Forced Oscillations and Resonance
~6 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable with the equation of motion for a damped harmonic oscillator and the definition of the damping ratio \(\zeta\). Familiarity with trigonometric identities and basic complex number arithmetic is helpful but not strictly required.
Why this matters
Resonance is the mechanism by which a small, periodic push can build up to enormous amplitudes.
From the tuning of a guitar string to the structural failure of bridges and the operation of radio receivers, forced oscillations are ubiquitous. Understanding how a system responds to an external drive frequency, and specifically why it "resonates" near its natural frequency, is fundamental to engineering design and physical intuition. This lecture establishes the steady-state response of a damped oscillator to a sinusoidal force, identifying the precise conditions for maximum amplitude and the phase relationship between the drive and the motion.
The Intuitive Picture
Imagine pushing a child on a swing. If you push at random times, the swing will not go very high. However, if you push exactly when the swing is moving away from you, your small force adds energy efficiently to the system. This is resonance. In a forced oscillator, the external force does work on the mass. When the driving frequency matches the system's natural tendency to oscillate, the force and the velocity of the mass remain in phase for much of the cycle, leading to maximum energy transfer. Damping acts as a brake, limiting the amplitude. Without damping, the amplitude would grow indefinitely at resonance. With damping, the amplitude reaches a finite maximum value determined by the strength of the damping. The phase of the motion relative to the drive also changes: at low frequencies, the mass moves in step with the force; at high frequencies, it moves in opposition; and at resonance, it lags by exactly \(90^\circ\).
The Steady-State Solution
Consider a mass \(m\) attached to a spring of stiffness \(k\) and subject to viscous damping coefficient \(c\). The system is driven by a force \(F(t) = F_0 \cos(\omega t)\). The equation of motion is \(m\ddot{x} + c\dot{x} + kx = F_0 \cos(\omega t)\). We seek a solution of the form \(x(t) = A \cos(\omega t - \phi)\), where \(A\) is the amplitude and \(\phi\) is the phase lag. Substituting this trial solution into the differential equation and matching coefficients of \(\cos(\omega t)\) and \(\sin(\omega t)\) yields the amplitude and phase given in the cited derivation D-061.
Here, \(\omega_0 = \sqrt{k/m}\) is the natural frequency and \(\zeta = \frac{c}{2\sqrt{mk}}\) is the damping ratio. The denominator represents the "impedance" of the system. The term \((\omega_0^2-\omega^2)\) reflects the restoring force and inertia mismatch, while \(2\zeta\omega_0\omega\) reflects the dissipative effect. The phase lag is given by:
Note that the phase is defined such that \(x(t)\) lags the driving force. At \(\omega = 0\), \(\phi = 0\). As \(\omega \to \infty\), \(\phi \to \pi\). The transition occurs smoothly through \(\pi/2\) at \(\omega = \omega_0\).
Resonance Frequency and Amplitude
Resonance is defined as the frequency at which the amplitude \(A(\omega)\) is maximized. To find this, we minimize the denominator of \(A(\omega)\) with respect to \(\omega\). Let \(D(\omega) = (\omega_0^2-\omega^2)^2+(2\zeta\omega_0\omega)^2\). Setting \(\frac{dD}{d\omega} = 0\) and solving for \(\omega\) yields the resonant frequency:
This result is valid only for \(\zeta < 1/\sqrt{2}\). If the damping is too high (\(\zeta \geq 1/\sqrt{2}\)), the amplitude is monotonically decreasing with frequency, and no peak exists. At the resonant frequency, the amplitude reaches its maximum value:
For light damping, the resonant frequency is very close to the natural frequency \(\omega_0\). The peak amplitude is inversely proportional to the damping coefficient \(c\). This explains why lightly damped systems exhibit sharp, high-amplitude resonance peaks, while heavily damped systems show broad, low-amplitude responses.
Phase Behavior and Energy Transfer
The phase lag \(\phi(\omega)\) provides a sensitive indicator of the system's state. At \(\omega = \omega_0\), the phase lag is exactly \(\pi/2\), regardless of the damping ratio. This is a robust feature of linear systems. Physically, this means the velocity \(\dot{x}(t)\) is in phase with the driving force \(F(t)\). Since power delivered is \(P = F \dot{x}\), the average power transfer is maximized when the force and velocity are in phase. At frequencies far from resonance, the phase is either near \(0\) or near \(\pi\), meaning the force and velocity are nearly out of phase, resulting in low average power transfer. The phase curve is a smooth, monotonic function that transitions from \(0\) to \(\pi\) as \(\omega\) increases from \(0\) to \(\infty\). This behavior is universal for all linear, time-invariant second-order systems.
The derivation
Driven Oscillator: Amplitude, Phase and Resonance
For a mass on a linear spring with viscous damping, driven by a sinusoidal force \(F_0\cos\omega t\), the steady-state motion is oscillation at the drive frequency \(\omega\) with amplitude \(A(\omega)=\dfrac{F_0/m}{\sqrt{(\omega_0^2-\omega^2)^2+(2\zeta\omega_0\omega)^2}}\) and phase lag \(\phi(\omega)=\arctan\!\dfrac{2\zeta\omega_0\omega}{\omega_0^2-\omega^2}\); the amplitude peaks at \(\omega_{\text{res}}=\omega_0\sqrt{1-2\zeta^2}\) while the phase sweeps monotonically from \(0\) through \(\pi/2\) at \(\omega=\omega_0\) to \(\pi\) as \(\omega\to\infty\).
A common misconception is that the resonant frequency is always equal to the natural frequency \(\omega_0\). This is only true in the limit of zero damping. For any finite damping, the amplitude peak occurs at \(\omega_{\text{res}} < \omega_0\). However, the phase lag reaches \(\pi/2\) exactly at \(\omega_0\). Therefore, if you measure resonance by maximum amplitude, you find a frequency slightly below \(\omega_0\); if you measure it by the phase crossing \(\pi/2\), you find exactly \(\omega_0\). This distinction is crucial in precision measurements and filter design. C
Worked example
Problem. A mass of \(m = 2.0\ \mathrm{kg}\) is attached to a spring with stiffness \(k = 800\ \mathrm{N/m}\). The system is damped with coefficient \(c = 4.0\ \mathrm{N\cdot s/m}\). It is driven by a force \(F(t) = 10 \cos(\omega t)\ \mathrm{N}\). Calculate the resonant frequency \(\omega_{\text{res}}\) and the maximum amplitude \(A_{\text{max}}\).
- Calculate the natural frequency: \(\omega_0 = \sqrt{k/m} = \sqrt{800/2.0} = \sqrt{400} = 20\ \mathrm{rad/s}\).
- Calculate the damping ratio: \(\zeta = \frac{c}{2\sqrt{mk}} = \frac{4.0}{2\sqrt{2.0 \times 800}} = \frac{4.0}{2\sqrt{1600}} = \frac{4.0}{80} = 0.05\).
- Calculate the resonant frequency: \(\omega_{\text{res}} = \omega_0\sqrt{1-2\zeta^2} = 20\sqrt{1-2(0.05)^2} = 20\sqrt{1-0.005} = 20\sqrt{0.995} \approx 20 \times 0.9975 = 19.95\ \mathrm{rad/s}\).
- Calculate the maximum amplitude. Minimising the denominator of the steady-state amplitude gives \(A_{\text{max}} = \dfrac{F_0/m}{2\zeta\omega_0^2\sqrt{1-\zeta^2}}\). Numerator: \(F_0/m = 10/2.0 = 5.0\ \mathrm{m/s^2}\). Denominator: \(2(0.05)(20)^2\sqrt{1-0.05^2} = 40\sqrt{0.9975} \approx 39.95\ \mathrm{s^{-2}}\). Thus \(A_{\text{max}} = 5.0/39.95 \approx 0.125\ \mathrm{m}\). (For light damping this is close to \(F_0/(c\,\omega_0) = 10/(4.0 \times 20) = 0.125\ \mathrm{m}\).)
Units check. \(\omega\) is in \(\mathrm{rad/s}\). Amplitude \(A\) is in \(\mathrm{m}\). The light-damping form \(A_{\text{max}} \approx \frac{F_0}{c\,\omega_0}\) gives \(\frac{\mathrm{N}}{(\mathrm{N\cdot s/m})(\mathrm{1/s})} = \frac{\mathrm{N}}{\mathrm{N/m}} = \mathrm{m}\). Correct.
Three quick tests
- What is the phase lag \(\phi\) when the driving frequency \(\omega\) is much smaller than the natural frequency \(\omega_0\)?
Answer
\(\phi \approx 0\). The mass moves in phase with the force. - If the damping ratio \(\zeta\) is increased, what happens to the resonant frequency \(\omega_{\text{res}}\)?
Answer
It decreases. \(\omega_{\text{res}} = \omega_0\sqrt{1-2\zeta^2}\), so as \(\zeta\) increases, the term under the square root decreases. - At what frequency does the phase lag equal \(\pi/2\)?
Answer
\(\omega = \omega_0\). This is independent of the damping ratio.
Problem set for PU-103 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.