Exercises — Resonance — physical consequences, design implications

WHAT tool: ${\omega }_0=\sqrt{k/m}$ — this comes straight from Simple Harmonic Motion, where the spring's restoring force $-kx$ and Newton's $F=ma$ combine into $\ddot{x}=-(k/m)x$, whose rhythm is $\sqrt{k/m}$. WHY: natural frequency is the "target" resonance sits near, so it is always step one. ${\omega }_0=\sqrt{\frac{200}{0.50}}=\sqrt{400}=20\text{rad/s}.$

The amber point at the top of the hump is resonance — the driving frequency that gives the largest swing $A$. Its height is set by how small the denominator of the master formula can get, which at the peak is dominated by the damping term $b\omega /m$. So the peak height depends on damping: less damping → taller, narrower peak.

WHAT tool: master formula. WHY at $\omega ={\omega }_0$: the term $({\omega }_0^2-{\omega }^2)$ becomes $0$, so the messy denominator collapses to just $b\omega /m$. $A=\frac{F_0/m}{\sqrt{0+(b\omega /m{)}^2}}=\frac{F_0/m}{b\omega /m}=\frac{F_0}{b\omega }.$ Numerically: $A=\frac{2.0}{0.50\times 20}=\frac{2.0}{10}=0.20\text{m}.$ Notice the mass cancelled — at resonance, amplitude is limited only by damping.

WHAT tool: $Q={\omega }_{0}m/b$, then $\Delta \omega ={\omega }_0/Q$. WHY: $Q$ says how many "clean" oscillations happen before damping eats the energy — bigger $Q$ = narrower peak. $Q=\frac{{\omega }_{0}m}{b}=\frac{20\times 0.50}{0.50}=20.$ $\Delta \omega =\frac{{\omega }_0}{Q}=\frac{20}{20}=1.0\text{rad/s}.$

WHAT tool: ${\omega }_{res}=\sqrt{{\omega }_0^2-b^2/(2m^2)}$ — obtained by setting $dA/d\omega =0$ in Forced Oscillations. WHY it's below ${\omega }_0$: the damping term $b\omega /m$ grows with $\omega$, so the minimum of the whole denominator is pushed to a slightly lower frequency than the point where $({\omega }_0^2-{\omega }^2)=0$. $\frac{b^2}{2m^2}=\frac{0.50^2}{2\times 0.50^2}=\frac{0.25}{0.50}=0.50{\text{rad}}^2/{\text{s}}^2.$ ${\omega }_{res}=\sqrt{400-0.50}=\sqrt{399.5}\approx 19.9875\text{rad/s}.$ Only about $0.06\%$ below ${\omega }_0$ — for this light damping the shift is tiny, confirming ${\omega }_{res}\approx {\omega }_0$ but not exactly equal.

WHY this frequency: the two half-power points sit roughly at ${\omega }_0\pm \Delta \omega /2$; there the amplitude should be about $1/\sqrt{2}\approx 0.707$ of the peak $0.20\text{m}$, i.e. about $0.1414\text{m}$. Compute each piece of the master formula at $\omega =20.5$:

• ${\omega }_0^2-{\omega }^2=400-420.25=-20.25$.
• $b\omega /m=(0.50\times 20.5)/0.50=20.5$.
• denominator $=\sqrt{(-20.25{)}^2+20.5^2}=\sqrt{410.06+420.25}=\sqrt{830.31}\approx 28.82$.
• numerator $F_0/m=2.0/0.50=4.0$. $A=\frac{4.0}{28.82}\approx 0.1388\text{m}.$ That's $0.1388/0.20=0.694$ of the peak — very near the ideal $0.707$, confirming ${\omega }_0\pm \Delta \omega /2$ marks (approximately) the half-power edges.

WHY this maps to our formula: the LC circuit is the electrical mirror of the mass–spring: $1/C$ plays the role of stiffness $k$, $L$ the role of mass $m$. Its natural frequency is ${\omega }_0=1/\sqrt{LC}$. Tuning the dial sets ${\omega }_0$ equal to the station's frequency, so only that station resonates. Solve ${\omega }_0=1/\sqrt{LC}$ for $C$: $C=1/({\omega }_0^{2}L)$. ${\omega }_0=2\pi (100\times 10^6)=6.2832\times 10^8\text{rad/s}.$ $C=\frac{1}{(6.2832\times 10^8{)}^2\times 2.0\times 10^{-6}}=\frac{1}{(3.9478\times 10^{17})(2.0\times 10^{-6})}.$ $C=\frac{1}{7.8957\times 10^{11}}\approx 1.266\times 10^{-12}\text{F}=1.27\text{pF}.$

Convert the marching rate: $120\text{steps/min}=120/60=2.0\text{steps/s}=2.0\text{Hz}$. WHY dangerous: the driving frequency $f=2.0\text{Hz}$ equals $f_0=2.0\text{Hz}$, so every footfall pushes in phase with the bridge's own sway — coherent pushes stack up (resonance) and the amplitude climbs toward its damping-limited maximum $F_0/(b{\omega }_0)$. The fix: "break step" randomises footfall timing. A random force has no single dominant frequency (see Fourier Analysis) — its energy spreads across many frequencies, so almost none lands exactly at $f_0$, and the resonant build-up never happens.

(a) $Q={\omega }_{0}M/b=(1.2\times 6.0\times 10^6)/(4.0\times 10^5)=7.2\times 10^6/4.0\times 10^5=18.$ A moderate $Q$ — the building rings fairly sharply, which is exactly the danger.

(b) At resonance $A=F_0/(b{\omega }_0)$ (the $({\omega }_0^2-{\omega }^2)$ term vanishes): $A=\frac{3.0\times 10^5}{4.0\times 10^5\times 1.2}=\frac{3.0\times 10^5}{4.8\times 10^5}=0.625\text{m}.$ A $0.625\text{m}$ sway — uncomfortable and structurally serious.

(c) Doubling $b$ to $8.0\times 10^5\text{kg/s}$: since $A\propto 1/b$ at resonance, the amplitude halves to $0.3125\text{m}$. The damper cuts sway by a factor of 2 and, by lowering $Q$, also broadens the peak so the building is less sensitive to hitting ${\omega }_0$ exactly. This is the physics behind Taipei 101's giant pendulum.

If heating relied on a high-$Q$ (sharp) resonance, the oven's frequency would have to match a water transition to a tiny fraction of a percent — impossible to hold, and any drift would stop the heating. Instead the oven exploits dielectric (dipole-relaxation) heating: water's permanent dipoles try to twist to follow the oscillating field but lag behind, and that lag dumps energy as heat. This is a broad, low-$Q$ process — it works over a wide band of frequencies (roughly $1$–$100\text{GHz}$), so $2.45\text{GHz}$ is chosen for regulatory and penetration reasons, not because it hits a resonant peak. Robustness comes precisely from low $Q$: no sharp tuning is needed.

Peak amplitude formula at resonance? ::: $A_{max}=F_0/(b{\omega }_0)$ $Q$ in terms of ${\omega }_0,m,b$? ::: $Q={\omega }_{0}m/b$ Bandwidth from $Q$? ::: $\Delta \omega ={\omega }_0/Q$ Displacement resonance frequency? ::: ${\omega }_{res}=\sqrt{{\omega }_0^2-b^2/2m^2}$, just below ${\omega }_0$ Doubling damping does what to resonant amplitude? ::: halves it (since $A\propto 1/b$) LC natural frequency? ::: ${\omega }_0=1/\sqrt{LC}$