..... difference between natural, fundamental, resonant, and forced frequencies? $\endgroup$
    – Farcher
    Jan 5 at 8:21
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I think that the confusion I have may stem from the simple question: why does the frequency that produces maximum power absorption (Ο‰=Ο‰0 when the driving force is in phase with the velocity) not lead to a maximum in the potential energy (i.e. peak in the amplitude response) of the oscillator?

That is exactly where your confusion stems from. The one-dimensional (damped) harmonic oscillator has a two-dimensional phase space, meaning that position and velocity (strictly speaking: momentum) are completely independent from one another, statically. They are only coupled dynamically.

While the elastic energy of the spring element (which you refer to as "the" potential energy, although it is a property of the spring of this non-conserved system) is only a function of position, power dissipation is only a function of velocity (for the given excitation force).

Just as the underlying phase space degrees of freedom $x$ and $v$ (read: $p$) are independent of each other at every single instant $t$, potential energy and power dissipation are independent quantities for each $t$.

So the proper question should be:

why, at all, should potential energy and power dissipation be related in just the right way so as to assume their excitation maxima under sinusoidal force at exactly the same frequency?

Asking for an intuition about why they don't behave like that is, in some sense, like asking for an intuition about why there are no fairies and elves. At least as long as you do not provide a good explanation for your counter-intuition, that fairies and elves should actually exist...

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You must absorb the natural frequency $\omega_0$ into the units of time, and expel them from $\gamma =2\zeta \omega_0$, non-dimensionalizing. Thus, you end up with the standard complex form, $$ \ddot{x} + 2\zeta \dot{x} + x = e^{i\omega t}. $$ Now ΞΆ will turn out to be the half-width of the resonance curve.

To start with, the maximum amplitude of the displacement has to be at a frequency smaller than the natural undamped frequency, here 1, since you pump/replenish energy into the system by prolonging its cycle, i.e. stretching it longer at the point of minimum kinetic energy. For optimal coupling, the velocity goes through zero and changes sign a bit earlier than the driving force. (Some people learn something from this.)

So the displacement amplitude is maximum at $\omega_d=\sqrt{1-2\zeta^2}<1$, and the power dissipated and absorbed is $\propto \omega^2 A_d^2= A_v^2$. If the system were not damped, the displacement amp would diverge at $\omega=1$.

The displacement amp directly worked out in the WP link provided is then
$$ A_d= {1\over \omega \sqrt{(\omega -1/\omega)^2+ 4\zeta^2}} \propto {1\over \sqrt{(\omega^2 -(1-2\zeta^2))^2 + O(\zeta^4) }} . $$

The three amps, as functions of Ο‰, are related to each other, and hence so are their different maximizing frequencies.

Note the acceleration amp is $A_a=-\omega^2 A_d$ and the right-angle-phased velocity amp is $A_v= \omega A_d$, hence $$ |A_d ~ A_a|= |A_v|^2, $$ while $$A_v(\omega)= A_v(1/\omega)= ((\omega -1/\omega)^2+ 4\zeta^2)^{-1/2}$$ peaks at $\omega=1$, by simple inspection of the above.

I don't have a "physical intuition" for $$ A_a(\zeta, \omega) = A_d (\zeta, 1/\omega) $$ which yields your respective resonant frequencies by inspection, since the displacement is maximally out of phase with the acceleration (and hence one reaches its minimum when the other achieves its maximum, and vice versa).

It is evident by the $\omega \to 1/\omega$ symmetry that the resonant frequencies are then the inverse of each other as functions of $\zeta$. The formulas are so gloriously more eloquent than physical "stories", here...


Addendum

If you bravely ventured into F Crawford's book Waves (Berkeley Physics Course, Vol. 3) ISBN-13:978-0070048607 to monitor all input power, friction dissipated power, and stored energy time averages, you might go in depth in Sec 3.2; in this language, you'd directly work out the averaged input (and, necessarily, friction dissipated) power to be $$ \langle P\rangle = \frac{\zeta \omega^2}{(1-\omega)^2 + 4\zeta ^2 \omega^2} ~~,\tag{21} $$ peaking at Ο‰=1, like the velocity; and, further, the stored energy, summing kinetic and potential terms, $$ \langle E \rangle =\tfrac{1}{4}\frac{\omega^2 +1}{(\omega^2-1)^2+4\zeta^2 \omega^2 } ~. \tag{23} $$ At Ο‰=1, the two are simply related.

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  • $\begingroup$ Could you clarify what you mean by "since you pump/replenish energy into the system by prolonging its cycle, i.e. stretching it longer at the point of minimum kinetic energy."? $\endgroup$
    – user246795
    Jan 4 at 10:02
  • $\begingroup$ The link you originally refer to is similarly close. I just don't understand what is meant by "For the former [maximum AMPLITUDE], you would like the frequency to be slightly lower (because you dissipate a certain amount of power per cycle)." $\endgroup$
    – user246795
    Jan 4 at 12:13
  • $\begingroup$ Apologies for the peremptoriness. It would hep if you schematically plotted the cycles for x, v, and F (and a, if you wanted, but it is irrelevant in the energetics) for Ο‰ near 1, with x(0)=0, so F is in phase with v, enhances it rather than damp it, and the power input is maximized: the oscillator is absorbing maximal power to offset its frictional losses. In the "optimal energy absorption" case, note F crosses 0 at t= Ο€/2, slightly before v, that is, it starts opposing it a bit and so prolongs the swing, thus increasing the period and ever so slightly decreasing Ο‰. $\endgroup$ Jan 4 at 16:08
  • $\begingroup$ That makes more sense. The last thing is: why does the arrangement that absorbs maximal power not give a maximum in the amplitude (potential energy). That is, why should the maximum amplitude of the displacement occur when $F$ opposes $v$ a little bit and increases the period? Thanks for your help so far. $\endgroup$
    – user246795
    Jan 4 at 20:09
  • $\begingroup$ I'm not sure I did not reverse the order of the above crossings! t= Ο€/2 we are discussing is the maximum of the displacement, at the very top of the cycle. v crosses 0 at t= Ο€/2 but F bit later, at t= Ο€/2Ο‰. So the arrangement favors Ο‰<1 , where 1 is the natural undamped frequency (your $\omega_0$). The swing wants to start falling, but the driver pulls it back, stretching the period and the potential energy, the surplus to be expended on matching frictional energy losses, in the steady state. $\endgroup$ Jan 4 at 20:50
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I think that the confusion I have may stem from the simple question: why does the frequency that produces maximum power absorption (Ο‰=Ο‰0 when the driving force is in phase with the velocity) not lead to a maximum in the potential energy (i.e. peak in the amplitude response) of the oscillator?

Intuitively, it is because the existence of damping makes the transformation of kinetic energy into potential energy not perfect. What I mean is the following:

  1. If there was no damping, then $\omega_{dis} = \omega_{vel}$, which you may see not only by substituting in the equation, but also by considering that no energy is lost, so kinetic energy is being transformed into potential energy and viceversa as the system oscillates. For that reason, if you want to get the maximum displacement, you may achieve that by setting $\omega = \omega_0$ because this yields the maximum amplitude in velocity, which in turn yields the maximum amplitude in displacement.

  2. If there is damping, and you still want to get the maximum displacement, the previous reasoning does not work anymore. There is energy dissipation because of damping, and furthermore, this damping is proportional to velocity. So, imagine you try to do the same thing as before, and set $\omega = \omega_0$, getting the maximum amplitude in velocity and let's say that we are talking about a mass attached to a spring horizontally, just to put a simple example. At some instant, the mass has maximum kinetic energy, and its "large" (again, because $\omega = \omega_0$) but also damping is large (its proportional to $v$) and so you have little efficiency in converting kinetic energy into potential energy, this means that you are not going to have the largest possible amplitude in potential energy, or equivalently, in displacement. What can you do about it? Improve the efficiency. How? Decreasing damping, and this you may achieve by having less speed, therefore $\omega \neq \omega_0$ which explains why the $\omega$'s are different.

    Of course there is a compromise: you cannot decrease too much the amplitude of $v$, because although you will get a very good efficiency of conversion of kinetic energy into potential energy, you won't have too much kinetic energy to transform!

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