Resonance and damping

When we discussed simple harmonic motion and simple harmonic systems, we were talking about free oscillations. This is what occurs when we put some energy into the system (by pulling a mass down on a spring, for instance) and then let the system continue by itself, with no extra energy being added and no energy being taken away.

You may notice that the above definition actually tells us immediately that, in the real world, a mass-spring system cannot ever oscillate freely, because there will always be some energy transfer to the surroundings: as the mass moves through the air, energy is lost to air resistance, and even if we performed the experiment in a vacuum there would still be some energy dissipated by the spring as heat as it is stretched and compressed. The system therefore loses energy over time. This effect is known as damping.

We know that the system will naturally oscillate, but we can also force it to oscillate manually. The result is, unsurprisingly, known as forced oscillation. If you keep pushing the mass to ‘help’ it move, you’re causing a forced oscillation.

You might think of forced oscillations as the result of cooperation between the external force (you pushing the mass) and the natural forces in the system (the restoring force caused by the spring). The natural force will ultimately cause the system to oscillate at its natural frequency, but the external force might not: you can push the mass as often as you like. However, if the two forces work together, the result is a much larger amplitude of oscillation than if they work against each other. This effect is known as resonance.

Damping

Oscillation in a system can only continue for as long as there is energy in the system in some form. This does not have to be kinetic energy: recall that at the extremes of oscillation, $v = 0$ , so the object oscillating will have zero kinetic energy. Instead, there is potential energy, and this causes the object to accelerate again, gaining kinetic energy. With no energy loss, $E_\text{kinetic} + E_\text{potential}$ will remain constant, although the balance between them is always changing.

Inevitably, however, energy will be lost, and the oscillation will stop. Resistive forces such as air resistance and friction are major factors that contribute to energy loss. There are other, smaller, losses – for example, a system that uses a rubber band would lose some energy as heat when the rubber band stretches.

The loss of energy from an oscillating system causes an effect known as ‘damping’. It can be observed as a reduction in amplitude over time. To see how this will affect a system, consider our equation for displacement during simple harmonic motion:

$x=A\cos \omega t.$

If we reduce $A$ (such as by increasing the amount of damping), the range of $x$ will also be reduced. However, the $\cos \omega t$ term is independent of $A$ and thus the frequency of the oscillation does not change.

The rate at which the amplitude decreases in a damped system is typically exponential. That is,

$A\paren{t}={A_0} e^{-rt}$

where $A\paren{t}$ is the amplitude at time $t$ , $A_0$ is the initial amplitude, and $r$ is some constant that describes the rate of decay.

Using damping

Damping is a key consideration in any system that involves oscillators, because it shows up wherever there’s energy dissipation, and that’s everywhere.

A system that has no damping can be described as ‘undamped’. Such systems are entirely theoretical, because energy will always escape somehow. A true perpetual motion machine would be undamped. Sometimes we can model systems as being undamped, but only under the right circumstances. For example, if a system loses energy very slowly and we only need to consider a small number of oscillations, we may be able to assume negligible energy loss over the duration we care about. This is what we’ve been doing in our analysis of oscillators up to this point.

Damping can form a key part of a system. A simple analogue weighing scale has a needle that points to whatever value the scale is currently reading, but as many such scales use a spring inside, the needle will naturally oscillate back and forth rather than pointing directly to the correct value. We can use damping to our advantage to reduce this effect, making the scale easier to use. It’s important that we choose the right amount of damping, though, as you’ll see.

If we designed the scales without paying much attention to the oscillation of the needle, we might end up with something like this. The needle moves really fast, but it moves too fast and actually overshoots the target. This causes it to oscillate back and forth, which makes it hard to tell what the exact reading is. Luckily, the needle does settle after a short while due to the frictional losses, but it’s still not ideal for this use case. This is an ‘underdamped’ system.

We can reduce this oscillation by intentionally designing the system to dissipate the needle’s kinetic energy more quickly, but this can lead to the needle moving too slowly. Sure, it doesn’t wobble around anymore, but it takes a very long time to reach the right value (and in fact, it technically never quite gets there). This is an ‘overdamped’ system.

If we get things just right, we get the best of both worlds: the needle moves as fast as possible to the right position, but without overshooting it. This means we don’t have to wait for it to stop wobbling around before we are able to read it. This is now a ‘critically damped’ system.

In the above example it’s pretty clear that critical damping is ideal. However, that’s not always the case. In any situation where we need oscillation, our only option is an underdamped system (whether we’re adding damping or just allowing the natural damping of the system to do the job for us). For instance, inside a piano there are dampers touching the strings; these serve primarily to prevent each note lasting too long. Underdamping is essential here because although we do want to increase the rate of amplitude decay, we do still need the strings to oscillate – otherwise, we wouldn’t hear any notes at all. By pressing the sustain pedal we can lift the dampers; when this happens, the only damping effect will be the natural damping of the strings themselves, so the notes will last longer.

Applications of overdamping are primarily related to ‘smoothness’ – for example, the suspension in a luxury car might be slightly overdamped so that the passengers feel as few bumps as possible. We’re not too bothered about how quickly the suspension returns to exactly its original position, so critical damping wouldn’t necessarily be ideal here. The faster movement that would result from critical damping might cause a decrease in ride quality.

Resonance

Definition

‘Resonance’ is the name given to the behaviour of a system when it oscillates at a higher amplitude at a particular frequency (or a set of frequencies) than it does at other frequencies.

Provided we can exert the necessary force, we can make a system oscillate at any frequency we want it to. This is true even for things that we wouldn’t normally consider oscillators – if I move a pencil back and forth on my desk, I could argue that I’m causing it to oscillate.

In the case of the pencil, we will not observe anything particularly interesting, no matter how I change the nature of the oscillation: at any frequency, the movement of the pencil is determined entirely by what I do to it. This is because there is no restoring force acting on the pencil.

If I did the same to a system that did have a restoring force, it’s no longer just me: the total force is the vector sum of the force I exert and the restoring force. If I suddenly take my hands away and leave the object to rest, it will tend to oscillate at a particular frequency. This tendency is still present when I exert a force on the object, and the result is that when I match my frequency up with the system’s preferred frequency, the amplitude of the oscillations becomes larger. This is resonance.

What’s the maths behind resonance?

Maths ahead!

The level of detail we’ll be going into here is far beyond the scope of any A level physics specification. Feel free to skip over everything inside this foldout. (Just make sure you have a solid understanding of everything outside it, because there’s a decent chance you will need to know that.)

As a starting point, we’ll take a simple harmonic system – that is, one which experiences a restoring force $F_h$ proportional to the displacement from some equilibrium position – and add two new forces: an external driving force $F_d$ , and a resistive force $F_r$ . The addition of these two forces mean that the system’s oscillation will no longer be simple harmonic motion, so we can no longer rely on the same equations we’ve been using.

By virtue of this being a simple harmonic system, the natural restoring force must be given by

$F_h = -kx$

for some constant $k$ .

Note

We’ll be dealing a lot with displacement, velocity and acceleration, and it’s often useful to express them as derivatives: $x$ , $\diffn{2}{x}{t}$ and $\diffn{3}{x}{t}$ respectively. To keep things neat, we’ll just write $x$ , $\dot{x}$ and $\ddot{x}$ .

We will assume that the resistive force is given by

$F_r = -r\dot{x},$

for some constant $r$ . Finally, we assume that the external force $F_d$ is sinusoidal, so we will write it as

$F_d = q \sin \paren{{\omega}_d t},$

where $q$ is the maximum magnitude of the force and $\omega_d$ is the angular frequency of the force, equal to $2\pi$ times the frequency of the force.

As resonance is characterised by the variation of amplitude with frequency, our aim will be a formula for the amplitude of the system’s oscillation in terms of the frequency of the driving force.

The overall force $F$ acting on the system at time $t$ is the sum of all the forces we’ve defined so far:

$F = -kx -r\dot{x} + q \sin \paren{{\omega}_d t}.$

As $F=m\ddot{x}$ , this becomes

$m\ddot{x} + r\dot{x} + kx = q \sin \paren{{\omega}_d t}.$

Dividing through by $m$ , we get

$\ddot{x} + \frac{r}{m}\dot{x} + \frac{k}{m}x = \frac{q}{m} \sin \paren{{\omega}_d t}.$

The first step to solving this equation is to solve the simpler case where $F_d=0$ ; i.e. when the driving force is zero.

$\ddot{x} + \frac{r}{m}\dot{x} + \frac{k}{m}x = 0.$

This turns out to be really easy. Consider what it would mean to solve the above differential equation: our solution would be some function $x\paren{t}$ that gives us the displacement of the system due to the damped natural oscillation. We have seen already that amplitude decays exponentially towards zero with time, so we can assume that for large values of $t$ the natural oscillation will have decayed so much that it is negligible. Our solution is thus $x=0$ for this first equation, and our overall solution will come entirely from the alternative case, where $F_d\ne 0$ .

Note

‘Large’ values of $t$ may only be a couple of seconds, or even less, depending on the system.

Given that our driving force is sinusoidal, we might expect a sinusoidal displacement. We will assume such a solution exists, and attempt to find it. If we find a solution, then we’ve solved the problem; if we can’t find one, we’ll have to try something else.

If a sinusoidal solution exists, it will take the form

$x=A \sin \paren{ \omega_d t + \phi }.$

We need to substitute this into our differential equation, but we don’t want the phase shift inside the $\sin$ , because that will make things messy. Using the addition formula for $\sin$ , we can obtain

$x=A \cos\paren{\phi} \sin \paren{\omega_d t} + A \sin\paren{\phi} \cos \paren{\omega_d t}.$

To clear things up we replace the constant coefficients with $B$ and $C$ such that

$x=B\sin\paren{\omega_d t}+ C\cos\paren{\omega_d t}.$

We need to find $\dot{x}$ and $\ddot{x}$ , too. Differentiating once gives us

$\dot{x}=B\omega_d\cos\paren{\omega_d t} - C\omega_d\sin\paren{\omega_d t},$

and twice,

$\ddot{x}=-B\omega_d^2\sin\paren{\omega_d t} - C\omega_d^2\cos\paren{\omega_d t}.$

After performing the substitutions, we get quite a lot of trig flying around. For brevity, we will use $s$ and $c$ respectively for the sine and cosine of $\omega_d t$ .

$\begin{align*} -B\omega_d^2s &- C\omega_d^2c \\ &+ \frac{r}{m}B\omega_dc - \frac{r}{m}C\omega_ds \\ &+ \frac{k}{m}Bs + \frac{k}{m}Cc \\ &= \frac{q}{m}s. \end{align*}$

It might look scary, but it’s nothing particularly complex – just a lot of stuff in one place. We can see that on the right-hand side, the coefficient of $c$ is zero (because there’s only one term, and it’s a term in $s$ ). Therefore, the sum of the coefficients of the $c$ terms on the left-hand side must be zero:

$- C\omega_d^2 + \frac{r}{m}B\omega_d + \frac{k}{m}C = 0.$

Equating the coefficients of $s$ , we get

$-B\omega_d^2- \frac{r}{m}C\omega_d+ \frac{k}{m}B= \frac{q}{m}.$

The algebra isn’t difficult, but there’s quite a lot to do so we won’t cover it here. However, solving the above two equations simultaneously gives us

$B=\frac{q}{R}\paren{k-m\omega_d^2}$

and

$C=\frac{q}{R}r\omega_d,$

where

$R=k^2-2km\omega_d^2+m^2\omega_d^4 + r^2\omega_d^2.$

Recall that

$C=A \sin{\phi}$

and

$B=A \cos{\phi}.$

Therefore,

$C^2=A^2 - B^2,$

and it follows that

$A = \sqrt{B^2 + C^2}.$

Hint

This can be deduced from the identity

$\sin^2{\phi} + \cos^2{\phi} = 1.$

We could keep going to find a complete solution to the differential equation, but we don’t need to: $A$ is the amplitude, which is what we’re looking for. All that remains is to get this in a single neat equation.

We can take a factor of $\frac{q}{R}$ out of the square root to get

$A = \frac{q}{R} \sqrt{\paren{k-m\omega_d^2}^2 + r^2\omega_d^2}.$

It’s not too difficult to see that the square root just evaluates to $\sqrt{R}$ , which becomes $\frac{1}{\sqrt{R}}$ when multiplied by $R$ . However, we’ll actually use the above form because it’s neater.

Hint

If you expand out the terms on the inside of the square root, you’ll find that they equal $R$ .

We now have

$A = \frac{q}{\sqrt{\paren{k-m\omega_d^2}^2 + r^2\omega_d^2}},$

which can be written

$A = \frac{q}{m\sqrt{\paren{\frac{k}{m}-\omega_d^2}^2 + \paren{\frac{r\omega_d}{m}}^2}}.$

This might seem arbitrary, but as we based this all on a simple harmonic system, there’s a neat substitution to make:

$\omega_n^2 = \frac{k}{m},$

where $\omega_n$ is the natural angular frequency of the system. Therefore,

$A = \frac{q}{m\sqrt{\paren{\omega_n^2-\omega_d^2}^2 + \paren{\frac{r\omega_d}{m}}^2}}.$

In many cases it is more meaningful to use ‘normal’ frequency instead of angular frequency, so we use

$\omega = 2\pi f$

to get (after some simplification)

$A = \frac{q}{4\pi^2 m\sqrt{\paren{f_n^2-f_d^2}^2 + \paren{\frac{rf_d}{2\pi m}}^2}}.$

The general relationship is thus

$A \propto \frac{1}{\sqrt{\paren{f_n^2-f_d^2}^2 + cf_d^2}}$

for some constant $c$ .

Resonance is a result of the relationship

$A \propto \frac{1}{\sqrt{\paren{f_n^2-f_d^2}^2 + cf_d^2}},$

where $A$ is the amplitude of the oscillation, $f_n$ is the natural frequency of the system, $f_d$ is the frequency of the periodic driving force, and $c$ is some constant that represents the strength of the damping on the system. Notice that this relationship is just one of proportionality: the actual constant of proportionality depends on the system in question. (If you’re interested in knowing the details of all this, check the above foldout.)

Resonance curves

When we need to know how a system responds to different driving frequencies, we can draw a graph, known as a ‘resonance curve’, that shows how the amplitude of the oscillation varies with the driving frequency. Resonance shows up very clearly on such diagrams:

A sketch of a typical resonance curve. The driving frequency at which the amplitude is greatest is known as the ‘resonant frequency’.

The resonant frequency can be identified as the value on the frequency axis, the $x$ ‑axis, which produces the greatest amplitude (on the $y$ ‑axis).

Be really, really careful!

The resonant frequency is not the same as the natural frequency. A common mistake is to assume the two are the same because they are typically fairly close to each other. Even many educational resources get this wrong; make sure you don’t get them mixed up too.

Here’s the same diagram as above, but with the two frequencies marked:

f_r

is the resonant frequency, and

f_n

is the natural frequency.

Sure, they’re close, but they’re definitely not the same. And look what happens when the system has a bit more damping:

Now they’re not even close to being equal.

Increasing the damping doesn’t affect the natural frequency, but it does affect the resonant frequency: the more damping there is, the lower the resonant frequency will be. In an undamped system the two are equal, but this is the only such case; in all other instances the resonant frequency will be below the natural frequency. Below is a diagram showing the resonance curve for a theoretical undamped system.

Around the resonant frequency, the amplitude of an undamped system goes to infinity.

The infinite amplitude we see is a result of there being zero energy loss, but non-zero energy input: the driving force is constantly adding energy to the system, but with zero damping there is no way for energy to escape, so at the resonant frequency (which, as mentioned, is equal to the natural frequency in the undamped case) the amount of energy in the system goes to infinity, which means the amplitude also goes to infinity.

How can we find the resonant frequency?

As the resonant frequency occurs at the point where the amplitude is greatest, the derivative of amplitude with respect to driving frequency must be zero at that point:

$\at{\diff{A}{f_d}}{f_d=f_r} = 0.$

To find $f_r$ , therefore, we just need to solve the above equation. We begin by writing $A$ in index form:

$A\propto \paren{\paren{f_n^2 - f_d^2}^2 + cf_d^2}^{-\frac{1}{2}}.$

Differentiating the above and cleaning things up shows us that

$\diff{A}{f_d} \propto A^3 f_d \paren{2\paren{f_n^2 - f_d^2}-c}.$

Note

We’re able to ignore any constants that multiply the whole of the right-hand side, because we’re representing this as a proportionality relationship rather than an equality.

When $f_d=f_r$ , the whole of the right-hand side becomes zero. We know that $A\ne 0$ and $f_d\ne 0$ , so we’re left with

$2\paren{f_n^2 - f_r^2}-c = 0.$

Rearranging, we get

$f_r = \sqrt{f_n^2 - \frac{c}{2}}.$

In a system where the amplitude and driving frequency are related by

$A \propto \frac{1}{\sqrt{\paren{f_n^2-f_d^2}^2 + cf_d^2}},$

the resonant frequency $f_r$ is given by

$f_r = \sqrt{f_n^2 - \frac{c}{2}}.$

This immediately shows us three key things:

$f_r$ does not depend on the constant of proportionality in the amplitude relationship (i.e. the ‘scale’ of the amplitude).
As $c$ increases, $f_r$ drops further below $f_n$ . $c$ is related to the amount of damping, so this means that increased damping reduces the resonant frequency. As $c$ can never be negative, $f_r$ can never be greater than $f_n$ .
The only time when $f_r=f_n$ is when $c=0$ ; that is to say, when the system is undamped.