Damping, Driving, 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 simple harmonic oscillator, the definition of angular frequency \(\omega_0\), and basic trigonometric identities for combining sine and cosine terms.
Why this matters
Real oscillators never oscillate forever; they lose energy to friction and can be forced to move at frequencies other than their natural one.
In the previous lecture, we studied ideal simple harmonic motion where energy is conserved. In the real world, however, dissipative forces like air resistance or internal friction always act to remove energy from the system. Furthermore, oscillators are rarely isolated; they are often subjected to external periodic forces. Understanding how damping affects the decay of free oscillations and how driving forces lead to resonance is fundamental to fields ranging from structural engineering (preventing bridge collapse) to electronics (tuning radio circuits) and quantum mechanics (line broadening).
The Physical Picture
Imagine a mass attached to a spring. If you pull it and let go, it oscillates. In a perfect vacuum with a perfect spring, it would oscillate forever. In reality, it slows down and stops. This slowing down is due to damping. Now, imagine someone pushing that mass back and forth with a regular rhythm. If the rhythm matches the mass's natural "preference" for speed, the amplitude of the motion grows significantly. This is resonance. If the rhythm is too fast or too slow, the mass moves only a little. The key insight is that the system's response depends critically on the relationship between the driving frequency and the natural frequency, modified by the amount of damping present.
The Equation of Motion
We model the system with a mass \(m\), a spring constant \(k\), and a damping coefficient \(b\). The damping force is assumed to be linear in velocity, \(F_d = -b\dot{x}\), which is a good approximation for low-speed motion in a viscous fluid. The driving force is sinusoidal, \(F(t) = F_0 \cos(\omega t)\). Newton's second law gives us the differential equation:
Dividing by \(m\) and defining the natural frequency \(\omega_0 = \sqrt{k/m}\) and the damping parameter \(\gamma = b/m\), we rewrite this as:
This is a linear, second-order, non-homogeneous differential equation. The solution consists of two parts: a transient solution (which decays over time due to damping) and a steady-state solution (which persists as long as the driving force is applied). We are primarily interested in the steady-state behavior because the transient dies out, leaving only the response to the driver.
Steady-State Solution and Phase Lag
Because the driving force is a cosine, we guess a solution of the form \(x(t) = A \cos(\omega t - \delta)\), where \(A\) is the amplitude and \(\delta\) is the phase lag. The phase lag \(\delta\) indicates that the mass does not move in perfect sync with the driver; it lags behind. Substituting this trial solution into the equation of motion and using trigonometric identities allows us to solve for \(A\) and \(\delta\) in terms of the system parameters.
The resulting amplitude is given by:
The phase lag is determined by:
Note that \(\delta\) is always between 0 and \(\pi\). At very low driving frequencies (\(\omega \ll \omega_0\)), \(\delta\) is close to 0, meaning the mass moves almost in phase with the driver. At very high frequencies (\(\omega \gg \omega_0\)), \(\delta\) approaches \(\pi\), meaning the mass moves almost exactly out of phase with the driver. At the natural frequency \(\omega = \omega_0\), the phase lag is exactly \(\pi/2\) (90 degrees), regardless of the damping.
Resonance and the Quality Factor
Resonance occurs when the amplitude \(A(\omega)\) is maximized. To find the resonant frequency \(\omega_{\text{res}}\), we minimize the denominator of the amplitude expression with respect to \(\omega\). This yields:
For light damping (\(\gamma \ll \omega_0\)), \(\omega_{\text{res}} \approx \omega_0\). However, strictly speaking, the peak occurs slightly below the natural frequency. The sharpness of this peak is characterized by the quality factor \(Q\), defined as:
A high \(Q\) value indicates low damping and a very sharp resonance peak, meaning the system is highly selective to frequency. A low \(Q\) value indicates high damping and a broad, flat response. The maximum amplitude at resonance is approximately \(A_{\text{max}} \approx \frac{F_0}{m \gamma \omega_0}\) for high \(Q\) systems. This shows that as damping \(\gamma\) decreases, the resonant amplitude increases, which can lead to structural failure if the driving force is large enough.
The derivation
Damped, Driven Oscillators and Resonance
For a mass on a linear spring subject to linear (viscous) damping and a sinusoidal driving force \(F(t)=F_0\cos\omega t\), the equation of motion \(m\ddot{x}+b\dot{x}+kx=F_0\cos\omega t\) has a long-time steady-state solution \(x(t)=A(\omega)\cos\!\big(\omega t-\delta(\omega)\big)\). We derive the amplitude \(A(\omega)\), the phase lag \(\delta(\omega)\), the driving frequency \(\omega_{\text{res}}\) that maximises \(A\), and show that the sharpness of the resonance peak is governed by the quality factor \(Q=\omega_0/\gamma\).
A common misconception is that the resonant frequency is always exactly \(\omega_0\). While this is a good approximation for light damping, the true peak of the amplitude curve shifts to lower frequencies as damping increases. If the damping is strong enough that \(\gamma^2 \ge 2\omega_0^2\) (that is, \(\gamma \ge \sqrt{2}\,\omega_0\)), the amplitude curve decreases monotonically with frequency and there is no resonance peak at all. Note that this happens before critical damping (\(\gamma = 2\omega_0\)), so a free oscillator can still ring while showing no driven resonance peak. C
Worked example
Problem. A mass \(m = 2.0\ \mathrm{kg}\) is attached to a spring with constant \(k = 800\ \mathrm{N/m}\). The system is damped with coefficient \(b = 4.0\ \mathrm{kg/s}\) and driven by a force \(F(t) = 10 \cos(\omega t)\ \mathrm{N}\). Calculate the natural frequency \(\omega_0\), the damping parameter \(\gamma\), the resonant frequency \(\omega_{\text{res}}\), and the amplitude at resonance.
- Calculate \(\omega_0 = \sqrt{k/m} = \sqrt{800/2.0} = \sqrt{400} = 20\ \mathrm{rad/s}\).
- Calculate \(\gamma = b/m = 4.0/2.0 = 2.0\ \mathrm{s^{-1}}\).
- Calculate \(\omega_{\text{res}} = \sqrt{\omega_0^2 - \frac{1}{2}\gamma^2} = \sqrt{400 - \frac{1}{2}(4)} = \sqrt{398} \approx 19.95\ \mathrm{rad/s}\).
- Amplitude at resonance: \(\omega_0^2 - \omega_{\text{res}}^2 = \tfrac12\gamma^2 = 2\ \mathrm{s^{-2}}\), so the denominator is \(\sqrt{(2)^2 + (2.0)^2(398)} = \sqrt{4 + 1592} = \sqrt{1596} \approx 39.95\ \mathrm{s^{-2}}\), and \(A = \frac{F_0/m}{39.95} = \frac{5.0}{39.95} \approx 0.125\ \mathrm{m}\). (The high-\(Q\) estimate \(F_0/(m\gamma\omega_0) = 10/(2.0 \times 2.0 \times 20) = 0.125\ \mathrm{m}\) agrees, since \(Q = 10\).)
Units check. \(F_0/m\) has units \(\mathrm{N/kg} = \mathrm{m/s^2}\). The denominator has units \((\mathrm{rad/s})^2 = \mathrm{s^{-2}}\). So \(A\) has units \(\mathrm{m/s^2} / \mathrm{s^{-2}} = \mathrm{m}\). Correct.
Three quick tests
- What happens to the phase lag \(\delta\) as the driving frequency \(\omega\) approaches zero?
Answer
It approaches 0 radians (in phase). - If the damping coefficient \(b\) is increased, does the resonant frequency \(\omega_{\text{res}}\) increase or decrease?
Answer
It decreases. - For a system with very high \(Q\), at what driving frequency is the power absorption maximum?
Answer
At \(\omega = \omega_0\) (approximately).
Problem set for PU-101 →
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.