Damped Oscillations: CCEA A2 Further Mathematics Mechanics 1
Welcome to your study notes on Damped Oscillations! In simple harmonic motion (SHM), we often imagine an idealized world where a pendulum swings forever or a spring bobs up and down endlessly. But in reality, friction and air resistance always step in to slow things down.
In this chapter from A2 2 Section A: Mechanics 1, you will learn how to model these real-world resistive forces using second-order differential equations, classify how systems respond, and solve initial value problems. Don't worry if differential equations have felt intimidating in the past—we will break down each step clearly!
1. Setting Up the Physical and Mathematical Model
Consider a body of mass \(m\) attached to a spring moving along a straight line. As it moves, it experiences two main forces acting towards or away from the equilibrium position \(x = 0\):
1. Restoring Force: Given by Hooke's Law, \(F_{\text{restore}} = -kx\) (where \(k > 0\) is the spring stiffness constant). The negative sign shows that the force always pulls the mass back towards \(x = 0\).
2. Damping (Resistive) Force: A friction or fluid drag force that opposes velocity, \(F_{\text{damp}} = -c\frac{\mathrm{d}x}{\mathrm{d}t}\) (or \(-\lambda v\), where \(c > 0\) is the damping coefficient). The faster the particle moves, the harder the medium pushes back.
Applying Newton’s Second Law (\(F_{\text{net}} = ma = m\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\)):
\(m\frac{\mathrm{d}^2x}{\mathrm{d}t^2} = -c\frac{\mathrm{d}x}{\mathrm{d}t} - kx\)
Rearranging this gives the Standard Governing Differential Equation for Damped Oscillations:
\(m\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + c\frac{\mathrm{d}x}{\mathrm{d}t} + kx = 0\)
Dividing through by the mass \(m\), you will often see this written in standard parameters:
\(\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 2k'\frac{\mathrm{d}x}{\mathrm{d}t} + \omega^2x = 0\)
where \(\omega^2 = \frac{k}{m} > 0\) and \(2k' = \frac{c}{m} > 0\).
Key Takeaway:
Both restoring force and damping force oppose displacement and velocity respectively. Therefore, when moved to the left-hand side of the differential equation, all coefficients (\(m\), \(c\), and \(k\)) must be positive.
2. The Auxiliary Equation and Three Damping Regimes
To solve the homogeneous linear equation \(a\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + b\frac{\mathrm{d}x}{\mathrm{d}t} + cx = 0\), we try a trial solution of the form \(x = e^{mt}\). This gives the auxiliary (characteristic) equation:
\(am^2 + bm + c = 0\)
The roots are determined by the quadratic formula, and the physical behavior of the system is entirely controlled by the discriminant \(\Delta = b^2 - 4ac\):
Regime 1: Light Damping (Underdamped) — \(\Delta < 0\)
Condition: \(b^2 - 4ac < 0\)
Roots: Complex conjugates, \(m = -\alpha \pm i\beta\) (where \(\alpha = \frac{b}{2a} > 0\) and \(\beta = \frac{\sqrt{4ac - b^2}}{2a}\)).
General Solution:
\(x(t) = e^{-\alpha t}\left(A\cos(\beta t) + B\sin(\beta t)\right) = Re^{-\alpha t}\sin(\beta t + \phi)\)
Physical Motion: The particle oscillates back and forth about the equilibrium position (\(x = 0\)), but the amplitude decays exponentially over time. The motion is bounded within the decay envelope curves \(x = \pm Re^{-\alpha t}\).
Pseudo-Period: The time taken between successive peaks is \(T = \frac{2\pi}{\beta}\).
Regime 2: Critical Damping — \(\Delta = 0\)
Condition: \(b^2 - 4ac = 0\)
Roots: Equal real negative root, \(m = -\alpha\) (where \(\alpha = \frac{b}{2a} > 0\)).
General Solution:
\(x(t) = (A + Bt)e^{-\alpha t}\)
Physical Motion: The system returns to equilibrium in the shortest possible time without ongoing oscillations. Depending on the initial push (initial velocity), the object may pass through \(x = 0\) at most once before dying away to rest.
Real-world analogy: Modern car shock absorbers and automatic closing doors use critical damping so that they snap shut smoothly and quickly without bouncing or slamming.
Regime 3: Heavy Damping (Overdamped) — \(\Delta > 0\)
Condition: \(b^2 - 4ac > 0\)
Roots: Two distinct real negative roots, \(m = -\alpha\) and \(m = -\beta\) (where \(\alpha, \beta > 0\)).
General Solution:
\(x(t) = Ae^{-\alpha t} + Be^{-\beta t}\)
Physical Motion: The resistive force is so dominant that no oscillations take place. The displacement decays monotonically towards zero as \(t \to \infty\), but it takes longer to reach equilibrium compared to a critically damped system.
Real-world analogy: Moving a spoon through thick honey or treacle.
Summary Comparison Table
• Light Damping (\(\Delta < 0\)): Roots are \(-\alpha \pm i\beta\); Solution: \(x(t) = e^{-\alpha t}(A\cos\beta t + B\sin\beta t)\); Oscillates with decaying amplitude.
• Critical Damping (\(\Delta = 0\)): Roots are \(-\alpha\) (repeated); Solution: \(x(t) = (A + Bt)e^{-\alpha t}\); Returns to rest in fastest time without oscillating.
• Heavy Damping (\(\Delta > 0\)): Roots are \(-\alpha, -\beta\) (distinct real); Solution: \(x(t) = Ae^{-\alpha t} + Be^{-\beta t}\); Sluggish decay to rest without oscillating.
3. Initial Value Problems: Finding Constants \(A\) and \(B\)
In examination questions, you are typically given initial conditions at \(t = 0\):
1. Initial displacement: \(x(0) = x_0\)
2. Initial velocity: \(\dot{x}(0) = v_0\) (where \(\dot{x} = \frac{\mathrm{d}x}{\mathrm{d}t}\))
Step-by-Step Method:
Step 1: Write down the general solution \(x(t)\) based on the roots of your auxiliary equation.
Step 2: Substitute \(t = 0\) into \(x(t)\) to set up your first equation involving \(A\) and \(B\).
Step 3: Differentiate \(x(t)\) with respect to \(t\) to find \(\dot{x}(t)\). Be careful to use the Product Rule and Chain Rule correctly!
Step 4: Substitute \(t = 0\) into \(\dot{x}(t)\) to set up your second equation.
Step 5: Solve the simultaneous equations for \(A\) and \(B\).
Worked Example: Critical Damping
Problem: A particle of mass \(1\text{ kg}\) moves along the \(x\)-axis such that \(\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 6\frac{\mathrm{d}x}{\mathrm{d}t} + 9x = 0\). At time \(t = 0\), the particle is released from \(x = 2\) with velocity \(v = -4\text{ ms}^{-1}\). Find the displacement \(x(t)\) in terms of \(t\).
Solution:
1. Auxiliary Equation:
\(m^2 + 6m + 9 = 0 \implies (m + 3)^2 = 0 \implies m = -3\) (repeated root).
Since \(\Delta = 6^2 - 4(1)(9) = 0\), this is critical damping.
2. General Solution:
\(x(t) = (A + Bt)e^{-3t}\)
3. Use \(x(0) = 2\):
\(2 = (A + B(0))e^{0} \implies A = 2\)
4. Differentiate using the Product Rule:
\(\dot{x}(t) = \frac{\mathrm{d}}{\mathrm{d}t}[(A + Bt)] \cdot e^{-3t} + (A + Bt) \cdot \frac{\mathrm{d}}{\mathrm{d}t}[e^{-3t}]\)
\(\dot{x}(t) = Be^{-3t} - 3(A + Bt)e^{-3t}\)
5. Use \(\dot{x}(0) = -4\) and \(A = 2\):
\(-4 = B(1) - 3(2 + 0)(1)\)
\(-4 = B - 6 \implies B = 2\)
6. Final Particular Solution:
\(x(t) = (2 + 2t)e^{-3t} = 2(1 + t)e^{-3t}\)
4. Common Pitfalls & Examiner Warnings
Avoid these frequent traps highlighted in CCEA examiner reports:
• Sign Errors in Newton’s Second Law: Remember that damping resists motion (\(-c\dot{x}\)) and springs pull back (\(-kx\)). Setting up \(m\ddot{x} = c\dot{x} - kx\) is a critical mistake that will flip signs throughout your auxiliary equation.
• Forgetting the Product Rule during Differentiation: When differentiating the critically damped solution \(x(t) = (A + Bt)e^{-\alpha t}\), students frequently write \(\dot{x} = B(-\alpha e^{-\alpha t})\). Remember: \(\frac{\mathrm{d}}{\mathrm{d}t}[(A+Bt)e^{-\alpha t}] = Be^{-\alpha t} - \alpha(A+Bt)e^{-\alpha t}\).
• Misunderstanding Boundary Crossings: Many students mistakenly believe that critically damped or overdamped systems can never cross \(x = 0\). If an object is given a large initial velocity directed back towards the origin, it can cross the equilibrium point \(x = 0\) once before dying out asymptotically.
• Confusing Terminology: Do not confuse critical damping (\(\Delta = 0\)) with heavy damping (\(\Delta > 0\)). Always calculate the discriminant carefully before naming the regime.
5. Quick Summary Revision Checklist
Before sitting your exam, make sure you can confidently:
• Set up \(m\ddot{x} + c\dot{x} + kx = 0\) directly from a worded physical description.
• Write down the auxiliary equation \(am^2 + bm + c = 0\) and evaluate the discriminant \(\Delta = b^2 - 4ac\).
• State whether the system exhibits light damping (\(\Delta < 0\)), critical damping (\(\Delta = 0\)), or heavy damping (\(\Delta > 0\)).
• Quote the correct general solution form for all three damping cases.
• Apply the Product Rule correctly to differentiate displacement and determine integration constants from initial conditions.