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Differential Equations Unit 6: Mechanical vibrations

Springs and circuits follow the same second-order equation. Damping and forcing decide whether the motion dies out, oscillates or blows up.

Covers: Spring-mass systems, damping, resonance

Study guide

A spring–mass system obeys my″ + cy′ + ky = F(t), from Newton's second law: m is the mass, c is the damping coefficient (friction or air resistance), k is the spring constant from Hooke's law, and F(t) is an external force. Measure y from the equilibrium position.

Without damping or forcing, my″ + ky = 0 gives simple harmonic motion, y = A cos(ω₀t) + B sin(ω₀t), with natural frequency ω₀ = √(k/m) and period 2π/ω₀. It can be written as a single cosine with amplitude R = √(A² + B²) and a phase shift.

With damping, the characteristic equation's discriminant c² − 4mk decides the behavior. Underdamped (c² 4mk): it returns slowly without oscillating. Shock absorbers are designed near critical damping.

With periodic forcing F₀ cos ωt, the solution is a transient part (decaying from the homogeneous solution if there's damping) plus a steady-state oscillation at the forcing frequency. Without damping, forcing at the natural frequency (ω = ω₀) causes resonance: the amplitude grows without bound, like t sin(ω₀t). With small damping, forcing near ω₀ causes very large but bounded oscillations. When ω is close to but not equal to ω₀, the result is beats.

The same equation models RLC electric circuits, with inductance, resistance and 1/capacitance in place of m, c and k.

For my″ + cy′ + ky = F₀cos ωt, the steady-state oscillation has amplitude F₀/√((k − mω²)² + (cω)²). The gain is this amplitude divided by F₀, which equals 1/|p(iω)| with p(r) = mr² + cr + k. The steady-state response is F₀·gain·cos(ωt − φ). The phase lag φ is the argument of p(iω), so tan φ = cω/(k − mω²), and φ lies between 0 and π. The response lags by exactly π/2 when ω = ω₀.

Underdamped motion e^(−ct/2m)(A cos μt + B sin μt) oscillates with quasi-frequency μ = √(4mk − c²)/(2m) and pseudoperiod 2π/μ. It crosses equilibrium twice per pseudoperiod, so π/μ apart. For fixed m, c and ω, the amplitude is largest when k = mω², which tunes the natural frequency to the forcing frequency. If an undamped system's response contains t·sin ωt or t·cos ωt, the forcing is at resonance, so ±iω are roots of the characteristic polynomial.

Key ideas

• Spring–mass: my″ + cy′ + ky = F(t).
• No damping (c = 0): simple harmonic motion with ω₀ = √(k/m).
• Underdamped: oscillates while decaying. Critically damped: fastest return without oscillating. Overdamped: slow return.
• Resonance: forcing at the natural frequency with no damping makes amplitude grow without bound.

Formulas and rules

Common mistakes

• Mixing up the damping categories.
• Using weight as mass (mass = weight / g).

Words to know

Spring–mass equation

my″ + cy′ + ky = F(t)

Natural frequency

ω₀ = √(k/m)

Condition for critical damping

c² = 4mk

Resonance

Unbounded growth when forcing matches the natural frequency (no damping).

Underdamped motion

Oscillation with decreasing amplitude.

Simple harmonic motion

Undamped, unforced oscillation: A cos ω₀t + B sin ω₀t.

Critically damped

c² = 4mk: returns fastest without oscillating.

Overdamped

c² > 4mk: returns slowly without oscillating.

Steady-state solution

The long-term oscillation at the forcing frequency.

Transient solution

The part that decays away with damping.

Beats

Interference when forcing is close to the natural frequency.

Sample practice questions

19 more questions like these, from level 1 to 5, are in the practice set.