Difference between resonance in undamped vs damped mass-spring-dashpot

1) "Undamped system is forced at the same frequency as one of its natural frequencies."

Consider the 2nd order differential equation

$$\ddot{x}+\omega_0^2x=F_0\cos{\omega t}\tag{1}$$

which models a mass attached to a spring (attached to a wall) with spring constant ##k## and ##\omega_0=\sqrt{\frac{k}{m}}##.

A particular solution is

$$\frac{F_0\cos{(\omega t)}}{\omega_0^2-\omega^2}\tag{2}$$

if ##\omega_0\neq \omega## and

$$\frac{F_0 t\sin{(\omega_0 t)}}{2\omega_0}\tag{3}$$

when ##\omega_0=\omega##.

The latter case is called resonance, and is characterized by an amplitude of oscillation that grows in time.

This is the case of a pendulum or a swing in which there is a force that is synchronized with the natural frequency such that the force is zero when the amplitude is at its peaks and acts to "push" the pendulum or swing with increasing magnitude towards the equilibrium point, and once that equilibrium point is reached the force continues to push but with decreasing intensity until it reaches zero again at the next peak.

This timing makes it so that each successive peak has a greater amplitude.

One doubt I have is about what it means that an undamped system has multiple natural frequencies.

2) If we consider a general mass-spring-dashpot system the equation that describes the system is

$$m\ddot{x}+c\dot{x}+kx=F_0\cos{\omega t}\tag{4}$$

The steady-state solution is

$$\frac{F_0}{\sqrt{(k-\omega^2m)^2+\omega^2c^2}}\cos{(\omega t-\phi)}\tag{6}$$

The amplitude of the steady state solution is

$$A=\frac{F_0}{\sqrt{(k-\omega^2m)^2+\omega^2c^2}}\tag{7}$$

Of this expression, Simmons book "Differential Equations with Applications and Historical Notes" says

This expression for the amplitude holds some interesting secrets, for it depends not only on ##\omega## and ##F_0## but also on ##k,c##, and ##m##. As an example, we note that if ##c## is very small and ##\omega## is close to ##\sqrt{k/m}## (so that ##k-\omega^2m## is very small), which means that the motion is lightly damped and the impressed frequency ##\omega/2\pi## is close to the natural frequency

$$\frac{1}{2\pi}\sqrt{\frac{k}{m}-\frac{c^2}{4m^2}}\tag{8}$$

then the amplitude is very large. This phenomenon is known as resonance.

This case does not involve a growing amplitude, but rather simply a large amplitude. It seems that one could make the amplitude arbitrarily large if we are allowed to tweak any of the parameters.

For a given set of parameters except ##\omega##, there seems to be a maximum amplitude and this is shown in part 3 below.

3) I found a third use of the term resonance in the following problem.

Consider the forced vibration in (6) in the underdamped case and find the impressed frequency for which the amplitude attains a maximum.

Will such an impressed frequency necessarily exist? This value of the impressed frequency when it exists) is called the resonance frequency. Show that it is always less than the natural frequency.

If we consider (7) as a function of ##\omega## and try to find a critical point, we find

$$\omega=\sqrt{\frac{2km-c^2}{2m^2}}=\sqrt{\frac{k}{m}-\frac{c^2}{2m^2}}\tag{8}$$

which is the resonance frequency. Compared to the natural frequency

$$\sqrt{\frac{k}{m}-\frac{c^2}{4m^2}}\tag{9}$$

we see that the resonance frequency is smaller.

The amplitude for this frequency is

$$\frac{2F_0m}{\sqrt{c^2(4km-c^2)}}\tag{10}$$

Once more, the amplitude does not grow in time, but we can make it arbitrarily large at the resonance frequency by tweaking the parameters.

In particular, for a damped system, the closer we are to critical damping (ie, the closer ##4km-c^2## is to zero) the larger the amplitude.

Here is an example of this latter case

$$\ddot{x}+\dot{x}+2x=\cos{(\omega t)}\tag{11}$$

Here is a plot of amplitude as a function of ##\omega##



In summary, I showed three cases.

2) and 3) actually seem to be the same thing.

I was under the impression that the concept of resonance was only case 1.

However, resonance frequency seems to be the frequency at which amplitude is maximized in a general 2nd order linear differential equation with constant coefficients and sinusoidal input.

One further comment.

If there is no external sinusoidal input, then the system has a steady state of zero. In this case, it does not seem there is any resonance frequency.

However, if there is sinusoidal input, then the amplitude of the steady-state has a max relative to the input angular frequency.

In practice, I think you'll find lots of different informal usages. IMO, the term resonance applies any time the solution contains complex roots, at least for linear (or linearized) systems. This would include low Q systems that don't appear to be "very resonant", but wouldn't include systems that have frequency peaks constructed of simple poles/zeros, like a bandpass filter for example.

Other people will undoubtedly give you a different definition.

DaveE said:

In practice, I think you'll find lots of different informal usages. IMO, the term resonance applies any time the solution contains complex roots, at least for linear (or linearized) systems. This would include low Q systems that don't appear to be "very resonant", but wouldn't include systems that have frequency peaks constructed of simple poles/zeros, like a bandpass filter for example.

Other people will undoubtedly give you a different definition.

A simple harmonic oscillator with no external input has complex roots but is not in resonance.

In addition, consider the system

$$\ddot{x}+\sqrt{2}\dot{x}+0.75x=\cos{\omega t}$$

The characteristic polynomial is ##r^2 +\sqrt{2}r+0.75## and the roots are complex.

Here is a plot of the amplitude of the steady-state relative to the input frequency ##\omega##



When ##\omega=0## there is no oscillation so the amplitude is a fixed long-term position that the system is in.

The roots being complex does not seem to be a sufficient condition.

In the OP, this point was made when comparing equations (8) and (9): the input angular frequency that maximizes amplitude is smaller than the natural angular frequency.

This maximizing input frequency does not always exist however since it is

$$\sqrt{\frac{2mk-c^2}{2m^2}}$$

and the natural frequency is

$$\sqrt{\frac{4mk-c^2}{4m^2}}$$

when ##4mk-c^2\geq 0## (ie, the discriminant of the characteristic polynomial is ##c^2-4mk\leq 0##).

The maximizing input frequency only exists when we have the more stringent condition that ##2mk-c^2\geq 0##.

Thus, if an underdamped system is nonetheless dampened too much, then no resonance occurs. There is a maximum level of damping, namely ##c\leq\sqrt{2mk}##, such that resonance can occur. By resonance, in this paragraph, I mean a local maximum for amplitude.

A system capable of resonance will not spontaneously resonate. To exhibit resonance it must be suitably excited.
The child on a swing will need to pushed at the appropriate frequency by external forces. The Tacoma Narrows bridge will need wind and some von Karmann shedding.

zenterix said:

does it also have meaning in a system that has damping?

All real systems above the quantum level have damping. Resonance happens.

zenterix said:

One doubt I have is about what it means that an undamped system has multiple natural frequencies.

Well, it's hard to find undamped systems in classical physics. But a vibrating string is an example of a system with multiple natural frequencies.

zenterix said:

A simple harmonic oscillator with no external input has complex roots but is not in resonance.

OK, but it is a resonant system. It could resonate if excited. A speaker is still a speaker when the stereo is turned off.

hutchphd said:

The Tacoma Narrows bridge

… was the result of aeroelastic flutter rather than mechanical resonance.

I think we say that Q must be greater than 0.5 for the circuit to be resonant. This is where a reversal of either displacement or force is seen, in other words we see a wave. I think Q=0.5 corresponds to critical damping. I might also mention that for low Q systems, the resonant frequency can be defined either as the frequency of maximum amplitude or the frequency of zero phase shift, and the two can differ slightly.

Orodruin said:

was the result of aeroelastic flutter rather than mechanical resonance.

Potato, PotAHto. The bridge was oscillating, and the details are in fact fascinating: call it what you will.

hutchphd said:

Potato, PotAHto. The bridge was oscillating, and the details are in fact fascinating: call it what you will.

Orodruin. Properly naming disastrous physical phenomena since 2014.

Your terminology is in fact more succinct, but I got to use "von Karmann shedding" which is more descriptive......

zenterix said:

The roots being complex does not seem to be a sufficient condition.

For what?

I guess your definition of resonance is a transfer function ##|H(ω_0)| > 1##? OK, be prepared for many engineers to use the presence of complex roots (or ##Q>\frac{1}{2}##) definition instead of your preferred ##Q>1## choice.

These notes seem to explain what I am getting at in my posts above.

They define more terms, practical resonance frequency and practical resonance.
They say

If the amplitude has a peak at ##\omega_r## then we call this the practical resonance frequency. If the damping ##b## gets too large then, for the system

$$mx''+bx'+kx=B\cos{\omega t}$$

there is no peak and hence no practical resonance.

and

Practical resonance occurs at the frequency ##\omega_r## where the gain has a maximum.

The gain of such a system is

$$g(\omega)=\frac{1}{|p(i\omega)|}=\frac{1}{\sqrt{(k-m\omega^2)^2+b^2\omega^2}}$$

where ##p## is the characteristic polynomial of the differential equation.

The max occurs when the denominator has a minimum.

This happens when

$$\omega_r=\sqrt{\frac{k}{m}-\frac{b^2}{2m^2}}$$

In my original post I found ##\omega_r## by maximizing amplitude, but this is equivalent to maximizing gain, which amounts to minimizing the denominator of gain, and the minimum occurs at a value of ##\omega>0## only when certain conditions are met in addition to the roots being complex.

DaveE said:

For what?

I guess your definition of resonance is a transfer function ##|H(ω_0)| > 1##? OK, be prepared for many engineers to use the presence of complex roots (or ##Q>\frac{1}{2}##) definition instead of your preferred ##Q>1## choice.

See the post above.

Why do you want to say the system in Fig 1b is displaying resonance?

It has complex roots but no resonance in the SHM sense of increasing amplitude, and also no maximal amplitude of actual oscillation. The gain is always smaller than ##1##.

When you say transfer function is this the equivalent of gain?

tech99 said:

I think we say that Q must be greater than 0.5 for the circuit to be resonant. This is where a reversal of either displacement or force is seen, in other words we see a wave. I think Q=0.5 corresponds to critical damping. I might also mention that for low Q systems, the resonant frequency can be defined either as the frequency of maximum amplitude or the frequency of zero phase shift, and the two can differ slightly.

What is Q?

Quality factor of oscillator. It is Q=2π(energy stored)/ (energy dissipated) per cycle of a nondriven (homogenious) harmonnic oscillator. You need to study this: any calculus based Freshman physics book. This is the good stuff.

The canonical form of a second order response can be expressed in terms of the quadratic polynomial:
$$ P(s) =(\frac{s}{\omega_o})^2 + \frac{1}{Q} (\frac{s}{\omega_o})+1 $$
Where ##\omega_o## is the resonant frequency, and ##s=j \omega## for sinusoids.

Honestly, there's too much to learn about the SHO to be done in one reply. Ask a more specific question for more info.