Part 1 · Oscillators and waves

1.4 Forced and damped oscillators: resonance

Source: Postorino, week 1 slides, pp. 15–20, and week 2 slides, p. 11

Objectives

Recipe. Solve the driven oscillator, with and without damping, by complex substitution in three lines.
Picture. The response curve: a peak at the natural frequency whose height and width are set by the damping; the phase lag runs from $0$ through $π /2$ to $π$ .
Theorem. $ω_{ris} = ω_{0}^{2} - γ^{2} /2$ ; for weak damping the peak is a Lorentzian of width $γ$ .
Vocabulary. Steady state, resonance, damping constant, phase lag, FWHM.

The forced oscillator

Add a periodic driving force $F_{0} cos (ω t)$ to the spring. The driving frequency $ω$ is the driver’s, not the spring’s: $m \overset{x}{¨} = - k x + F_{0} cos (ω t) ⟹ \overset{x}{¨} + ω_{0}^{2} x = \frac{F _{0}}{m} cos (ω t) .$

Theorem 1.4.1Forced oscillator, steady stateslides w1, p. 15–16

With $ω_{0} = k / m$ , the steady-state complex amplitude and solution are $\overset{x}{^} = \frac{F ^ _{0} / m}{ω _{0}^{2} - ω ^{2}}, x (t) = Re {\frac{F ^ _{0} / m}{ω _{0}^{2} - ω ^{2}} e^{iω t}} = \frac{F _{0} / m}{ω _{0}^{2} - ω ^{2}} cos (ω t) .$

Proof

Complex substitution: $x \to \overset{x}{^} e^{iω t}$ and $\frac{F _{0}}{m} cos (ω t) \to \frac{F ^ _{0}}{m} e^{iω t}$ , with $\hat{F}_{0}$ the force’s complex amplitude ( $\hat{F}_{0} = F_{0}$ , real, for the force as written).

In the steady state (a regime) the time dependence is dictated by the driver, $e^{iω t}$ : the oscillator ends up moving at the driver’s frequency $ω$ , not at its own $ω_{0}$ .
After differentiating, $e^{iω t}$ appears on both sides and cancels: $[(iω)^{2} + ω_{0}^{2}] \overset{x}{^} = \frac{F ^ _{0}}{m} ⟹ \overset{x}{^} = \frac{F ^ _{0} / m}{ω _{0}^{2} - ω ^{2}} .$ Take the real part.

The slide writes the driver’s substitution as $cos (ω)$ , dropping the $t$ .

RemarkReading the resultbeyond the lectures

$\overset{x}{^}$ is real here. For $ω < ω_{0}$ it is positive: the displacement is in phase with the force, as the slide says. For $ω > ω_{0}$ it is negative: antiphase. Pushed faster than it can follow, the mass moves opposite to the push.
At $ω \to 0$ , $x \to F_{0} / (m ω_{0}^{2}) = F_{0} / k$ : the static stretch of a spring under a constant force.
This is the steady-state (particular) solution. The full solution adds the free oscillation $x_{0} cos (ω_{0} t + φ)$ , by superposition. Without damping it never dies out; with damping it decays (Exercise 1.3.4), which is why the slide speaks of the steady state.

Resonance. The squared maximum amplitude depends on $ω$ : $(Amp)^{2} = \frac{( F _{0} / m ) ^{2}}{[ ω _{0}^{2} - ω ^{2} ] ^{2}} .$ In units of the static stretch $F_{0} / k$ , $∣ A (ω) ∣$ starts at 1, diverges at $ω = ω_{0}$ , and falls towards 0 beyond.

Key observation 1.4.2Resonanceslides w1, p. 16

The natural frequency of the harmonic oscillator is the system’s resonance frequency: the frequency at which the transfer of energy from the driving field to the oscillator is maximum. Without friction it diverges.

Recall that energy goes as the squared amplitude (Key observation 1.3.8): $(Amp)^{2}$ is directly how much energy the driver has pumped in.

The forced damped oscillator

The realistic case adds a viscous (dissipative) force $- β \overset{x}{˙}$ , opposite to the velocity: $m \overset{x}{¨} = - k x - β \overset{x}{˙} + F_{0} cos (ω t) ⟹ \overset{x}{¨} + γ \overset{x}{˙} + ω_{0}^{2} x = \frac{F _{0}}{m} cos (ω t), γ = \frac{β}{m} .$ $γ$ is the damping constant, a rate (units 1/s).

Theorem 1.4.3Forced damped oscillatorslides w1, p. 17–18

$\overset{x}{^} = \frac{F ^ _{0} / m}{ω _{0}^{2} - ω ^{2} + iγω}, ∣ \overset{x}{^} ∣^{2} = \frac{( F _{0} / m ) ^{2}}{( ω _{0}^{2} - ω ^{2} ) ^{2} + ( γω ) ^{2}} .$

Proof

Substitute: $\overset{x}{˙} \to iω \overset{x}{^} e^{iω t}$ , $\overset{x}{¨} \to - ω^{2} \overset{x}{^} e^{iω t}$ , so $- ω^{2} \overset{x}{^} + iγω \overset{x}{^} + ω_{0}^{2} \overset{x}{^} = \frac{F ^ _{0}}{m} .$ Solve for $\overset{x}{^}$ . The squared amplitude, proportional to the energy, is $∣ \overset{x}{^} ∣^{2} = \overset{x}{^} \overset{x}{^}^{*} = \frac{F ^ _{0} / m}{ω _{0}^{2} - ω ^{2} + iγω} \cdot \frac{F ^ _{0}^{*} / m}{ω _{0}^{2} - ω ^{2} - iγω} .$ The denominator is $(a + ib) (a - ib) = a^{2} + b^{2}$ with $a = ω_{0}^{2} - ω^{2}$ , $b = γω$ ; the numerator is $∣ \hat{F}_{0} ∣^{2} / m^{2} = (F_{0} / m)^{2}$ .

Damping puts an imaginary part in the denominator, which can never vanish now: the divergence is cured.

Loading the demo…

Figure 1.4.1 The driven damped oscillator. Left: the amplitude

∣ \overset{x}{^} ∣

in units of

F_{0} / k

against

ω / ω_{0}

, with the

reference curves

for

γ / ω_{0} = 1, 0.5, 0.1

. Drag the driver’s frequency

ω

along the axis, or change

γ

. Middle: the phase of

x

relative to

F

. Right: force and displacement against time, so you can see the lag. It snaps at the

peak

and at

ω_{0}

Key observation 1.4.4The width depends on the dampingslides w1, p. 18–19

All curves start at 1 at $ω = 0$ and agree far from resonance. Smaller $γ$ gives a taller, narrower peak; at $γ = ω_{0}$ the peak is barely there and sits below $ω_{0}$ ; at $γ = 0$ it diverges again. The natural frequency is very close to the resonance frequency, the divergence is cured, and the width of the resonance depends on $γ$ .

Where is the peak?

Proposition 1.4.5Frequency of maximum responseslides w1, p. 18

$ω_{ris} = ω_{0}^{2} - \frac{γ ^{2}}{2} .$ It is smaller than $ω_{0}$ , and moves further down as damping grows. For weak damping, $γ ≪ ω_{0}$ , it is practically $ω_{0}$ . If $γ^{2} > 2 ω_{0}^{2}$ there is no peak at all.

Proofbeyond the lectures

The numerator of $∣ \overset{x}{^} ∣^{2}$ is constant, so the amplitude is largest where the denominator is smallest (the slide sets this up): $\frac{\partial}{\partial ω} [(ω_{0}^{2} - ω^{2})^{2} + (γω)^{2}] = 2 (ω_{0}^{2} - ω^{2}) (- 2 ω) + 2 γ^{2} ω = 2 ω [γ^{2} - 2 (ω_{0}^{2} - ω^{2})] = 0.$ Apart from $ω = 0$ this gives $ω^{2} = ω_{0}^{2} - γ^{2} /2$ . If $γ^{2} > 2 ω_{0}^{2}$ the right side is negative: the only stationary point is $ω = 0$ and the amplitude only falls.

Phase

$\overset{x}{^}$ is now genuinely complex, so the displacement is no longer in phase with the force.

Proposition 1.4.6Phase of the responseslides w1, p. 19

The phase is set by $tan φ = \frac{γω}{ω _{0}^{2} - ω ^{2}} .$ At resonance, force and oscillation are out of phase by $π /2$ : at $ω = ω_{0}$ the real part of the denominator vanishes and $tan φ \to \infty$ .

The slide’s figure plots amplitude and phase difference against $ω / ω_{0}$ for two dampings. The phase goes from $0$ well below resonance, through $- 90°$ at $ω_{0}$ , to $- 180°$ well above. With less damping the switch is sharper.

WarningSlide slip

The caption says $γ_{2} > γ_{1}$ , but the figure labels the tall, narrow peak $γ_{2}$ and the low, broad one $γ_{1}$ . Larger damping gives the lower, broader peak and the more gradual phase change, so the labels are swapped.

RemarkWhat the phase means, and why the graph is negativebeyond the lectures

The slide’s $tan φ$ is Im/Re of the denominator $D = ω_{0}^{2} - ω^{2} + iγω$ . Dividing by $D$ subtracts its angle (Proposition 1.2.5), so the displacement lags the force by $δ = ar g D$ , with $tan δ = \frac{γω}{ω _{0}^{2} - ω ^{2}}, 0 < δ < π .$ $δ$ is the slide’s $φ$ ; the figure plots the phase of $x$ relative to $F$ , which is $- δ$ . Since $Im D = γω > 0$ , $D$ is in the upper half plane: $δ$ is in the first quadrant below resonance (lag under $90°$ ) and in the second above it. $tan δ$ alone doesn’t tell these apart (Proposition 1.2.3).

Physically: below resonance the mass follows the force; far above, it moves against it. At resonance the velocity, $90°$ ahead of the displacement (Key observation 1.2.6), is exactly in phase with the force, so the force always pushes the way the mass moves and does the most work.

Width of the resonance

Proposition 1.4.7Lorentzian lineshapebeyond the lectures

For weak damping, near the peak, $∣ \overset{x}{^} ∣^{2} \approx \frac{( F _{0} / m ) ^{2}}{4 ω _{0}^{2} [ ( ω _{0} - ω ) ^{2} + ( γ /2 ) ^{2} ]}, FWHM \approx γ, ∣ \overset{x}{^} ∣_{max}^{2} \approx \frac{( F _{0} / m ) ^{2}}{( γ ω _{0} ) ^{2}} .$

Proofbeyond the lectures

Near $ω \approx ω_{0}$ : $ω_{0}^{2} - ω^{2} = (ω_{0} - ω) (ω_{0} + ω) \approx 2 ω_{0} (ω_{0} - ω)$ and $γω \approx γ ω_{0}$ . Substitute and factor out $4 ω_{0}^{2}$ . The bracket doubles its minimum $(γ /2)^{2}$ at $ω_{0} - ω = \pm γ /2$ , so the curve is at half height there, a full width $γ$ apart.

Halving the damping halves the width and quadruples the peak. The same lineshape describes absorption lines (§3.2 The complex refractive index and absorption).

The method, summarised

Complex substitution: $x \to \overset{x}{^} e^{iω t}$ and $\frac{F _{0}}{m} \to \frac{F ^ _{0}}{m} e^{iω t}$ , with $ω_{0} = k / m$ .

	equation	solution
harmonic oscillator	$\overset{x}{¨} + ω_{0}^{2} x = 0$	$x (t) = x_{0} cos (ω_{0} t + φ)$
forced	$\overset{x}{¨} + ω_{0}^{2} x = \frac{F _{0}}{m} cos (ω t)$	$x (t) = \frac{F _{0} / m}{ω _{0}^{2} - ω ^{2}} cos (ω t)$
forced and damped, $γ = β / m$	$\overset{x}{¨} + γ \overset{x}{˙} + ω_{0}^{2} x = \frac{F _{0}}{m} cos (ω t)$	$x (t) = \frac{F _{0} / m}{( ω _{0}^{2} - ω ^{2} ) ^{2} + ( γω ) ^{2}} cos (ω t + φ (ω))$

In the last row the amplitude is $∣ \overset{x}{^} ∣$ and $φ (ω)$ is the phase of $\overset{x}{^}$ , which is $- δ$ , running from $0$ to $- π$ . On the slide the first row reads $Re {e^{(i ω_{0} t + φ)}}$ : as on p. 14 the $x_{0}$ is missing, and the $i$ should multiply $φ$ too.

Why oscillators matter

Harmonic approximation. Disturb a system a little from stable equilibrium and it makes small harmonic oscillations about it (Theorem 1.2.9).
Resonance. Small periodic forces can produce very large oscillations if their frequency is close to the system’s own. Every system at equilibrium has characteristic frequencies and can be driven into resonance. Example: tuning a radio changes the resonance frequency of a circuit until it equals the station’s; the station’s waves then drive forced oscillations in the circuit.
Because of the harmonic approximation, oscillations are everywhere. Resonance will later be generalised to quantum resonances.
Fourier’s theorem. Any function, periodic or not, can be written in terms of sines and cosines, so any motion is a superposition of simple harmonic motions (§1.6 Fourier series and the elementary wave). With superposition, solving the single-frequency case solves them all.

Exercises

Exercise 1.4.1beyond the lectures

A damped oscillator has $ω_{0} = 10$ rad/s and $γ = 2$ s⁻¹. Find $ω_{ris}$ , the phase lag at $ω_{ris}$ , and the FWHM.

Solution

$ω_{ris} = 100 - 2 = 9.90$ rad/s. $tan δ = γω / (ω_{0}^{2} - ω^{2}) = 2 \cdot 9.90/2 = 9.90$ , so $δ = 84.2°$ , just short of $90°$ . FWHM $\approx γ = 2$ rad/s.

Exercise 1.4.2beyond the lectures

Show that at $ω = ω_{0}$ the driving force does positive work on average, and the damping force removes the same amount.

Solution

At $ω_{0}$ , $δ = π /2$ , so $x = ∣ \overset{x}{^} ∣ sin (ω_{0} t)$ and $v = ω_{0} ∣ \overset{x}{^} ∣ cos (ω_{0} t)$ , in phase with $F = F_{0} cos (ω_{0} t)$ . Average power in: $⟨ F v ⟩ = \frac{1}{2} F_{0} ω_{0} ∣ \overset{x}{^} ∣$ . With $∣ \overset{x}{^} ∣ = F_{0} / (mγ ω_{0})$ this is $F_{0}^{2} / (2 mγ)$ . Average power lost: $⟨ β v^{2} ⟩ = \frac{1}{2} β ω_{0}^{2} ∣ \overset{x}{^} ∣^{2} = \frac{1}{2} mγ F_{0}^{2} / (m^{2} γ^{2}) = F_{0}^{2} / (2 mγ)$ . They balance.

Exercise 1.4.3beyond the lectures

Without damping, at what driving frequency is the amplitude half the static stretch $F_{0} / k$ , with the mass in antiphase?

Solution

$∣ A ∣ = ω_{0}^{2} /∣ ω_{0}^{2} - ω^{2} ∣ = \frac{1}{2}$ with $ω > ω_{0}$ : $ω^{2} - ω_{0}^{2} = 2 ω_{0}^{2}$ , so $ω = 3 ω_{0}$ .

Exercise 1.4.4beyond the lectures

Show that the peak of $∣ \overset{x}{^} ∣$ in units of $F_{0} / k$ is $ω_{0} / γ$ for weak damping. (This ratio is the quality factor $Q$ .)

Solution

At $ω \approx ω_{0}$ , $∣ \overset{x}{^} ∣ \approx (F_{0} / m) / (γ ω_{0})$ . Divide by $F_{0} / k = F_{0} / (m ω_{0}^{2})$ : $ω_{0}^{2} / (γ ω_{0}) = ω_{0} / γ$ .

Exercise 1.4.5beyond the lectures

Find the phase lag at $ω = ω_{0} /2$ and at $ω = 2 ω_{0}$ for $γ = 0.2 ω_{0}$ .

Solution

At $ω_{0} /2$ : $tan δ = 0.1/0.75 = 0.133$ , $δ = 7.6°$ . At $2 ω_{0}$ : $tan δ = 0.4/ (- 3) = - 0.133$ , and $δ$ is in the second quadrant: $δ = 180° - 7.6° = 172.4°$ .