Resonance: Peak Response at the Natural Frequency

A driven, damped harmonic oscillator responds most strongly when the driving frequency matches its natural frequency $\omega_0$. This phenomenon is called resonance, and it's behind everything from singers shattering wine glasses to the Tacoma Narrows Bridge collapse.

The Setup

A mass-spring system with damping coefficient $\gamma$ and natural frequency $\omega_0 = \sqrt{k/m}$, driven by a sinusoidal force $F\sin(\omega t)$, has a steady-state response amplitude:

$A(\omega) = \dfrac{F/m}{\sqrt{(\omega_0^{2} - \omega^{2})^{2} + (\gamma\omega)^{2}}}$

This is a Lorentzian-like peak in $\omega$, centered near $\omega_0$. The width of the peak depends on the damping: small damping gives a tall, narrow peak; large damping gives a short, broad one.

Step-by-Step Solution

Given: A driven oscillator with $\omega_0 = 5\;\text{rad/s}$, $\gamma = 0.5\;\text{s}^{-1}$, and unit driving amplitude $F/m = 1$.

Find: The response amplitude at $\omega = 3$, $\omega = 5$ (resonance), and $\omega = 8$.

---

Step 1 — Apply the formula at $\omega = 3$ (below resonance).

$A(3) = \dfrac{1}{\sqrt{(25 - 9)^{2} + (0.5 \cdot 3)^{2}}}$

$= \dfrac{1}{\sqrt{256 + 2.25}}$

$= \dfrac{1}{\sqrt{258.25}}$

$\approx \dfrac{1}{16.07}$

$\approx 0.0622$

A modest response — driving below resonance gives a small amplitude.

Step 2 — At $\omega = 5$ (resonance).

$A(5) = \dfrac{1}{\sqrt{(25 - 25)^{2} + (0.5 \cdot 5)^{2}}}$

$= \dfrac{1}{\sqrt{0 + 6.25}}$

$= \dfrac{1}{2.5}$

$= 0.4000$

A massive response — about 6.4 times the off-resonance value at $\omega = 3$. The first term in the denominator vanishes, leaving only the damping term.

Step 3 — At $\omega = 8$ (above resonance).

$A(8) = \dfrac{1}{\sqrt{(25 - 64)^{2} + (0.5 \cdot 8)^{2}}}$

$= \dfrac{1}{\sqrt{1521 + 16}}$

$= \dfrac{1}{\sqrt{1537}}$

$\approx 0.0255$

Even smaller — driving above the natural frequency also gives a small response. The mass can't keep up with the rapid forcing.

Step 4 — The peak height.

At exact resonance, the amplitude is:

$A_{\text{peak}} = \dfrac{F/m}{\gamma\,\omega_0} = \dfrac{1}{(0.5)(5)} = 0.4$

Smaller damping → larger peak. In the limit $\gamma \to 0$ (no damping), the peak goes to infinity — the oscillator's amplitude grows without bound. This is what destroyed the Tacoma Narrows Bridge in 1940 (well, it was a more complex aeroelastic effect, but the principle is the same).

Step 5 — The "Q factor".

The sharpness of the resonance peak is measured by the quality factor:

$Q = \dfrac{\omega_0}{\gamma}$

For our system: $Q = 5/0.5 = 10$. A pendulum clock might have $Q \sim 100$, a quartz crystal $Q \sim 10^{6}$, an atomic clock $Q \sim 10^{14}$. Higher Q = sharper resonance = more accurate timekeeping.

---

Answer:

• $A(3) \approx 0.062$ (off-resonance, small)
• $A(5) = 0.400$ ← resonance peak
• $A(8) \approx 0.026$ (above-resonance, small)

The amplitude peaks sharply at the natural frequency $\omega_0 = 5$ — about 6× larger than even nearby frequencies. The peak height is $1/(\gamma\omega_0)$, which goes to infinity as damping vanishes. The width of the peak is set by the quality factor $Q = \omega_0/\gamma$.

Try It

• Slide the driving frequency through the natural frequency at 5 rad/s.
• Watch the response amplitude grow dramatically near $\omega = \omega_0$.
• Adjust the damping $\gamma$ — smaller damping makes a sharper, taller peak.
• The plot shows $A(\omega)$ across the entire frequency range.