Damping & Resonance

Imagine you have a spring. You attach a mass to the bottom end of the spring, and let it hang until equilibrium is reached (the spring stops moving up and down). You then get bored, so you decide to pull the mass downwards, and the spring begins oscillating up and down in a constant time interval - Simple Harmonic Motion (SHM). Over time, you start to notice that the amplitude of the oscillation (how far displaced the mass gets from its position at the equilibrium point) decreases. This is due to damping.

In a perfect physical world, with no friction and the system kept in a vacuum, no energy would be lost from the system, so the spring would keep oscillating at its initial amplitude forever and ever (as the blue plot shows). Unfortunately, we don't live in physicist's Utopia, so there's gonna be some energy lost at various points during motion, whether that's through the heat produced when the mass and spring rub against each other, or the sound that the spring makes while stretching, or just the force of fluid resistance slowing oscillations down (imagine a spring in water - it'll move a lot slower than a spring in air due to the higher viscosity of the former). This means that eventually, like everything in this world, the oscillations will wither away into oblivion. We can quantify all the damping effects on an oscillating system nice and conveniently by chucking it all into one number called the "Damping Coefficient", which describes how quickly the oscillations die out (shown by the orange plot).

Now let's make things a bit more interesting. You want to increase the amplitude of the oscillations, so the obvious way to do this is to apply a periodic force to the mass (the force shouldn't be constant, because then the mass would just get stuck at the top or bottom): F = F₀ sin(ωt). If you ever go outside (unlikely because you're reading this), you may have noticed that when pushing someone on a swing, there is a particular rate (angular frequency) at which you push them so that they can go as high as possible (peak amplitude). As there is usually only one person pushing the swing, this push occurs when the swing has reached its crest. This can be thought of as a periodic force, with a period equal to the period of oscillation of the swing.

Going back to our example of the mass on the spring, we use the swing example to say that our resonant frequency (the frequency at which we push the spring up and down) is the same as the natural frequency (the frequency at which the spring oscillates). We can actually work out the natural frequency of the system analytically using the spring constant of the spring and the mass, as the amplitude of oscillation doesn't affect its natural frequency (derivation below if you love maths). The green plot shows the amplitude of the oscillation with respect to the frequency of the driving force (which has been divided by the natural frequency), and sure enough, when there's no damping, the peak amplitude is basically when the normalised driving frequency is 1. This matches with what we said about the swing - the frequency of pushing is the same as the frequency of oscillation to get maximum amplitude.

Hang on a minute. When the damping is increased, the peak amplitude becomes less than the natural frequency. This is because damping removes energy from the system, and so the frequency starts to decrease. Imagine you start rubbing something very quickly, but then you get tired, so the frequency of the rubbing decreases. In the same way, reducing energy decreases frequency. This effect of damping decreasing oscillation frequency can also be seen in the orange plot.

A common rote learned fact in school is that damping decreases the peak amplitude, decreases the resonant frequency and "broadens the peak". The reasons for the two facts have been explained conceptually, but I think that the third requires a little mathematical derivation.

Undamped

Let's begin with the simple undamped oscillator. The force equation on the spring is:

F = - k x

This equation does not involve the weight of the mass because we only care about the changing 'restoring force' (the force moving the mass back to equilibrium). At equilibrium, the weight of the mass is cancelled out by the spring force upwards, meaning the total force is zero. Due to the weight force being constant, we need not involve it in the force equation as we only care about the oscillatory forces.

\begin{matrix} F & = & m a \\ m a & = & - k x \\ m \overset{..}{x} & = & - k x \\ \overset{..}{x} & + & \frac{k}{m} x & = & 0 \end{matrix}

Now we have a second-order differential equation, we can generate our auxiliary equation and then use this to find our solution for x.

Auxiliary Equation:

\begin{matrix} λ^{2} & + & \frac{k}{m} & = & 0 \\ λ^{2} & = & - \frac{k}{m} \\ λ & = & \pm \sqrt{- \frac{k}{m}} & = & \pm i \sqrt{\frac{k}{m}} \end{matrix}

This yields a complementary function of the form:

\begin{matrix} α \cos (\sqrt{\frac{k}{m}} t) & + & β \sin (\sqrt{\frac{k}{m}} t) \end{matrix}

Using the addition formula for cosine, we can convert this to our solution for x:

\begin{matrix} x & = & A & \cos & ( & \sqrt{\frac{k}{m}} & t & + & φ & ) \end{matrix}

where

α = A \cos (φ)

and

β = - A \sin (φ)

Some eagle-eyed readers may have noticed that it is also possible to use the sine addition formula to get an equation of the form:

\begin{matrix} x & = & A \sin (\sqrt{\frac{k}{m}} t + φ) \end{matrix}

where

α = A \sin (φ)

and

β = A \cos (φ)

As shown, it is actually possible to use a sine wave as the function for our position, but generally, the practice is to use the cos function, as it fits with the initial conditions (these being that when t = 0, x is not 0, as it has been displaced to start the oscillatory motion).

Using our solution for x, we can see that the angular velocity,

ω_{0} = \sqrt{\frac{k}{m}}

, and the phase difference =

φ

. To simplify things, we can take the phase difference to be 0, so we have

x = A \cos (\sqrt{\frac{k}{m}} t)

as our solution for x, as shown on the blue plot.

Damped

Our damped oscillator has an extra term for damping in the force equation. This time, the term is proportional to velocity. The reason is because damping is the sum of the forces resisting an object's motion. Hence, if an object wasn't moving, there would be no damping at all, and so the damping force is proportional to velocity. Now, our force equation is:

F = - k x - c v

Now, we have to solve our differential equation again, but it is quite a bit more long-winded this time.

\begin{matrix} F & = & m a \\ m a & = & - k x - c v \\ m \overset{..}{x} & = & - k x - c \dot{x} \\ \overset{..}{x} & + & \frac{c}{m} \dot{x} & + & \frac{k}{m} x & = & 0 \end{matrix}

We now know that

\frac{k}{m} = {ω_{0}}^{2}

, and to simplify the equation, we'll say that

γ = \frac{c}{m}

. Our auxiliary equation is as follows:

\begin{matrix} λ^{2} & + & γ λ & + & {ω_{0}}^{2} & = & 0 \end{matrix}

Using the quadratic formula on the above yields:

\frac{- γ \pm \sqrt{γ^{2} - 4 {ω_{0}}^{2}}}{2}

We now have 3 possible solution sets for our auxiliary. These are when the radicand is positive (two real roots), zero (one real root), or negative (two complex roots). These 3 solutions all represent different types of damping. To simplify, let's say that

z = γ^{2} - 4 {ω_{0}}^{2}

Two Real Solutions: Overdamped

This set occurs when $z > 0$ . Two real roots for our auxiliary means that the solution for x only involves exponentials. The solution for x is therefore:

x = C e^{\frac{- γ + \sqrt{z}}{2} t} + D e^{\frac{- γ - \sqrt{z}}{2} t}

Now, as this solution involves two exponentials (and not sine or cosine), we can tell that the spring will not oscillate. Intuitively, this means that the spring must be heavily damped - if you try your hardest to stop as spring's motion, it will not oscillate, it will just go striaght to equilibrium and stop. This also joins up with our

z

requirement. For

z

to be greater than 0,

γ

must be big, which in turn means

c

must be big - meaning high damping. Try this out by watching the orange plot as you increase

c

One Real Solution : Critically Damped

This set occurs when $z = 0$ . Our solution for x is:

x = (C + D t) e^{\frac{- γ}{2} t}

and this shows Critical Damping. Critical Damping is when the stopping time is at a minimum. This is due to the linear term. As x increases, the linear term becomes more significant than the negative exponential term, and so x reaches 0 faster (due to a negative linear coefficient). Remember, x will not oscillate here, as there are no sines or cosines in the solution.

Conceptually, this makes sense, because the critical damping solution is in between the underdamped and overdamped solutions (0 discriminant), meaning that critical damping takes the shortest time.

Two Complex Solutions: Underdamped

This set occurs when $z < 0$ . Two complex roots for our auxiliary means that the solution for x will contain an exponential and oscillatory term. You can try deriving this from the solution for the real roots using Euler's formula and exponential rules. Our x solution looks like:

x = e^{\frac{- γ}{2} t} * (C \cos (\frac{\sqrt{γ}}{2} t) + D \sin (\frac{\sqrt{γ}}{2} t))

We can see that this function will oscillate due to the trigonometric functions involved. This solution conceptually makes sense because low

z

means low

γ

which means low

c

. It follows that a low damping coefficient means less damping, hence the spring will oscillate. Turn the damping knob to show this effect.

Resonance Curve

Alright, get your bomb suit on. This is about to get explosive.

Let's start with our force equation, this time including our periodic driving force. The aim of the game here is to find the relationship between a particular driving frequency and the peak amplitude that the spring reaches due to this driving force.

\begin{matrix} F & = & - k x - c v + F_{0} \cos ω t \\ m \overset{..}{x} & = & - k x - c \dot{x} + F_{0} \cos ω t \\ m \overset{..}{x} + c \dot{x} + k x & = & F_{0} \cos ω t \\ \overset{..}{x} + \frac{c}{m} \dot{x} + \frac{k}{m} x & = & \frac{F_{0}}{m} \cos ω t \end{matrix}

We can now use our definitions (for

γ

and

ω_{0}

) from the left. It is essential to note that

ω_{0}

refers to the NATURAL FREQUENCY of the spring (or the angular frequency of oscillation), while

ω

refers to the angular frequency of the DRIVING FORCE - these are different (very important).

\begin{matrix} \overset{..}{x} + γ \dot{x} + {ω_{0}}^{2} x & = & \frac{F_{0}}{m} \cos ω t \end{matrix}

We've arrived at something of a sticky situation. We could continue in the same way as the damped derivation, but as there are 3 different possibilities, a single general solution is not possible (from which we obtain our amplitude expression). Hence, we can include the complex numbers in our solution for

x

to find an equation for amplitude. Let's move the problem into the complex plane, by using exponentials and complex numbers to represent our oscillatory functions. We'll now use

z

to represent our complex

x

. We can rewrite the problem as:

\begin{matrix} \overset{..}{z} + γ \dot{z} + {ω_{0}}^{2} z & = & \frac{F_{0}}{m} e^{i ω t} \end{matrix}

z

is oscillatory, our "guess" for what

z

is can be written as the general oscillator function:

\begin{matrix} z & = & A e^{i (ω t - δ)} \end{matrix}

Where A is amplitude,

δ

is the phase difference and

ω

is the angular velocity of the spring. It is important to note that, due to the applied driving force, the spring will now oscillate at the frequency of the driving force (not the natural frequency), because the driving force overcomes the spring's natural tendencies. Hence, the

ω

used here is the same as the driving frequency.

\begin{matrix} z & = & A e^{i (ω t - δ)} \\ \dot{z} & = & i ω A e^{i (ω t - δ)} & = & i ω z \\ \overset{..}{z} & = & - ω^{2} A e^{i (ω t - δ)} & = & - ω^{2} z \\ - ω^{2} z + i γ ω z + {ω_{0}}^{2} z & = & \frac{F_{0}}{m} e^{i ω t} \\ (- ω^{2} + i γ ω + {ω_{0}}^{2}) z & = & \frac{F_{0}}{m} e^{i ω t} \\ (- ω^{2} + i γ ω + {ω_{0}}^{2}) A e^{i (ω t - δ)} & = & \frac{F_{0}}{m} e^{i ω t} \\ (- ω^{2} + i γ ω + {ω_{0}}^{2}) A e^{- i δ} & = & \frac{F_{0}}{m} \\ A (- ω^{2} + i γ ω + {ω_{0}}^{2}) & = & \frac{F_{0}}{m} e^{i δ} \end{matrix}

The end is in sight. All we have to do is to separate the exponential into a real equation and a complex one (using Euler's formula) and then piece the parts back together by eliminating

δ

\begin{matrix} Re (\frac{F_{0}}{m} e^{i δ}) & = & \frac{F_{0}}{m} \cos δ & = & A (- ω^{2} + {ω_{0}}^{2}) \\ \cos δ & = & \frac{A m ({ω_{0}}^{2} - ω^{2})}{F_{0}} \\ Im (\frac{F_{0}}{m} e^{i δ}) & = & \frac{F_{0}}{m} \sin δ & = & A γ ω \\ \sin δ & = & \frac{A m γ ω}{F_{0}} \end{matrix}

To get rid of the

δ

terms, the Pythagorean trigonometry identity can be used:

\begin{matrix} \cos^{2} δ & + & \sin^{2} δ & = & 1 \\ {(\frac{A m ({ω_{0}}^{2} - ω^{2})}{F_{0}})}^{2} & + & {(\frac{A m γ ω}{F_{0}})}^{2} & = & 1 \\ A^{2} & = & \frac{{F_{0}}^{2}}{m^{2} ({({ω_{0}}^{2} - ω^{2})}^{2} + {(ω γ)}^{2})} \\ A & = & \frac{F_{0}}{m \sqrt{{({ω_{0}}^{2} - ω^{2})}^{2} + {(ω γ)}^{2}}} \end{matrix}

And finally, there's an expression for

A

in terms of

ω

and a bunch of other constants. Let's unpack the changes on the resonance curve as c (damping) is increased.

Firstly, the peak gets smaller. This one's obvious - more damping means lower amplitude.

The second change is how the peak gets further away from 1. This was explained above as the frequency of oscillation decreasing when damping is added. Hence, the graph essentially shifts to the left slightly.

Finally, we have the big dog. The peak broadens, but why? We can explain this mathematically by inspecting our equation for $A$ . Peak broadening, in this mathematical context, means that the maximum amplitude (and values very close to it) are reached by a range of $ω$ values as opposed to only a couple of $ω$ values. This means that the effect of changing $ω$ must decrease as peak broadening occurs. As $c$ increases, $γ$ also increases, meaning that the second term under the radicand becomes dominant. As a result, any small differences between $ω$ and $ω_{0}$ become irrelevant, because this is minute compared to the second term. Hence, a range of $ω$ values return similar amplitudes - peak broadening. I learnt this derivation from this video by Walter Lewin. I highly recommend giving the entire thing a watch.