Driven Oscillations §

Decomposition into transient + steady-state. Because the equation is linear, the full solution is the sum of two independent parts:

• The transient $x_{\text{trans}}$ solves the homogeneous (free) equation $m\ddot{x} + c\dot{x} + kx = 0$. It is set by the initial conditions and decays as $e^{-\gamma t}$ (see Damped oscillations).
• The steady-state $x_{ss}$ is a particular solution that oscillates at the drive frequency $\omega$, not at $\omega_0$. It persists forever as long as the drive is on.

After a few damping time constants ($t \gtrsim 1/\gamma$), the transient is gone and only the steady-state remains. This is why a driven system “locks” to the drive frequency: whatever the initial motion, the long-time motion is a pure sinusoid at $\omega$.

Authoritative source for this section: Landau & Lifshitz, Mechanics, §14: Forced Oscillations

Three frequency regimes.

• Low frequency ($\omega \ll \omega_0$): the spring dominates, $X \approx F_0/k$ (the static deflection) and $\delta \approx 0$ (in phase). The mass barely moves before the force reverses.
• Resonance ($\omega \approx \omega_0$): the inertial and spring terms nearly cancel, so the damping term alone balances the drive. The amplitude is large and $\delta \approx \pi/2$ (velocity in phase with force).
• High frequency ($\omega \gg \omega_0$): inertia dominates, $X \approx F_0/(m\omega^2) \to 0$ and $\delta \approx \pi$ (out of phase). The mass is effectively rigid and barely responds.

Amplitude response curve. The plot of $X(\omega)$ versus $\omega$ is the frequency response of the oscillator: a resonance peak centered near $\omega_0$. The width of the peak at half its maximum height is the bandwidth $\Delta\omega$.

Phase-lag curve. The phase lag $\delta(\omega)$ rises smoothly from $0$ to $\pi$, passing through $\pi/2$ exactly at $\omega = \omega_0$.

Authoritative source for this section: OpenStax University Physics Volume 1, §15.6: Forced Oscillations and Resonance

Displacement vs. velocity resonance. The amplitude $X(\omega)$ is maximized at

not exactly at $\omega_0$. The displacement peak only coincides with $\omega_0$ in the weak-damping limit. By contrast, the velocity (and hence the power) response is maximized exactly at $\omega = \omega_0$, because there the velocity is in phase with the driving force. For high-$Q$ systems the two frequencies are nearly identical, which is why the distinction is often glossed over.

Quality factor. $Q = \omega_0/(2\gamma)$ quantifies the sharpness of resonance. A high-$Q$ system (tuning fork, $Q \sim 10^3$; superconducting cavity, $Q \sim 10^{10}$) has a narrow peak and stores energy for many cycles; a low-$Q$ system (shock absorber, $Q \sim 1$) has a broad, flat response. Equivalently, $Q$ is the number of radians of phase over which the free (undriven) oscillation decays by $e$; see Damped oscillations for the free-decay analysis.

Bandwidth. The full width at half maximum of the amplitude peak is

so a higher-$Q$ system has a narrower resonance. (Note $\Delta\omega$ is a frequency in s⁻¹, whereas $1/Q$ is dimensionless; the two are related through $\omega_0$.)

Practical resonance. In engineering, resonance is exploited (musical instruments, radio tuning, MRI) or avoided (bridge design, building seismic isolation, turbine-blade fatigue). The 1940 Tacoma Narrows Bridge collapse is a classic—though often oversimplified—example of wind-driven resonance.

Authoritative source for this section: Landau & Lifshitz, Mechanics, §14: Forced Oscillations

Maximum power transfer at $\omega = \omega_0$. The average power is proportional to $\sin\delta$. Since $\delta = \pi/2$ exactly at $\omega = \omega_0$, the power is maximized there—not at the displacement peak $\omega_{res}$. Physically, at $\omega = \omega_0$ the velocity $\dot{x}$ is in phase with the force, so the drive does positive work on every half-cycle. This is why velocity resonance (and power resonance) sits at $\omega_0$ while displacement resonance sits slightly below it.

At resonance the average power is $\langle P\rangle_{\omega_0} = F_0^2/(2c)$, and the stored energy is $E = \tfrac{1}{2}k X_{\max}^2$. The quality factor can be read as the number of radians of phase over which the stored energy decays by $e$: $Q \approx \omega_0 E / \langle P\rangle$.

Authoritative source for this section: OpenStax University Physics Volume 1, §15.6: Forced Oscillations and Resonance

The mapping between the mechanical and electrical systems is exact:

Under this map $\omega_0 = 1/\sqrt{LC}$, $\gamma = R/(2L)$, and $Q = \omega_0 L/R = \sqrt{L/C}/R$. The resonance peak becomes the tuned circuit: at $\omega = \omega_0$ the inductive and capacitive reactances cancel, the impedance is purely resistive ($Z = R$), and the current is maximal. This is the basis of radio tuning and band-pass filters, and it is the electrical realization of everything in the resonance analysis above.

Authoritative source for this section: OpenStax University Physics Volume 2, §16.5: RLC Series Circuits

This is a Mathieu equation (when $\varepsilon$ is small and $c = 0$). The key feature: energy is pumped into the system by modulating a parameter, not by applying a direct force.

Parametric resonance occurs when the modulation frequency is near twice the natural frequency, $\Omega \approx 2\omega_0$. Within the instability band the amplitude grows exponentially, provided the modulation depth $\varepsilon$ exceeds a threshold set by damping. The instability region in the $(\Omega, \varepsilon)$ plane is a “tongue” opening out from $\Omega = 2\omega_0$:

Examples:

• A child on a swing pumps by periodically changing their effective length (standing/sitting).
• Parametric amplifiers in superconducting circuits modulate a Josephson junction’s inductance.
• MEMS resonators use parametric pumping for frequency multiplication.
• The Faraday wave experiment: a fluid surface modulated vertically develops standing waves at half the driving frequency.

Parametric excitation is a subcase of driven oscillations: the “drive” enters through a coefficient rather than a source term. The distinction matters because parametric resonance occurs at $\Omega = 2\omega_0$ (not $\omega_0$) and can amplify signals without adding a direct force.

Authoritative source for this section: Landau & Lifshitz, Mechanics, §14: Forced Oscillations

Problem: A driven damped oscillator has $\omega_0 = 10$ rad/s, $\gamma = 0.5$ s⁻¹, and $F_0/m = 2$ m/s². Find the steady-state amplitude at (a) $\omega = \omega_0$, (b) $\omega = 2\omega_0$, and (c) the ratio $X(\omega_0)/X(2\omega_0)$.