# Resonance

## Alexei Gilchrist

When an oscillator is driven at just the right frequency it hits resonance and absorbs energy from the driving. The details of resonance for a driven and damped harmonic oscillator are explored.

### 1 Introduction

On a long enough time scale the solution to the driven and damped oscillator (driven-damped-oscillator) becomes dominated by what is known as the “steady state” solution, which looks like

\begin{align*}x_p(t) &= A_p\cos(\omega_f t + \phi_p) \\ A_p &= \frac{f_0}{\sqrt{(\omega_0^2-\omega_f^2)^2+\gamma^2\omega_f^2}} \\ \phi_p &= \tan^{-1}\left(\frac{-\gamma\omega_f}{\omega_0^2-\omega_f^2}\right).\end{align*}
In these equations $$\omega_0$$ and $$\gamma$$ are determined by the physics of the problem, but we are free to choose how we drive the oscillator with $$f_0$$ and $$\omega_f$$.

If $$\gamma$$ was zero and we drove the oscillator so that $$\omega_f=\omega_0$$, $$A_p$$ would become infinite! Something interesting is happening here and this is resonance.

### 2 Amplitude resonance

First consider the limits of $$A_p(\omega_f)$$ as a functions of $$\omega_f$$. When $$\omega_f\rightarrow\infty$$ then $$A_p\rightarrow 0$$ and at the other limit $$A_p(0) = f_0/\omega_0^2$$. In between, the amplitude peaks at some value of $$\omega_f$$.

We can find where the maximum occurs by solving $$dA_p/d\omega_f=0$$ for $$\omega_f$$,

$\begin{equation*}\frac{d A_p}{d \omega_p} = \frac{f_0 \omega_f(\gamma^2+2(\omega_f^2-\omega_0^2))}{[(\omega_f^2-\omega_0^2)^2+\gamma^2\omega_f^2]^{3/2}} = 0,\end{equation*}$
which yields three possible solutions: (i) $$\omega_f=0$$, (ii) $$\omega_f=-\sqrt{\omega_0^2-\gamma^2/2}$$, and (iii) $$\omega_f=\sqrt{\omega_0^2-\gamma^2/2}$$. The first solution is trivial, the second unphysical (negative frequencies), which leaves the third so the resonance angular frequency is
$\begin{equation*}\omega_r\equiv\omega_f^\mathrm{max} = \sqrt{\omega_0^2-\gamma^2/2}.\end{equation*}$
Comparing with the angular frequency of the damped harmonic oscillator $$\omega_d = \sqrt{\omega_0^2-\gamma^2/4}$$, and the undamped angular frequency $$\omega_0$$ we see
$\begin{equation*}\omega_r \le \omega_d \le \omega_0 ,\end{equation*}$
and the difference becomes smaller and smaller as the damping is reduced.

How big will the response be? Now we know at which frequency resonance occurs, we can evaluate $$A_p$$ at that point, $$A_p^\mathrm{max} = A_p(\omega_r) = f_0 / \gamma \omega_d$$. This makes it clear that the response becomes large when $$\gamma$$ becomes small! In fact, for weak damping $$\omega_d$$ is approximately $$\omega_0$$, i.e. $$\omega_d=\sqrt{\omega_0^2-\gamma^2/4}\approx \omega_0$$, and the ratio of the amplitude driven at resonance to the amplitude at zero frequency is given by the $$Q$$ of the oscillator:

$\begin{equation*}\frac{A_p(\omega_r)}{A_p(0)} = \frac{f_0 / \gamma \omega_d}{f_0/\omega_0^2} \approx \frac{\omega_d}{\gamma} = Q.\end{equation*}$
All of this can be summarised by the figure below.

The phase shift

$\begin{equation*}\phi_p = \tan^{-1}\left(\frac{-\gamma\omega_f}{\omega_0^2-\omega_f^2}\right),\end{equation*}$
also undergoes changes around resonance. The phase is close to zero for low driving frequencies so the object moves in phase with the driving, and it quickly changes to a $$-\pi$$ phaseshift around the resonance as in the plot below

This behaviour can be quickly verified with a simple experiment—hang a small weight onto the end of a rubber band. If you drive the weight sinusoidally very slowly you will see that the response is some fixed amplitude and in-phase. If you increase the driving frequency, at some value you will hit resonance and it will bob up and down like crazy. Above that, the weight settles into periodic motion that is out of phase with your driving. If you drive it as fast as you can (keeping the motion even) then the weight will hardly move despite the driving force, a result of $$A_p\rightarrow 0$$.

The following simulation shows how the resonance peak broadens out with increasing damping.

### 3 Velocity resonance

It may be surprising to discover that the velocity has a different resonance to the amplitude but this is indeed the case! The velocity in the steady state is given by

$\begin{equation*}\dot{x}_p(t) = -A_p\omega_f \sin(\omega_f t + \phi_p),\end{equation*}$
and now we want to find the $$\omega_f$$ where
$\begin{equation*}\frac{d}{d \omega_p}(A_p\omega_f) = \omega_f\frac{d A_p}{d \omega_p}+ A_p = 0,\end{equation*}$
which will yield $$\omega_f = \omega_0$$ as the only physically valid solution (i.e. not negative or imaginary).

In the limit of weak damping both these resonances, and that of the energy discussed below, become $$\omega_0$$.

### 4 The steady state energy

We already have the position $$x(t)$$, so we can get the velocity by differentiating: $$\dot{x} = -\omega_f A_p\sin(\omega_f + \phi_p)$$, consequently the kinetic ($$E_k$$), potential ($$E_p$$), and the total mechanical energy ($$E_m$$) is

\begin{align*}E_k &= \frac{1}{2}m\dot{x}^2 = \frac{1}{2}m A_p^2 \omega_f^2 \sin^2(\omega_f + \phi_p) \\ E_p &= \frac{1}{2}m\omega_0^2 x^2 = \frac{1}{2}m A_p^2\omega_0^2 \cos^2(\omega_f t + \phi_p) \\ E_m &= E_k+E_p = \frac{1}{2}m A_p^2\left[\omega_0^2\cos^2(\omega_f t + \phi_p) + \omega_f^2\sin^2(\omega_f + \phi_p) \right].\end{align*}
Note that this is an open system, it’s gaining energy from the driving and losing energy due to the damping, so we shouldn’t expect the mechanical energy to be conserved.

If we examine the mechanical energy of the oscillator averaged over a cycle we find

$\begin{equation*}\langle E_m \rangle_c = \frac{1}{4}m A_p^2 (\omega_0^2+\omega_f^2).\end{equation*}$
In this expression the angle brackets $$\langle \cdot \rangle_c$$ denote the averaging over a cycle, and we’ve used $$\langle \cos^2\theta \rangle_c = \langle \sin^2\theta \rangle_c = 1/2$$. So if we look at the average energy after transients have died off, the oscillator settles to a constant (per cycle) mechanical energy with the energy gained from the driving balancing the losses from the damping. The ratio of cycle-averaged kinetic to potential energy is the ratio of the squares of the frequencies:
$\begin{equation*}\frac{\langle E_k \rangle_c}{\langle E_p \rangle_c} = \frac{\omega_f^2}{\omega_0^2},\end{equation*}$
so below resonance most of the energy is potential energy and above resonance the energy is mostly kinetic. When the oscillator is driven at resonance the total mechanical energy is constant even within a cycle and the system behaves as an undamped oscillator.

The simulation above shows that the energy in a driven oscillator depends strongly on the driving frequency. Examining this more closely we can expand $$\langle E_m \rangle_c$$ to show the dependence on $$\omega_f$$ explicitly. Remember that $$A_p$$ depends on $$\omega_f$$ too.

$\begin{equation*}\langle E_m \rangle_c = \frac{m f_0^2(\omega_0^2+\omega_f^2)}{4\left[(\omega_0^2-\omega_f^2)^2+\gamma^2\omega_f^2\right]} .\end{equation*}$