2.4 Driven oscillations and resonance

Push the damped oscillator sinusoidally. The equation of motion becomes

m \ddot x + b \dot x + k x \;=\; F_0 \cos(\omega t),

or, in canonical form,

\ddot x + 2\gamma \dot x + \omega_0^2 x \;=\; (F_0 / m) \cos(\omega t).

The drive has its own frequency $\omega$ , which need not match the natural frequency $\omega_0$ . Things get interesting when the two are close.

Phasor solution

Replace the real drive by the complex one $\tilde F_0\, e^{i\omega t}$ (with $\tilde F_0 = F_0$ ) and try $x = \tilde X\, e^{i\omega t}$ . Substituting:

\big[-\omega^2 + 2 i \gamma \omega + \omega_0^2\big]\, \tilde X\, e^{i\omega t} \;=\; (F_0/m)\, e^{i\omega t}.

Cancel the exponentials and solve:

\boxed{\;\;\tilde X(\omega) \;=\; \frac{F_0/m}{(\omega_0^2 - \omega^2) + 2 i \gamma \omega}\;\;}

A single complex number, derived in one line. From it we read off everything:

Amplitude: $|\tilde X(\omega)| = \dfrac{F_0/m}{\sqrt{(\omega_0^2 - \omega^2)^2 + (2\gamma\omega)^2}}$
Phase relative to drive: $\varphi(\omega) = -\arctan\!\dfrac{2\gamma\omega}{\omega_0^2 - \omega^2}$

Resonance

The amplitude $|\tilde X(\omega)|$ is small when $\omega \ll \omega_0$ (the drive moves the oscillator quasi-statically: $\tilde X \to F_0/k$ ), and small again when $\omega \gg \omega_0$ (the oscillator can’t keep up: $\tilde X \to -F_0/(m\omega^2)$ ). In between, near $\omega = \omega_0$ , the denominator gets small and the amplitude peaks. This is resonance.

For weak damping the peak occurs at $\omega \approx \omega_0$ and has height

|\tilde X|_\text{peak} \;\approx\; \frac{F_0}{2 m \gamma \omega_0} \;=\; \frac{F_0}{b\, \omega_0}.

The smaller the damping, the taller and narrower the peak.

▶ Why the peak occurs at $\omega \approx \omega_0$ (and not exactly there)

The peak of $|\tilde X(\omega)|^2$ as a function of $\omega^2$ is found by differentiating the denominator with respect to $\omega^2$ and setting to zero:

\frac{d}{d(\omega^2)}\Big[ (\omega_0^2 - \omega^2)^2 + 4\gamma^2 \omega^2\Big] = -2(\omega_0^2 - \omega^2) + 4\gamma^2 = 0,

giving $\omega_\text{peak}^2 = \omega_0^2 - 2\gamma^2$ . For underdamped systems ( $\gamma \ll \omega_0$ ) this is essentially $\omega_0$ ; the correction is second order in $\gamma/\omega_0$ .

Phase

Look at $\varphi(\omega)$ . For $\omega \ll \omega_0$ , the response is in phase with the drive ( $\varphi \approx 0$ ). For $\omega = \omega_0$ exactly, $\varphi = -\pi/2$ — the response lags the drive by a quarter cycle. For $\omega \gg \omega_0$ , $\varphi \to -\pi$ — the response is exactly out of phase with the drive. The phase transit through $-\pi/2$ at resonance is sharp for weakly damped systems and is one of the experimental signatures of being on resonance.

The interactive

driving frequency f

700 Hz

quality factor Q

10.0

f / f₀: 0.700
|H(f)| / |H(0)|: 1.94
phase lag: -7.8°

Slide $\omega_0$ , $\gamma$ , and the driving frequency. The interactive shows the amplitude curve, the time trace, and the phasor diagram simultaneously. Watch what happens to the phase as the drive passes through resonance.

Why this matters for sound

Every acoustic resonance — the modes of an organ pipe, a wine glass driven by a finger, a room mode coupled to a speaker, the cochlea responding to a sinusoid — is mathematically this same driven-oscillator response. Different system, same equation; different $\omega_0$ , $\gamma$ , and $F_0$ , but the same Lorentzian peak in the frequency-response curve. The next lesson sharpens that intuition by introducing the quality factor $Q$ , which captures the trade between peak height and bandwidth in a single dimensionless number.