Forced harmonic vibration and resonance

Why it matters

Aircraft structures are driven by periodic forces all the time: engine and propeller imbalance, rotor blade passage, pump pulsations and buffet with a dominant frequency. If a forcing frequency sits close to a natural frequency the response is amplified many times, and fatigue cracks or failed equipment follow. Forced-response theory tells you how large the response will be, how it lags the force, and how much margin you need between excitation and natural frequencies.

Key ideas

Equation of motion. A damped SDOF system under a harmonic force F₀·sin(ω·t) obeys

m·ẍ + c·ẋ + k·x = F₀·sin(ω·t)

The full solution is the decaying free vibration (transient) plus a particular solution (steady state). After the transient dies out, the system vibrates at the forcing frequency ω, not at its natural frequency: x(t) = X·sin(ω·t − φ).

Frequency ratio and magnification. Write r = ω/ω_n and X_st = F₀/k (the deflection the same force would cause if applied statically). The dynamic magnification factor is

M = X/X_st = 1/√((1 − r²)² + (2·ζ·r)²)

Three regions follow from it:

• r ≪ 1 (stiffness controlled): M ≈ 1, response in phase with the force.
• r ≈ 1 (damping controlled): M ≈ 1/(2ζ), large; phase lag 90°.
• r ≫ 1 (mass controlled): M ≈ 1/r², small; response nearly 180° out of phase.

Resonance. Strictly, resonance is forcing at ω = ω_n, where the phase lag is exactly 90° for any damping and M = 1/(2ζ). The peak of M occurs slightly below ω_n, at r = √(1 − 2ζ²) (only if ζ < 1/√2), with M_max = 1/(2ζ·√(1 − ζ²)). For light damping the two are practically equal. With ζ = 0 the steady amplitude at r = 1 is unbounded and grows linearly with time.

Phase. tan φ = 2ζr/(1 − r²), with φ between 0 and 180°. Use the quadrant correctly: above resonance the denominator is negative and φ is greater than 90°. A calculator's arctan returns a negative angle there; add 180°. The 90° phase lag at r = 1 is the most reliable way to locate a natural frequency in a test.

Quality factor and bandwidth. Q = 1/(2ζ) is the resonant magnification. The half-power points, where the amplitude is M_peak/√2, are separated by Δω ≈ 2ζ·ω_n, so ζ ≈ Δω/(2ω_n). This is how damping is measured from a frequency-response sweep (topic on vibration testing).

Force transmitted and energy. At resonance the spring and inertia forces cancel and the applied force is balanced only by the damping force c·ω·X; the work done by the force per cycle equals the energy dissipated, π·c·ω·X².

Rotating unbalance. A mass m₀ at eccentricity e rotating at ω gives F₀ = m₀·e·ω², so the force grows with speed. Then m·X/(m₀·e) = r²/√((1 − r²)² + (2ζr)²), which tends to 1 at high speed.

Assumptions. Linear spring and viscous damper, steady state (transient decayed), single harmonic force. A periodic non-harmonic force is handled by superposing the response to each Fourier harmonic, any of which may be resonant.

Formulas

m·ẍ + c·ẋ + k·x = F₀·sin(ω·t) — forced SDOF; F₀ force amplitude (N), ω forcing frequency (rad/s).

X = F₀/√((k − m·ω²)² + (c·ω)²) — steady-state amplitude (m).

r = ω/ω_n, X_st = F₀/k, M = X/X_st = 1/√((1 − r²)² + (2ζr)²) — magnification factor (dimensionless).

tan φ = c·ω/(k − m·ω²) = 2ζr/(1 − r²) — phase lag of displacement behind force, 0 ≤ φ ≤ 180°.

M(r = 1) = 1/(2ζ) — magnification when forcing at ω_n; Q = 1/(2ζ).

r_peak = √(1 − 2ζ²), M_max = 1/(2ζ·√(1 − ζ²)) — true amplitude peak, valid for ζ < 0.707.

Δω = ω₂ − ω₁ ≈ 2ζ·ω_n — half-power bandwidth (rad/s), light damping.

F₀ = m₀·e·ω² — rotating-unbalance force (N); m₀ unbalanced mass (kg), e eccentricity (m).

Worked examples

Example 1 (standard). m = 10 kg, k = 1000 N/m, c = 50 N·s/m, F₀ = 100 N. Find the amplitude and phase for ω = 10 rad/s and ω = 8 rad/s.

1. ω_n = √(k/m) = √100 = 10 rad/s; ζ = c/(2·√(k·m)) = 50/200 = 0.25.
2. At ω = 10 rad/s: k − m·ω² = 1000 − 1000 = 0; c·ω = 500 N/m. X = F₀/√(0² + 500²) = 100/500 = 0.20 m; φ = 90°.
3. Check with M: X_st = 100/1000 = 0.10 m; M = 1/(2 × 0.25) = 2; X = 0.20 m.
4. At ω = 8 rad/s: k − m·ω² = 1000 − 640 = 360 N/m; c·ω = 400 N/m. X = 100/√(360² + 400²) = 100/538.1 = 0.186 m. tan φ = 400/360, φ = 48.0°.

Answer: X = 0.20 m with φ = 90° at 10 rad/s; X = 0.186 m with φ = 48.0° at 8 rad/s.

Example 2 (GATE level). An equipment rack of mass 50 kg on mounts of total stiffness 2 × 10⁵ N/m and damping ratio 0.1 is shaken by a harmonic force of amplitude 500 N. Find (a) ω_n, (b) the frequency and value of the maximum amplitude, (c) the amplitude and phase at r = 1.5, (d) the half-power bandwidth in Hz.

1. ω_n = √(k/m) = √(200 000/50) = √4000 = 63.25 rad/s. X_st = 500/200 000 = 2.5 mm.
2. r_peak = √(1 − 2ζ²) = √0.98 = 0.990, so ω_peak = 62.6 rad/s. M_max = 1/(2ζ·√(1 − ζ²)) = 1/(0.2 × 0.99499) = 5.025. X_max = 5.025 × 2.5 = 12.56 mm.
3. At r = 1.5: (1 − r²) = −1.25, 2ζr = 0.30. M = 1/√(1.5625 + 0.09) = 1/1.2855 = 0.778, X = 1.94 mm. tan φ = 0.30/(−1.25): second quadrant, φ = 180° − 13.5° = 166.5°.
4. Δω ≈ 2ζ·ω_n = 2 × 0.1 × 63.25 = 12.65 rad/s = 2.01 Hz.

Answer: (a) 63.2 rad/s; (b) 62.6 rad/s, 12.6 mm; (c) 1.94 mm lagging by 166.5°; (d) about 2.0 Hz.

Common mistakes

• Squaring c·ω wrongly: (c·ω)² for c = 50 N·s/m and ω = 10 rad/s is 250 000, not 2500.
• Taking arctan of a negative number above resonance and reporting a negative phase; the lag lies between 90° and 180° there.
• Saying the system vibrates at its natural frequency in steady state; it vibrates at the forcing frequency.
• Believing damping always lowers response a lot: away from resonance (r < 0.5 or r > 2) damping has little effect; it controls the peak.
• Confusing the amplitude peak (r = √(1 − 2ζ²)) with the 90° phase point (r = 1).
• Using M with a rotating unbalance, where the force itself grows as ω².

For GATE AE

Expect amplitude and phase calculations from m, c, k, F₀ and ω; the magnification at resonance (1/2ζ); identifying stiffness-, damping- and mass-controlled regions; frequency of peak response; half-power damping estimates; and rotating-unbalance problems. Practise the quadrant of φ and checking answers with the static deflection F₀/k.

Quick check

1. ζ = 0.05. What is the magnification factor at r = 1?
2. What is the phase lag at r = 1, for any damping?
3. At r = 3 with light damping, approximately what is M?
4. Half-power points are at 98 and 102 rad/s. Estimate ζ.
5. In steady state, at what frequency does the system respond?

Answers: 1. 10; 2. 90°; 3. about 1/(r² − 1) = 0.125; 4. ζ ≈ 4/(2 × 100) = 0.02; 5. at the forcing frequency ω.