Section 3.3 Resonance
The discussion Section 3.2 about the similarity between forcing terms and the homogeneous solutions leads into the subject of resonance in harmonic sequences. The question of resonance is this: is there a particular frequency for an external force on a harmonic system which produces the strongest effect?
This is an important question in a number of situations. In audio design and acoustics, an engineer may want to design explicitly for resonance. In the safety of structures, an engineer would like to ensure that resonant behaviour is impossible.
I’ll start with the SOCCLDE describing an underdamped harmonic system. (Underdamped is necessary to allow for the possibility of resonance, as the calculations will show). Recall that for harmonic systems, the coefficients can be identified as mass \(m\text{,}\) spring constant \(k\text{,}\) coefficient of friction \(b\) and forcing \(f(t)\text{.}\)
\begin{equation*}
m y^{\prime \prime} + b y^{\prime} + ky = f(t)
\end{equation*}
Look at the characterstic equation \(mr^2 + br + k=0\text{.}\) It has solutions
\begin{equation*}
r = \frac{-b}{2m} \pm \frac{\sqrt{b^2-4km}}{2m}\text{.}
\end{equation*}
I can define a new constant, called the damping constant, to keep track of the behaviour.
\begin{equation*}
\zeta \defeq \frac{b}{2\sqrt{km}}
\end{equation*}
The damping constant gives a nice measure of the friction. If \(\zeta \lt 1\) then the system is underdamped and there is sinusoidal behaviour. If \(\zeta = 1\) the situation is critically damped and if \(\zeta > 1\text{,}\) the situation is overdamped; in both cases, there is exponential decay without oscillations. The frictionless case is \(\zeta = 0\text{.}\) Let me return to the frictionless case for a moment. The solutions when \(\zeta = 0\) are
\begin{equation*}
y = A \cos \sqrt{\frac{k}{m}} t + B \sin \sqrt{\frac{k}{m}} t\text{.}
\end{equation*}
Now I will define a second important constant.
\begin{equation*}
\omega \defeq \sqrt{\frac{k}{m}}
\end{equation*}
This constant is called the natural frequency. It represents the frequency of sinusoidal oscillation in a perfect system without friction. (Note that all of the frequencies in this section are not true frequencies: they are off by a factor of \(\frac{1}{2\pi}\text{.}\) However, I’ll ignore this fact and keep referring to them as frequencies.)
Finally, I’ll define a third important constant.
\begin{equation*}
\lambda \defeq \frac{b}{2m}
\end{equation*}
If I look back to the underdamped case, the exponential decay term can be written \(e^{-\lambda t}\text{.}\) Therefore, \(\lambda\) is called the decay coefficient. With these new constants, the roots of the characteristic equation become \(r
= -\lambda \pm \imath \sqrt{\omega^2 - \lambda^2}\) and I can rewrite the homogeneous differential equation as
\begin{equation*}
y^{\prime \prime} + 2\lambda y^\prime + \omega^2 y = 0 \text{.}
\end{equation*}
Now return to the idea of a sinusoidal forcing term \(f(t) = F\sin \gamma t\) with some frequency \(\gamma\text{.}\) I want to look at four situations to understand the effect of this force and the possibility of resonance.
Subsection 3.3.1 Situation 1: No Friction, No Forcing
This is the trivial base case, where the system just oscillates forever with frequency \(\omega\text{.}\) This frequency is called the natural frequency because it describes the dynamics of this trival base case, even if the ideal frictionless situation isn’t particularly natural.
Subsection 3.3.2 Situation 2: No Friction, Forcing
The differential equation in this case has the form
\begin{equation*}
y^{\prime \prime} + \omega^2 y = F \sin \gamma t\text{.}
\end{equation*}
This is something that I can solve with undetermined coefficients in two subcases. In the first sub-case, I assume that \(\gamma \neq \omega\text{.}\) Then the forcing term is unlike the homogeneous solutions and I can make the following guess for the particular solution.
\begin{align*}
y_p \amp = C \sin \gamma t + D \cos \gamma t
\end{align*}
I’ll take the derivatives of the particular solution, put them in the differential equation, and compare coefficients to get a system of two equations.
\begin{align*}
y_p^\prime \amp = \gamma C \cos \gamma t - \gamma D \sin
\gamma t\\
y_p^{\prime \prime} \amp = -\gamma^2 C \sin \gamma t -
\gamma^2 D \cos \gamma t\\
Ly_p \amp = (-\gamma^2 C + \omega^2 C) \sin \gamma t +
(-\gamma^2 D + \omega^2 D) \cos \gamma t = F \sin \gamma t\\
-\gamma^2 C + \omega^2 C \amp = F \implies C =
\frac{F}{\omega^2 - \gamma^2 }
\end{align*}
The two equations in the system are independent, since the first only \(C\) and the second only involves \(D\text{.}\) I’ll solve them each and use the results to write the particular solution. The full solution is then the homogeneous solution (which is case 1 without forcing, so uses the natural frequency \(\omega\)) plus this new particular solution.
\begin{align*}
-\gamma^2 D + \omega^2 D \amp = 0 \implies D = 0\\
y_p \amp = \frac{F}{\omega^2 - \gamma^2} \sin \gamma t\\
y \amp = A \sin \omega t + B \cos \omega t +
\frac{F}{\omega^2 - \gamma^2} \sin \gamma t
\end{align*}
The particular solution is a sine wave with amplitutde \(\frac{F}{\omega^2 - \gamma^2}\text{.}\) The closer the forcing frequency is to the natural frequency, the greater the amplitude of the resulting oscillation. (The restriction \(\gamma \neq \omega\) is also obviously important here to avoid division by zero.) If I impose the initial conditions \(y(0) = 0\) and \(y^\prime(0) = 0\text{,}\) then the system is initially at rest and the only energy in the system comes from the external forcing. In this case, I can calculate that \(B=0\) from the first initial condition. The calculation for the second initial condition is as follows.
\begin{align*}
y^{\prime} \amp = \omega A \cos \omega t + \frac{\gamma
F}{\omega^2 - \gamma^2} \cos \gamma t\\
y^\prime (0) \amp = A \omega + \frac{\gamma F}{\omega^2 -
\gamma^2} = 0 \implies A = \frac{-\gamma F}{\omega(\omega^2
- \gamma^2)}
\end{align*}
Then I can write the unique solution for this initial value problems. I do a bit of algebra to write it in common denominator.
\begin{align*}
y \amp = \frac{-\gamma F}{\omega( \omega^2 - \gamma^2)} \sin
\omega t + \frac{F}{\omega^2 - \gamma^2} \sin \gamma t =
\frac{-\gamma F \sin \omega t + \omega F \sin \gamma
t}{\omega (\omega^2 - \gamma^2)}
\end{align*}
Now, to answer the question of resonance, I’ll take the limit of this solution as \(\gamma \rightarrow \omega\text{.}\) The denominator is undefined, but the numerator also evaluates to \(0\text{,}\) so I can apply L’Hôpital’s rule to calculate the limit. After the derivative of the numerator and the denominator, the resulting limit can be calculated directly.
\begin{align*}
\lim_{\gamma \rightarrow \omega} \frac{-\gamma F \sin \omega
t + \omega F \sin \gamma t}{\omega (\omega^2 - \gamma^2)}
\amp = \lim_{\gamma \rightarrow \omega} \frac{-F \sin \omega
t + \omega t F \cos \gamma t}{-2\gamma \omega}\\
\amp = \frac{-F \sin \omega t + \omega t F \cos \omega t}{-2
\omega^2}\\
\amp = \frac{F}{2\omega^2} \sin \omega t - \frac{Ft \cos
\omega t}{2\omega}
\end{align*}
The result of the limit has a \(t \cos \omega t\) term, which is a sinusoidal wave with linear growth in amplitude. The amplidute of the system will grow linearly without bound. This is the ideal (frictionless) understanding of resonance, where the oscillations of the system continue to grow with larger and larger amplitude.
This is a justification, if you want, of why multiplying by \(t\) gives the particular solutions when there is a forcing term similar to the homogeneous solutions. When \(\gamma =
\omega\text{,}\) the forcing is already a homogeneous solution That’s precisely what happens in the resonant case — in the limit \(\gamma \rightarrow \omega\text{.}\) It is also what we would have found if we with \(\omega = \gamma\) and used undetermined coefficients, instead of using the limit process. When the forcing term lines up perfectly with the natural frequency, there is an additional linear growth term in the resulting solution, due to resonance.
Subsection 3.3.3 Section 3: Friction, No Forcing
In the third case, I finally include friction. Using the new constant, I can write the general homogeneous SOCCLDE as folllows
\begin{equation*}
y^{\prime \prime} + 2\lambda y^\prime + \omega^2 y = 0\text{.}
\end{equation*}
The characteristic equation is \(r^2 + 2\lambda r +
\omega^2\) with roots \(-\lambda \pm \imath \sqrt{\omega^2 -
\lambda^2}\text{.}\) If I let \(\omega_d = \sqrt{\omega^2 -
\lambda^2}\text{,}\) then the homogeneous solutions are
\begin{equation*}
y= e^{-\lambda t} (A \cos \omega_d t + B \sin \omega_d t)\text{.}
\end{equation*}
The new constant \(\omega_d\) is called the damped frequency. It is not the same as the natural frequency, since friction changes the ‘natural’ frequency of oscillations of the system. I can explicitly state the relationship between damped and natural frequency via rewriting the constants. From the definitions of the constants, the following identity is true.
\begin{align*}
\amp \lambda^2 = \frac{\zeta^2}{\omega^2} \amp \amp \implies
\amp \amp \lambda^2 \omega^2 = \zeta^2\text{.}
\end{align*}
Using this, I can write the relationship between the damped frequency and the natural frequency.
\begin{equation*}
\omega_d = \omega \sqrt{1 - \zeta^2}\text{.}
\end{equation*}
With friction involved and no forcing, the system will display sinusoidally decaying oscillations where the frequency is this \(\omega_d\text{.}\) As an interesting aside, as \(\zeta\) approaches \(1\) and the system approaches the critical damped situation with simple exponential decay, this damped frequency approaches zero. This explains the transition from sinusoidal to exponential behaviour, which might have seemed like a strange transition. As friction increases and the system approaches the critically damped situation, the wavelength of the sine wave (which is a reciprocal of frequency) grows to \(\infty\text{.}\) The wave stretches out further and further until there isn’t any wave left at all, just an exponential decay.
In this third case, the damped frequency completely described the behaviour (along with the decay coefficient). There isn’t resonance until a forcing term is added.
Subsection 3.3.4 Situation 4: Friction and Forcing
If I finally include both friction and a forcing term, I can use the new constant to rewrite the DE as
\begin{equation*}
y^{\prime\prime} + 2\lambda y^\prime + \omega^2 y = F \sin
\gamma t
\end{equation*}
where \(\gamma\) again is the forcing frequency. The homogeneous solutions are known from the previous case and depend on the damped frequency.
\begin{equation*}
y_h = e^{-\lambda t} (A \cos \omega_d t + B \sin \omega_d t)
\end{equation*}
Since the forcing lacks the exponential term, it is not the same as the homogeneous solutions even if \(\gamma = \omega_d\text{.}\) I don’t need any subcases here: I can use undetermined coefficients without any alteration. Like case 2, I use undetermined coefficients to solve for the particular solution. I start with the guess for the particular solution based on the forcing term.
\begin{align*}
y_p \amp = C \sin \gamma t + D \cos \gamma t
\end{align*}
Then I differentiate and put the results into the differential equation.
\begin{align*}
y_p^\prime \amp = C \gamma \cos \gamma t - D \gamma \sin
\gamma t\\
y_p^{\prime \prime} \amp = - C \gamma^2 \cos \gamma t - D
\gamma^2 \sin \gamma t\\
Ly_p \amp = (-C\gamma^2 - 2\lambda D \gamma + C \omega^2)
\sin \gamma t + (-D\gamma^2 + 2\lambda C \gamma + D
\omega^2) \cos \gamma t \\
\amp = F \sin \gamma t
\end{align*}
Then I compare the coefficients to get two equations.
\begin{align*}
-C\gamma^2 - 2\lambda D\gamma + C \omega^2 \amp = F\\
-D\gamma^2 + 2\lambda C\gamma + D \omega^2 \amp = 0
\end{align*}
This is a bit tricky to manipulate in general. I’ll set it up as a matrix system and invert the matrix to solve the system.
\begin{align*}
\left( \begin{matrix}
\omega^2 - \gamma^2 \amp -2\lambda \gamma \\
2\lambda \gamma \amp \omega^2 - \gamma^2
\end{matrix} \right)
\left( \begin{matrix}
C \\ D
\end{matrix} \right) \amp =
\left( \begin{matrix}
F \\ 0
\end{matrix} \right)\\
M \amp = \left( \begin{matrix}
\omega^2 - \gamma^2 \amp -2\lambda \gamma \\
2\lambda \gamma \amp \omega^2 - \gamma^2
\end{matrix} \right)\\
\det M \amp = (\omega^2 - \gamma^2)^2 + 4 \lambda^2
\gamma^2\\
M^{-1} \amp = \frac{1}{ \det M} \left( \begin{matrix}
\omega^2 - \gamma^2 \amp 2 \lambda \gamma \\
-2 \lambda \gamma \amp \omega^2 - \gamma^2
\end{matrix} \right)\\
\left( \begin{matrix}
C \\ D
\end{matrix} \right) \amp = \frac{1}{\det M} \left(
\begin{matrix}
\omega^2 - \gamma^2 \amp 2 \lambda \gamma \\
-2 \lambda \gamma \amp \omega^2 - \gamma^2
\end{matrix} \right) \left( \begin{matrix}
F \\ 0
\end{matrix} \right)\\
C \amp = \frac{(\omega^2 - \gamma^2) F}{(\omega^2 -
\gamma^2)^2 + 4 \lambda^2 \gamma^2}\\
D \amp = \frac{-2\gamma\lambda F}{(\omega^2 - \gamma^2)^2 +
4 \lambda^2 \gamma^2}
\end{align*}
Now that I have the coefficients, I can write the particular solution and then the general solution including the homogeneous solution. In the particular solution, the coefficients already have the same denominator, so I can combine the two terms into one fraction.
\begin{align*}
y_p \amp = \frac{(\omega^2 - \gamma^2) F\sin \gamma t - 2
\gamma \lambda F \cos \gamma t}{(\omega^2 - \gamma^2)^2 + 4
\lambda^2 \gamma^2}\\
y \amp = e^{-\lambda t} (A \cos \omega_d t + B \sin \omega_d
t) + \frac{(\omega^2 - \gamma^2) F\sin \gamma t - 2 \gamma
\lambda F \cos \gamma t}{(\omega^2 - \gamma^2) + 4 \lambda^2
\gamma^2}
\end{align*}
As time passes, the homogeneous solutions fall out and only the term with the forcing frequency \(\gamma\) remains. What is its amplitude? The term is a linear combination of a sine and cosine wave with the same frequencies, so I can use Proposition 3.1.4 to combine the result into one sinusoidal wave. What I care about, for the purposes of resonance, is just the amplitude of that wave. Using the proposition, I calculate that amplitude.
\begin{equation*}
a = F \frac{\sqrt{(\omega^2 - \gamma^2)^2 + 4 \gamma^2
\lambda^2}}{(\omega^2 - \gamma^2)^2 + 4 \gamma^2 \lambda^2} =
\frac{F}{\sqrt{(\omega^2 - \gamma^2)^2 + 4 \gamma^2
\lambda^2}}
\end{equation*}
The question of resonance is this: if the magnitude of the force \(F\) is fixed, what is the maximum amplitude can be achieved by altering the forcing frequency? Note this amplitude is always finite — with friction, there is no infinite growth of amplitude as there was in the second case. (This is expected predicted, since I didn’t need to multiply by \(t\) when we setup the undetermined coefficients.) However, the amplitude can be quite large. This is a optimization problem, so I differentiate the expression for amplitude and find its critical points. (Not that I am looking to optimize by amplitude, so the function and the derivative are in the variable \(\gamma\)).
\begin{align*}
a(\gamma) \amp = \frac{F}{\sqrt{(\omega^2 - \gamma^2)^2 + 4
\gamma^2 \lambda^2}}\\
a^\prime(\gamma) \amp = \frac{-F(2 (\omega^2 - \gamma^2)
(-2\gamma) + 8 \gamma \lambda^2)}{2((\omega^2 - \gamma^2)^2
+ 4 \gamma^2 \lambda^2)^{\frac{3}{2}}} = 0\\
4 \gamma (\omega^2 - \gamma^2) \amp = 8 \gamma \lambda^2\\
\omega^2 - \gamma^2 \amp = 2 \lambda^2\\
\gamma^2 \amp = \omega^2 - 2 \lambda^2\\
\gamma \amp = \sqrt{\omega^2 - 2 \lambda^2}
\end{align*}
I’ll skip the testing of intervals to classify this critical point; this is, indeed, a maximum for amplitude. Now I want to write this result is a slightly different way. Recall that \(\lambda^2 = \omega^2 \zeta^2\text{.}\) I can use that to factor \(\omega\) out of the expression for \(\gamma\text{.}\)
\begin{equation*}
\gamma = \sqrt{\omega^2 - 2 \zeta^2 \omega^2} = \omega \sqrt{1
- 2 \zeta^2}\text{.}
\end{equation*}
This is the resonant frequency. However, how can I be certain that it exists? I need \(1-2\zeta^2\) to be positive to define this square root. This implies, certainly, that \(\zeta \lt 1\text{,}\) which I assumed when I decided to work with the underdamped case (that assumption is now justified). However, the inequality is stricter.
\begin{equation*}
\zeta^2 \leq \frac{1}{2} \implies \zeta \leq
\frac{1}{\sqrt{2}}
\end{equation*}
The bound for the damping coefficient is smaller than simply for the underdamped case. I conclude that resonant frequency only exists is the friction is small enough, measured by this \(\zeta \lt \frac{1}{\sqrt{2}}\text{.}\) In particular, if I am concerned about safety and want to avoid the situation of resonant frequency, this calculation lets me know the minimum friction I need to build into the system to avoid the possibility of resonance.
