Worked examples — Resonance — physical consequences, design implications

This page is the "hit every case" companion to the parent note on Resonance. Before we compute anything, let us lay out every kind of situation the resonance formulas can hand you, so no example surprises you later.

Every symbol we use is built in the parent note, but here is the one-line reminder you will lean on constantly:

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The scenario matrix

Think of resonance problems as living in a grid. Each cell is a different regime, and each worked example below is tagged with the cell(s) it covers.

Cell	Case class	What is special about it
C1	Driving far below resonance ($\omega \ll {\omega }_0$)	Amplitude ≈ static; spring term dominates
C2	Driving at resonance ($\omega ={\omega }_0$)	Denominator collapses to $b\omega /m$; amplitude peaks
C3	Driving far above resonance ($\omega \gg {\omega }_0$)	Inertia dominates; amplitude dies as $1/{\omega }^2$
C4	Degenerate: zero damping ($b\to 0$)	Amplitude blows up to infinity — the limiting catastrophe
C5	Degenerate: zero drive frequency ($\omega =0$, a steady push)	No oscillation; pure static displacement $F_0/k$
C6	The peak is not at ${\omega }_0$: exact ${\omega }_{res}$	Displacement peak sits slightly below ${\omega }_0$
C7	Sharpness & bandwidth: $Q$, half-power width $\Delta \omega$	The two frequencies where power halves
C8	Real-world word problem	Bridge / soldiers / earthquake — translate words to $\omega$
C9	Electrical twin (LC circuit)	Same math, different symbols: ${\omega }_0=1/\sqrt{LC}$
C10	Exam-style twist: "microwave is NOT resonance"	Recognise when the formula does not apply

We now walk all ten cells across eight examples.

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Forecast: guess which of the three gives the biggest swing before reading on. Most people correctly guess (b) — but by how much?

1. Find ${\omega }_0$. ${\omega }_0=\sqrt{k/m}=\sqrt{100/1}=10\text{ rad/s}$. Why this step? Every case is measured relative to ${\omega }_0$; it is our yardstick.

2. (a) Far below, $\omega =1$. Plug in: $A=\frac{10/1}{\sqrt{(100-1{)}^2+(2\cdot 1/1{)}^2}}=\frac{10}{\sqrt{99^2+2^2}}=\frac{10}{99.02}\approx 0.101\text{ m}.$ Why this step? When $\omega \ll {\omega }_0$, the term ${\omega }_0^2-{\omega }^2\approx {\omega }_0^2=k/m$, so $A\approx F_0/k=10/100=0.1$ — the system barely notices it's being shaken; it just sits at its static stretch. This is C1.

3. (b) On resonance, $\omega =10$. Now ${\omega }_0^2-{\omega }^2=0$: $A=\frac{10}{\sqrt{0+(2\cdot 10{)}^2}}=\frac{10}{20}=0.5\text{ m}.$ Why this step? The big square-root term vanishes, leaving only the damping term $b\omega /m$. This is C2 — amplitude jumps 5× above the static value.

4. (c) Far above, $\omega =100$. $A=\frac{10}{\sqrt{(100-10000{)}^2+(2\cdot 100{)}^2}}=\frac{10}{\sqrt{9900^2+200^2}}=\frac{10}{9902}\approx 1.01\times 10^{-3}\text{ m}.$ Why this step? When $\omega \gg {\omega }_0$, the inertia term $-m{\omega }^2$ dominates; $A\approx F_0/(m{\omega }^2)$. The mass is too heavy to keep up — it hardly moves. This is C3.

Verify: the three amplitudes are $0.101>$ … no — ordered by size: $0.5\text{ (resonance)}\gg 0.101\text{ (low)}\gg 0.00101\text{ (high)}$. Resonance wins, exactly as the peak in the figure predicts. Units: $F_0/k=\text{N}/(\text{N/m})=\text{m}$. ✓

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Forecast: finite or infinite? Guess.

1. Set $b=0$ in the formula, then $\omega ={\omega }_0$. $A=\frac{F_0/m}{\sqrt{({\omega }_0^2-{\omega }^2{)}^2+0}}=\frac{F_0/m}{|{\omega }_0^2-{\omega }^2|}.$ Why this step? With no damping term, the denominator is only the spring-minus-inertia part.

2. Now push $\omega \to {\omega }_0$. The denominator $\to 0$, so $A\to \infty$. Why this step? This is the mathematical statement of "resonance disaster" — with nothing to bleed energy, each push adds coherently forever. This is C4.

Verify: compare to Example 1(b): there, $b=2$ capped the answer at $0.5\text{ m}$. Take $b\to 0$: $A_{max}=F_0/(b{\omega }_0)=10/(b\cdot 10)$ blows up as $b\to 0$. So the finite $0.5\text{ m}$ was entirely the doing of damping. ✓ Reality always has some $b>0$, so true infinity never occurs.

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Forecast: what displacement do you expect from a steady $10\text{ N}$ pull on a $100\text{ N/m}$ spring?

1. Set $\omega =0$. $A=\frac{F_0/m}{\sqrt{({\omega }_0^2-0{)}^2+0}}=\frac{F_0/m}{{\omega }_0^2}=\frac{F_0/m}{k/m}=\frac{F_0}{k}.$ Why this step? At $\omega =0$ the driver never reverses; both damping ($b\omega$) and inertia contribute nothing. We are left with Hooke's law: force = stiffness × stretch.

2. Number: $A=10/100=0.1\text{ m}$. Why this step? This is the C5 degenerate case and it's the anchor of the whole curve — the amplitude at the far left of the graph. Notice it matches Example 1(a) closely (that's why $\omega \ll {\omega }_0$ looks "static").

Verify: rearrange Hooke's law: $F=kx\Rightarrow x=F/k=10/100=0.1\text{ m}$. Units: $\text{N}/(\text{N/m})=\text{m}$. ✓

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Forecast: by how much below $10$ — a lot or a whisker?

1. Use the exact peak formula. ${\omega }_{res}=\sqrt{{\omega }_0^2-\frac{b^2}{2m^2}}$. Why this step? This comes from setting $dA/d\omega =0$; the damping term shifts the maximum of $A$ slightly left because friction "eats" more at higher $\omega$.

2. Plug in: ${\omega }_{res}=\sqrt{100-\frac{2^2}{2\cdot 1^2}}=\sqrt{100-2}=\sqrt{98}\approx 9.899\text{ rad/s}$. Why this step? This is C6: the peak sits at $9.899$, not $10$ — a tiny $\sim 0.1\text{ rad/s}$ shift here because damping is light.

3. Sanity on the direction: $9.899<10={\omega }_0$. ✓ Below, as promised.

Verify: the velocity/power resonance is exactly at ${\omega }_0=10$; the displacement peak is at $\sqrt{98}\approx 9.899$. Also check the limiting rule: as $b\to 0$, ${\omega }_{res}\to \sqrt{100-0}=10={\omega }_0$. ✓ Light damping ⇒ shift almost vanishes.

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Forecast: is this a "sharp radio" system or a "soft shock-absorber" system? Guess high or low $Q$.

1. Compute $Q$. $Q=\frac{{\omega }_{0}m}{b}=\frac{10\cdot 1}{2}=5$. Why this step? $Q$ measures peak sharpness: high $Q$ = tall narrow peak, low $Q$ = broad. $Q=5$ is moderate — a real spring, not a quartz crystal (which has $Q\sim 10^4$–$10^6$).

2. Bandwidth from $Q$. For a resonator, $\Delta \omega =\frac{{\omega }_0}{Q}$. Why this step? $\Delta \omega$ is the distance between the two frequencies where power has dropped to half its peak (amplitude to $1/\sqrt{2}$). $Q$ literally is ${\omega }_0/\Delta \omega$, so we just invert it.

3. Number: $\Delta \omega =\frac{10}{5}=2\text{ rad/s}$. Why this step? This is C7: the peak is $2\text{ rad/s}$ wide. To the light-damping approximation, damping $b/m=2$ equals $\Delta \omega$ here — a neat check.

4. The two half-power frequencies sit roughly at ${\omega }_0\pm \Delta \omega /2=10\pm 1$, i.e. about $9$ and $11\text{ rad/s}$. Why this step? This tells a radio engineer which neighbouring stations leak through.

Verify: $\Delta \omega \approx b/m$ for light damping: $b/m=2/1=2$. ✓ Matches ${\omega }_0/Q=10/5=2$. ✓ Units of $Q$: $(\text{rad/s})(\text{kg})/(\text{kg/s})$ = dimensionless. ✓

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Forecast: is $120$ steps/min dangerous or safe? Convert first, then guess.

1. Convert marching cadence to Hz. $120\text{ steps/min}=120/60=2.0\text{ steps/s}=2.0\text{ Hz}$. Why this step? Each footfall is one push cycle, so cadence in Hz is the drive frequency $f$. We must compare like-for-like with $f_0$.

2. Compare. Drive $f=2.0\text{ Hz}$ equals $f_0=2.0\text{ Hz}$ — this is dead-on resonance (C8, and the same physics as C2). Break step immediately. Why this step? A coherent $2\text{ Hz}$ push at a $2\text{ Hz}$ resonance is exactly the Millennium/soldier-bridge scenario.

3. How far is a full bandwidth? $\Delta f=f_0/Q=2.0/20=0.10\text{ Hz}$. Why this step? Stepping outside the half-power band drops the resonant amplification. We want $f$ at least $\Delta f/2=0.05\text{ Hz}$ off — or better, a full $\Delta f$.

4. Percentage slow-down for a full bandwidth: need $f=2.0-0.10=1.90\text{ Hz}$, a change of $0.10/2.0=0.05=5\%$. Why this step? Answers the "by what percentage" ask; but note reality is easier — breaking step randomises the force so no single $f$ dominates at all.

Verify: cadence conversion $120/60=2.0$. ✓ $\Delta f=2.0/20=0.10\text{ Hz}$. ✓ Percentage $=0.10/2.0=5\%$. ✓

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Forecast: picoFarads or Farads? Guess the order of magnitude.

1. Use the electrical natural frequency. ${\omega }_0=\frac{1}{\sqrt{LC}}$, and ${\omega }_0=2\pi f_0$. Why this step? The LC circuit obeys the same differential equation as the mass–spring; $L$ plays the role of mass, $1/C$ the role of stiffness $k$. Same math, new costume — this is C9.

2. Solve for $C$. From ${\omega }_0^2=1/(LC)$: $C=\frac{1}{{\omega }_0^{2}L}=\frac{1}{(2\pi f_0{)}^{2}L}$. Why this step? Algebraically isolate the unknown $C$.

3. Plug numbers. ${\omega }_0=2\pi (10^6)=6.283\times 10^6\text{ rad/s}$. $C=\frac{1}{(6.283\times 10^6{)}^2(2\times 10^{-4})}=\frac{1}{(3.948\times 10^{13})(2\times 10^{-4})}=\frac{1}{7.896\times 10^9}\approx 1.27\times 10^{-10}\text{ F}.$ Why this step? Now we have a real component value: $\approx 127\text{ pF}$ (picoFarads).

Verify: back-substitute: ${\omega }_0=1/\sqrt{LC}=1/\sqrt{(2\times 10^{-4})(1.27\times 10^{-10})}=1/\sqrt{2.53\times 10^{-14}}\approx 6.28\times 10^6\text{ rad/s}$, and $f_0={\omega }_0/2\pi \approx 1.0\text{ MHz}$. ✓ Order of magnitude: $\sim 100\text{ pF}$, typical for a tuning cap. ✓

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Forecast: decide true/false before reasoning.

1. What would resonance require? A sharp, high-$Q$ peak at a specific frequency, so tiny frequency drift would kill the heating. Why this step? Resonance (C2) is inherently frequency-selective — that is its whole nature.

2. What actually happens? Water heating is dielectric / dipole-relaxation heating: polar water molecules rotate to follow the oscillating field, lag behind, and dump energy over a broad (low-$Q$) band spanning many GHz. Why this step? This is a non-resonant, robust process — it works whether the field is $2.4$ or $2.5\text{ GHz}$; the frequency is chosen for engineering (magnetron cost, penetration depth, regulation), not a resonance.

3. Conclusion: the claim is FALSE (C10). If it were resonance, ovens would need laboratory-grade frequency stability — they don't. Why this step? Recognising when the resonance formula does not apply is as important as applying it.

Verify (consistency, not arithmetic): a true molecular rotational resonance of gaseous water lies elsewhere and is extremely sharp; a $2.45\text{ GHz}$ sharp resonance would make ovens fragile to drift — contradicting reality that ovens are cheap and tolerant. The broad low-$Q$ picture matches observation. ✓

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Active recall

Recall Which cell does each answer belong to?

• Amplitude $\approx F_0/k$ when $\omega \ll {\omega }_0$? ::: C1 (low-frequency / static-like limit)
• Denominator collapses to $b\omega /m$? ::: C2 (on resonance)
• Amplitude $\propto 1/{\omega }^2$? ::: C3 (far above resonance, inertia-dominated)
• Amplitude $\to \infty$? ::: C4 (zero damping)
• Amplitude $=F_0/k$ exactly? ::: C5 ($\omega =0$, static push)
• Peak sits below ${\omega }_0$? ::: C6 (${\omega }_{res}=\sqrt{{\omega }_0^2-b^2/2m^2}$)
• $\Delta \omega ={\omega }_0/Q$? ::: C7 (bandwidth / half-power)
• Microwave-oven trap? ::: C10 (dielectric heating, NOT resonance)

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Connections

• Forced Oscillations — supplies the $A(\omega )$ machinery every example uses.
• Damped Oscillations — provides $b$ and thus $Q$ (Examples 2, 4, 5).
• Simple Harmonic Motion — defines ${\omega }_0=\sqrt{k/m}$ (all examples).
• LC Circuits & AC Resonance — the electrical twin of Example 7.
• Standing Waves & Normal Modes — resonance of the bridge modes in Example 6.
• Fourier Analysis — why a random ("break-step") force spreads over frequency and avoids the peak.