Lesson 1: Spring-Mass: Free Vibrations
Estimated time: 40-50 minutes
Learning Objectives
By the end of this lesson, you will be able to:
- Model a spring-mass system with the ODE mx'' + cx' + kx = 0
- Solve the undamped case and identify natural frequency, amplitude, and phase
- Classify damped systems as overdamped, critically damped, or underdamped
- Solve IVPs for all three damping cases
- Interpret physical behavior from the mathematical solution
The Spring-Mass Model
A mass m attached to a spring with stiffness k, subject to damping with coefficient c, satisfies Newton's second law:
Spring-Mass Equation: mx'' + cx' + kx = F(t), where x(t) is displacement from equilibrium, c is the damping coefficient, k is the spring constant (Hooke's law), and F(t) is an external force. When F(t) = 0, we have the free vibration equation.
Undamped Free Vibrations (c = 0)
With no damping: mx'' + kx = 0, or x'' + ω0²x = 0 where ω0 = √(k/m) is the natural frequency.
Undamped Solution: x(t) = c1cos(ω0t) + c2sin(ω0t) = A cos(ω0t - φ), where A = √(c1² + c2²) is the amplitude and φ = arctan(c2/c1) is the phase angle. Period T = 2π/ω0.
Example 1: Undamped Spring
A 2-kg mass on a spring with k = 18 N/m is released from x(0) = 0.3 m with x'(0) = 0.
Solution: 2x'' + 18x = 0 ⇒ x'' + 9x = 0. ω0 = 3 rad/s.
x(t) = c1cos 3t + c2sin 3t. x(0) = c1 = 0.3. x'(0) = 3c2 = 0, c2 = 0.
Answer: x(t) = 0.3 cos(3t). Period = 2π/3 ≈ 2.09 seconds.
Damped Free Vibrations (c > 0)
The characteristic equation mr² + cr + k = 0 has discriminant D = c² - 4mk. Three cases arise:
Overdamped (c² > 4mk): Two negative real roots. x(t) = c1er1t + c2er2t. No oscillation; sluggish return to equilibrium.
Critically Damped (c² = 4mk): Repeated root r = -c/(2m). x(t) = (c1 + c2t)ert. Fastest return to equilibrium without oscillation.
Underdamped (c² < 4mk): Complex roots α ± βi. x(t) = eαt(c1cos βt + c2sin βt). Decaying oscillations with quasi-frequency β.
Example 2: Classifying Damping
Classify: (a) m=1, c=6, k=9; (b) m=1, c=4, k=5; (c) m=1, c=10, k=9.
(a) D = 36-36 = 0: critically damped. (b) D = 16-20 = -4: underdamped. (c) D = 100-36 = 64: overdamped.
Example 3: Underdamped Vibration
Solve x'' + 2x' + 5x = 0, x(0) = 1, x'(0) = 0.
r² + 2r + 5 = 0 ⇒ r = -1 ± 2i. α = -1, β = 2.
x = e-t(c1cos 2t + c2sin 2t). x(0) = c1 = 1. x' = e-t[(-c1+2c2)cos 2t + (-c2-2c1)sin 2t]. x'(0) = -1+2c2 = 0, c2 = 1/2.
Answer: x(t) = e-t(cos 2t + (1/2)sin 2t).
Example 4: Critically Damped
Solve x'' + 6x' + 9x = 0, x(0) = 2, x'(0) = 1.
(r+3)² = 0, r = -3 repeated. x = (c1 + c2t)e-3t.
x(0) = c1 = 2. x' = (c2 - 3c1 - 3c2t)e-3t. x'(0) = c2 - 6 = 1, c2 = 7.
Answer: x(t) = (2 + 7t)e-3t.
Physical Interpretation Summary
| Type | Condition | Physical Behavior |
|---|---|---|
| Undamped | c = 0 | Perpetual oscillation at ω0 |
| Underdamped | c² < 4mk | Decaying oscillation (most common in practice) |
| Critically damped | c² = 4mk | Fastest non-oscillatory return (door closers) |
| Overdamped | c² > 4mk | Sluggish non-oscillatory return |
Check Your Understanding
1. A spring has k = 50 N/m and m = 2 kg. Find the natural frequency ω0.
2. For m = 2, c = 8, k = 8, classify the damping.
3. For an underdamped system, what determines how quickly oscillations decay?
4. A door closer is designed to be critically damped. Why is this the preferred choice?
Key Takeaways
- mx'' + kx = 0 (undamped): simple harmonic motion at natural frequency ω0 = √(k/m).
- Damping classification depends on c² vs 4mk: overdamped (>), critical (=), underdamped (<).
- Underdamped is the most common physical case: decaying oscillations with quasi-frequency β.
- Critically damped returns to equilibrium fastest without overshooting.
- The natural frequency ω0 depends only on k and m, not on damping.