Resonance

Push a child on a swing at random moments and not much happens; time your pushes to match the swing's own natural rhythm and the arcs grow dramatically larger. That is resonance — the sharp peak in a driven oscillator's response when the driving frequency lines up with the system's own natural frequency. It is why a bridge can be set swaying by rhythmic footsteps, why a wine glass can shatter from one particular musical note, and why a radio dial picks out a single station from a sea of broadcast frequencies. This node sits at the end of the "Damped and Forced Oscillations" module: it takes the steady-state amplitude $A(\omega)$ and phase lag $\delta(\omega)$ already derived for a driven damped oscillator and asks a sharper question — at exactly which frequency is the response biggest, how narrow is that peak, and how much power does the driving force actually deliver at steady state?

The core insight

A driven damped oscillator's amplitude peaks near — but not exactly at — its natural frequency, while the power delivered peaks exactly at $\omega_0$ , and the sharpness of both peaks is set by the same damping rate that also controls free-oscillation decay, summarised by the quality factor $Q$ .

The core idea

Two different kinds of resonance. Start from the steady-state amplitude of a driven damped oscillator,

A(\omega) = \frac{F_0/m}{\sqrt{(\omega_0^2-\omega^2)^2+(2\gamma\omega)^2}},

where: $F_0$ is the driving force amplitude, $m$ is the mass, $\omega_0$ is the natural (undamped) angular frequency, $\gamma$ is the damping rate, and $\omega$ is the driving angular frequency. Amplitude resonance is the frequency at which $A(\omega)$ itself is largest. Differentiating $A(\omega)^{-2}$ with respect to $\omega^2$ and setting the result to zero gives

\omega_{\text{res}} = \sqrt{\omega_0^2-2\gamma^2}.

This peak exists only when $\gamma < \omega_0/\sqrt2$ ; for heavier damping $A(\omega)$ falls monotonically from its $\omega=0$ value and there is no interior maximum at all. Note that $\omega_{\text{res}}$ sits slightly below both $\omega_0$ and the damped free-oscillation frequency $\omega_1=\sqrt{\omega_0^2-\gamma^2}$ — three distinct numbers that only merge in the undamped or lightly-damped limit.

Power resonance is a different frequency. The instantaneous power delivered by the driving force is $P(t)=F(t)\dot x(t)$ ; its steady-state time average is

\langle P(\omega)\rangle = \frac12 F_0\,\omega\,A(\omega)\sin\delta = \frac{F_0^2}{2m}\cdot\frac{2\gamma\omega^2}{(\omega_0^2-\omega^2)^2+(2\gamma\omega)^2}.

where: $\delta$ is the phase lag of displacement behind the driving force. This is maximised exactly at $\omega=\omega_0$ , regardless of how strong the damping is. At that point $\delta=90^\circ$ , so velocity $\dot x$ is exactly in phase with the force $F$ , and every instant the force does positive work on the mass — no part of the cycle wastes the push. This is why $\omega=\omega_0$ is called velocity resonance or power resonance, and why it is a cleaner landmark than the amplitude peak, which shifts with damping.

Sharpness and the quality factor. Near $\omega_0$ the power curve has a Lorentzian-like shape, and its full width at half maximum works out to

\Delta\omega \approx 2\gamma,

so lighter damping (smaller $\gamma$ ) gives a taller, narrower peak. This links directly to the quality factor already defined for free damped oscillations, $Q=\omega_0/2\gamma=\omega_0/\Delta\omega$ : a high- $Q$ oscillator is frequency-selective — exactly what a radio tuner needs — while a low- $Q$ one responds broadly across many drive frequencies, which is the behaviour wanted in a shock absorber. The same $\gamma$ that sets how fast free oscillations die away also sets how sharp the forced-resonance peak is; they are not two separate damping stories but one.

Figure

Amplitude and power response curves against driving frequency showing the amplitude peak slightly left of omega zero and the power peak exactly at omega zero

Derivation of the amplitude-resonance frequency and its width

Derivation AThe amplitude-resonance frequency and its condition for existence

The goal is to locate the driving frequency that maximises the steady-state amplitude $A(\omega)=(F_0/m)/\sqrt{(\omega_0^2-\omega^2)^2+(2\gamma\omega)^2}$ , find the algebraic condition under which this peak exists at all, and connect the sharpness of the resulting peak to the quality factor $Q$ that already governs free damped oscillations.

Step 1. Maximising $A(\omega)$ is the same as minimising the quantity under the square root, since $A$ is a decreasing function of it. Define $D(\omega)=(\omega_0^2-\omega^2)^2+(2\gamma\omega)^2$ . It is easier to treat $u=\omega^2$ as the variable, writing $D(u)=(\omega_0^2-u)^2+4\gamma^2 u$ .

Step 2. Differentiate with respect to $u$ and set the result to zero: $\dfrac{dD}{du}=-2(\omega_0^2-u)+4\gamma^2=0$ , which gives $\omega_0^2-u=2\gamma^2$ , so $u=\omega_0^2-2\gamma^2$ .

Step 3. Since $u=\omega^2$ , the resonance frequency is $\omega_{\text{res}}=\sqrt{\omega_0^2-2\gamma^2}$ . For this to be a real, physically meaningful frequency, the quantity under the square root must be non-negative: $\omega_0^2-2\gamma^2\geq0$ , i.e. $\gamma\leq\omega_0/\sqrt2$ . If damping exceeds this bound, $D(u)$ is minimised at $u=0$ instead (the boundary of the allowed range), meaning $A(\omega)$ simply decreases from its static value as $\omega$ rises, with no interior peak.

Step 4. Near $\omega_0$ , expanding $D(\omega)$ for $\omega$ close to $\omega_0$ shows the power curve $\langle P(\omega)\rangle$ behaves like a Lorentzian in $(\omega-\omega_0)$ , and standard analysis of that Lorentzian gives a full width at half maximum $\Delta\omega\approx2\gamma$ . Combining this with $Q=\omega_0/2\gamma$ gives $\Delta\omega=\omega_0/Q$ — a sharp (high- $Q$ ) resonance has small $\Delta\omega$ , and the same $\gamma$ that appears in the free-oscillation decay envelope $e^{-\gamma t}$ reappears here as the resonance width.

Worked example

An oscillator has $\omega_0=50\ \text{rad/s}$ and $\gamma=3\ \text{rad/s}$ , driven with force amplitude $F_0=5.0\ \text{N}$ on a mass $m=0.10\ \text{kg}$ . Find (a) the amplitude-resonance frequency, (b) $Q$ and the FWHM of the power curve, and (c) the maximum average power and the amplitude at velocity resonance, compared with the true amplitude-resonance amplitude.

Step 1. Amplitude-resonance frequency: $\omega_{\text{res}}=\sqrt{\omega_0^2-2\gamma^2}=\sqrt{2500-18}=\sqrt{2482}\approx49.82\ \text{rad/s}$ , just below $\omega_0=50\ \text{rad/s}$ because $\gamma\ll\omega_0$ here (lightly damped).

Step 2. Quality factor and width: $Q=\omega_0/2\gamma=50/6\approx8.33$ , and $\Delta\omega\approx2\gamma=6\ \text{rad/s}$ , matching $\Delta\omega=\omega_0/Q=50/8.33\approx6\ \text{rad/s}$ . The power drops to half its peak once the drive is detuned by about $3\ \text{rad/s}$ either side of $\omega_0$ .

Step 3. At velocity resonance ( $\omega=\omega_0$ ), $(\omega_0^2-\omega^2)=0$ , so $A(\omega_0)=(F_0/m)/(2\gamma\omega_0)$ . With $F_0/m=5.0/0.10=50\ \text{N/kg}$ , $A(\omega_0)=50/(2\times3\times50)=50/300\approx0.1667\ \text{m}$ . The maximum average power is $\langle P\rangle_{\max}=\tfrac12 F_0\,\omega_0\,A(\omega_0)=\tfrac12\times5.0\times50\times0.1667\approx20.8\ \text{W}$ .

Step 4. At $\omega_{\text{res}}\approx49.82\ \text{rad/s}$ : $\omega_0^2-\omega_{\text{res}}^2=2500-2482=18$ and $2\gamma\omega_{\text{res}}=2\times3\times49.82\approx298.9$ , so the denominator is $\sqrt{18^2+298.9^2}\approx\sqrt{89665}\approx299.4$ , giving $A(\omega_{\text{res}})=50/299.4\approx0.1670\ \text{m}$ . Final answer: $\omega_{\text{res}}\approx49.82\ \text{rad/s}$ , $Q\approx8.33$ , $\Delta\omega\approx6\ \text{rad/s}$ , $\langle P\rangle_{\max}\approx20.8\ \text{W}$ , and $A(\omega_0)\approx0.1667\ \text{m}$ versus $A(\omega_{\text{res}})\approx0.1670\ \text{m}$ — under $0.2\%$ apart, confirming that for a lightly damped system the amplitude peak and the power peak are almost indistinguishable even though they occur at formally different frequencies.

Competitive stretch

If the same oscillator were damped more heavily, with $\gamma=40\ \text{rad/s}$ (still $\omega_0=50\ \text{rad/s}$ ), check resonance existence: the condition is $\gamma\leq\omega_0/\sqrt2\approx35.4\ \text{rad/s}$ . Since $40>35.4$ , $\omega_{\text{res}}$ would be imaginary — there is no amplitude peak at all, and $A(\omega)$ falls monotonically as $\omega$ increases from zero, even though power resonance still technically sits at $\omega=\omega_0$ in the formula. This is the regime examiners use to test whether you actually check the existence condition instead of blindly applying the formula.

Common traps

Treating $\omega_0$ , $\omega_1=\sqrt{\omega_0^2-\gamma^2}$ (damped free-oscillation frequency), and $\omega_{\text{res}}=\sqrt{\omega_0^2-2\gamma^2}$ (amplitude-resonance frequency) as the same number — they coincide only as $\gamma\to0$ .
Assuming amplitude resonance and power resonance sit at the same frequency in general — power resonance is always exactly at $\omega_0$ ; amplitude resonance is at the slightly lower $\omega_{\text{res}}$ , merging only for small $\gamma$ .
Forgetting to check whether resonance exists at all — once $\gamma\geq\omega_0/\sqrt2$ , $\omega_{\text{res}}$ becomes imaginary and there is no interior amplitude peak.
Missing that $Q$ ties together two seemingly separate ideas: how fast free damped oscillations decay, and how sharp the forced-resonance peak is — both are governed by the same $\gamma$ .

Ask the tutor: ask if you would like a side-by-side plot of how the amplitude and power resonance curves change shape as $Q$ is varied.

Why are there two different resonance frequencies instead of just one

Because amplitude and power are different quantities with different maximisation conditions. Amplitude resonance maximises $A(\omega)$ directly, which after calculus gives $\omega_{\text{res}}=\sqrt{\omega_0^2-2\gamma^2}$ — a frequency that depends on damping. Power resonance maximises $\langle P(\omega)\rangle$ , which turns out to peak exactly at $\omega_0$ for any damping, because that is where the phase lag is $90^\circ$ and the force does maximal positive work each cycle. They only merge when $\gamma$ is small.

Why does maximum power occur exactly at phase lag ninety degrees

Average power is $\langle P\rangle=\tfrac12F_0\omega A\sin\delta$ . Both $A$ and $\delta$ depend on $\omega$ , but the combination $\omega A\sin\delta$ turns out to be maximised precisely when $\sin\delta=1$ , i.e. $\delta=90^\circ$ , which by the phase-lag formula for a driven damped oscillator happens exactly at $\omega=\omega_0$ . Physically, at $90^\circ$ lag the velocity is exactly in phase with the driving force, so the force never opposes the motion during any part of the cycle — every instant it pushes in the direction the mass is already moving, which is the condition for delivering the most work per cycle.

Does a very high Q always mean a stronger response

It means a taller and much narrower peak near $\omega_0$ , not necessarily a stronger response everywhere. A high- $Q$ system barely responds at all unless you drive it very close to $\omega_0$ ; a low- $Q$ system responds more modestly but over a much wider band of frequencies. Which is "better" depends on the application: a radio tuner wants high $Q$ so it rejects nearby stations, while a car's shock absorber wants low $Q$ so it does not resonate violently at any one bump frequency and instead damps a broad range of disturbances smoothly.