Damped Oscillations
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You should be comfortable solving second-order linear differential equations with constant coefficients and recall the solution for the undamped harmonic oscillator. Familiarity with the exponential trial solution method is essential.
Why this matters
Real oscillators never oscillate forever; understanding how energy dissipates is key to predicting system behaviour.
In the previous lecture, we solved the ideal case where a mass on a spring oscillates indefinitely. In reality, friction, air resistance, or internal material losses remove energy from the system. This lecture introduces the mathematical framework for damped oscillations, showing how the damping coefficient determines whether the system oscillates while decaying, returns to equilibrium as quickly as possible without overshooting, or returns slowly. These three regimes are fundamental to engineering applications, from car suspension systems to electronic circuits.
The Physical Picture
Imagine a mass attached to a spring, but now moving through a viscous fluid like honey. The spring still pulls the mass back towards the equilibrium position, but the fluid exerts a drag force that opposes the motion. This drag force is proportional to the velocity of the mass. Unlike the spring force, which depends on position, the drag force depends on how fast the mass is moving. When the mass is at rest, there is no drag. When it moves quickly, the drag is large. This velocity-dependent force continuously removes mechanical energy from the system, converting it into heat. The result is that the amplitude of the oscillation decreases over time. Depending on how strong the drag is relative to the spring stiffness and the mass, the system can behave in three distinct ways: it can wobble back and forth while slowing down, it can slide back to the centre as fast as possible without crossing it, or it can creep back to the centre very slowly.
The Equation of Motion
We model the viscous drag force as \( F_{\text{drag}} = -c\dot{x} \), where \( c \) is the damping coefficient and \( \dot{x} \) is the velocity. The restoring force remains \( F_{\text{spring}} = -kx \). Applying Newton's second law, \( F = m\ddot{x} \), we sum the forces:
Rearranging terms gives the standard form of the damped harmonic oscillator equation:
This is a homogeneous second-order linear ordinary differential equation with constant coefficients. To solve it, we assume an exponential trial solution of the form \( x(t) = e^{\lambda t} \). Substituting this into the differential equation yields the characteristic equation:
The roots of this quadratic equation determine the time evolution of the system. Using the quadratic formula, the roots are:
It is conventional to define the damping rate \( \gamma = \frac{c}{2m} \) and the natural angular frequency \( \omega_0 = \sqrt{\frac{k}{m}} \). In these terms, the roots become:
The nature of these roots depends entirely on the sign of the discriminant \( \Delta = c^{2} - 4mk \), or equivalently \( \gamma^{2} - \omega_0^{2} \).
The Three Regimes of Damping
The value of the discriminant partitions the motion into three distinct physical regimes.
1. Under-damping (\( \Delta < 0 \)): Here, \( c < 2\sqrt{mk} \) or \( \gamma < \omega_0 \). The term under the square root is negative, so the roots are complex conjugates: \( \lambda_{\pm} = -\gamma \pm i\omega_d \), where \( \omega_d = \sqrt{\omega_0^{2} - \gamma^{2}} \) is the damped angular frequency. The general solution is:
This represents oscillatory motion with a frequency slightly lower than the natural frequency \( \omega_0 \), and an amplitude that decays exponentially with time constant \( 1/\gamma \). The system crosses the equilibrium position multiple times before settling.
2. Critical Damping (\( \Delta = 0 \)): Here, \( c = 2\sqrt{mk} \) or \( \gamma = \omega_0 \). The roots are real and equal: \( \lambda_{\pm} = -\gamma \). The general solution is:
This is the fastest return to equilibrium without oscillation. The system approaches \( x=0 \) monotonically. This regime is often desirable in engineering, such as in door closers or galvanometer needles, to avoid overshooting.
3. Over-damping (\( \Delta > 0 \)): Here, \( c > 2\sqrt{mk} \) or \( \gamma > \omega_0 \). The roots are real and distinct: \( \lambda_{\pm} = -\gamma \pm \sqrt{\gamma^{2} - \omega_0^{2}} \). Both roots are negative. The general solution is:
Since \( \lambda_+ > \lambda_- \) (both negative), the term with \( \lambda_+ \) (the less negative root) decays more slowly and dominates at late times. The motion is non-oscillatory and returns to equilibrium more slowly than in the critically damped case. The system creeps back to the centre.
Physical Interpretation of Parameters
The parameter \( \gamma = c/2m \) has units of \( \mathrm{s^{-1}} \) and represents the rate at which the amplitude decays in the under-damped case. The parameter \( \omega_0 = \sqrt{k/m} \) is the natural frequency of the undamped system. The ratio \( \zeta = \gamma / \omega_0 = \frac{c}{2\sqrt{mk}} \) is known as the damping ratio. It is a dimensionless quantity that characterises the system's damping level relative to its natural frequency.
If \( \zeta < 1 \), the system is under-damped. If \( \zeta = 1 \), it is critically damped. If \( \zeta > 1 \), it is over-damped. The critical damping coefficient is \( c_c = 2m\omega_0 = 2\sqrt{mk} \). Engineers often design systems to have \( \zeta \) close to 1 for rapid settling, or slightly less than 1 to allow a small amount of oscillation if faster response is needed.
The derivation
The Damped Harmonic Oscillator
For a mass \(m\) subject to a linear restoring force \(-kx\) and a linear viscous drag \(-c\dot{x}\), the equation of motion \(m\ddot{x}+c\dot{x}+kx=0\) is a homogeneous second-order linear ODE. The exponential trial solution \(x=e^{\lambda t}\) reduces it to the characteristic equation \(m\lambda^{2}+c\lambda+k=0\), with roots \(\lambda_{\pm}=-\gamma\pm\sqrt{\gamma^{2}-\omega_0^{2}}\) where \(\gamma\equiv c/2m\) and \(\omega_0^{2}\equiv k/m\). The sign of the discriminant \(\Delta=c^{2}-4mk\) (equivalently \(\gamma^{2}-\omega_0^{2}\)) partitions the motion into exactly three regimes: under-damped (\(\Delta<0\), oscillatory decay), critically damped (\(\Delta=0\), fastest non-oscillatory return), and over-damped (\(\Delta>0\), slow non-oscillatory return), with the boundary at the critical damping \(c_{c}=2\sqrt{mk}=2m\omega_0\).
Critical damping is the fastest return to equilibrium without overshoot, not the fastest approach overall: a lightly under-damped system reaches \( x=0 \) sooner but then overshoots. When a small, quick overshoot is acceptable, designers often choose a damping ratio somewhat below 1 (around 0.7) so that the system settles within a tolerance band sooner. 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 immersed in a fluid providing a damping coefficient \( c = 20\ \mathrm{N\cdot s/m} \). Determine the regime of damping and the damped angular frequency.
- Calculate the natural angular frequency \( \omega_0 \): \[ \omega_0 = \sqrt{\frac{k}{m}} = \sqrt{\frac{800}{2.0}} = \sqrt{400} = 20\ \mathrm{rad/s} \]
- Calculate the damping rate \( \gamma \): \[ \gamma = \frac{c}{2m} = \frac{20}{2 \times 2.0} = \frac{20}{4.0} = 5.0\ \mathrm{s^{-1}} \]
- Compare \( \gamma \) and \( \omega_0 \) to determine the regime: Since \( \gamma = 5.0\ \mathrm{s^{-1}} < \omega_0 = 20\ \mathrm{rad/s} \), the system is under-damped. Alternatively, check the discriminant: \[ \Delta = c^2 - 4mk = 20^2 - 4(2.0)(800) = 400 - 6400 = -6000 < 0 \]
- Calculate the damped angular frequency \( \omega_d \): \[ \omega_d = \sqrt{\omega_0^2 - \gamma^2} = \sqrt{20^2 - 5.0^2} = \sqrt{400 - 25} = \sqrt{375} \approx 19.36\ \mathrm{rad/s} \]
Units check. \( \omega_0 \) is in \( \mathrm{rad/s} \) (dimensionless radian per second). \( \gamma \) is in \( \mathrm{s^{-1}} \). The term \( \omega_0^2 - \gamma^2 \) has units \( \mathrm{s^{-2}} \), so \( \omega_d \) is in \( \mathrm{s^{-1}} \), which is consistent with angular frequency.
Three quick tests
- What is the condition for critical damping in terms of \( c \), \( m \), and \( k \)?
Answer
\( c = 2\sqrt{mk} \) or \( c = 2m\omega_0 \). - If the discriminant \( \Delta > 0 \), what is the nature of the roots \( \lambda_{\pm} \)?
Answer
Real and distinct (both negative). - In the under-damped case, how does the damped frequency \( \omega_d \) compare to the natural frequency \( \omega_0 \)?
Answer
\( \omega_d < \omega_0 \) because \( \omega_d = \sqrt{\omega_0^2 - \gamma^2} \).
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.