Forced vibration, resonance and vibration isolation

Why it matters

Motors, pumps, compressors, fans, presses and vehicles all apply periodic forces to their supports. Whether that produces a gentle hum or a destructive shake depends on how the forcing frequency compares with the natural frequency, and on damping. The same equations tell you how to mount a machine so that little force reaches the floor, or how to protect a sensitive instrument from a vibrating floor.

Key ideas

Harmonic forcing. A force F₀·sin ωt on a spring–mass–damper gives m·ẍ + c·ẋ + k·x = F₀·sin ωt. The response is a transient (the damped free vibration, which dies away) plus a steady-state vibration at the forcing frequency ω, not at ω_n: x = X·sin(ωt − φ).

Frequency ratio and magnification factor. With r = ω/ω_n, the steady amplitude divided by the static deflection F₀/k is the magnification factor (dynamic amplification).

• r ≪ 1: the spring controls; X ≈ F₀/k, response in phase with the force.
• r ≈ 1: resonance; only damping limits the amplitude, X = (F₀/k)/(2ζ), and the response lags the force by 90°.
• r ≫ 1: the mass controls; X falls roughly as 1/r², response nearly 180° out of phase. With damping the peak occurs slightly below ω_n, at r = √(1 − 2ζ²). An undamped system at resonance grows without limit.

Rotating unbalance. A machine of total mass M with an unbalanced mass m₀ at eccentricity e is excited by m₀·e·ω²·sin ωt, a force that grows with speed. Its amplitude ratio M·X/(m₀·e) starts at zero, peaks near resonance and tends to 1 at high speed (the machine then spins about the common centre of mass).

Transmissibility. The force passed to the foundation is the sum of the spring and damper forces. The force transmissibility TR = F_T / F₀ has the same form as the motion transmissibility for a machine or instrument on a vibrating base (X/Y).

• TR > 1 for r < √2 (amplification); TR = 1 at r = √2 for any damping.
• TR < 1 only for r > √2: this is the isolation region.
• In the isolation region more damping increases TR, but damping is still needed to limit the amplitude while the machine passes through resonance at start-up and shut-down.

Designing isolation. Choose soft mounts so that ω_n is well below the forcing frequency (typically r = 2.5–5, giving 80–95% isolation). Softer mounts mean larger static deflection (δ_st = g/ω_n²), so very low frequencies need heavy inertia blocks (to increase m without raising k), air springs or active isolation. Stiffening the mounts helps only if the machine runs well below resonance.

Dynamic vibration absorber. A small spring–mass tuned to the forcing frequency (√(k₂/m₂) = ω) and attached to the main mass makes the main mass's steady amplitude zero at that frequency; the absorber mass vibrates instead. It works for a constant-speed machine; tuned mass dampers in buildings use the same idea with added damping.

Formulas

ω_n = √(k/m) ; r = ω/ω_n ; ζ = c / (2√(km))

X = (F₀/k) / √((1 − r²)² + (2ζr)²) (steady amplitude, m) MF = X / (F₀/k) = 1 / √((1 − r²)² + (2ζr)²) tan φ = 2ζr / (1 − r²) (phase lag of displacement behind force) At resonance MF = 1/(2ζ).

Rotating unbalance: M·X / (m₀·e) = r² / √((1 − r²)² + (2ζr)²)

TR = F_T / F₀ = √(1 + (2ζr)²) / √((1 − r²)² + (2ζr)²) (also X/Y for base excitation) Undamped: TR = 1 / |r² − 1| ; for a target TR (r > √2): r² = 1 + 1/TR

Isolation efficiency = 1 − TR

Absorber tuning: k₂/m₂ = ω²

Worked examples

Example 1 (standard): machine on damped mounts. Given: m = 100 kg, k = 400 kN/m, ζ = 0.1, harmonic force amplitude 500 N at 1200 rpm.

1. ω_n = √(400 000/100) = 63.25 rad/s ; ω = 2π × 1200/60 = 125.7 rad/s ; r = 1.987.
2. (1 − r²)² = (1 − 3.948)² = 8.691 ; (2ζr)² = 0.3974² = 0.158.
3. MF = 1/√(8.691 + 0.158) = 0.336 → X = (500/400 000) × 0.336 = 0.42 mm.
4. TR = √(1 + 0.158) × 0.336 = 0.362 → F_T = 0.362 × 500 = 181 N. Answer: X ≈ 0.42 mm; about 181 N (36%) reaches the floor; softer mounts would isolate better.

Example 2 (GATE level): sizing mounts for a motor with unbalance. Given: motor and base 50 kg, unbalance m₀·e = 0.01 kg·m, 1500 rpm. Target TR = 0.1 (neglect damping for sizing); the mounts actually have ζ = 0.05.

1. ω = 2π × 1500/60 = 157.1 rad/s ; r² = 1 + 1/0.1 = 11 → r = 3.317.
2. ω_n = 157.1/3.317 = 47.36 rad/s → k = M·ω_n² = 50 × 2243 = 112.2 kN/m (total).
3. Static deflection δ_st = M·g/k = 490.5/112 200 = 4.37 mm.
4. With ζ = 0.05: TR = √(1 + 0.110)/√(100 + 0.110) = 1.054/10.006 = 0.105, so 89.5% isolation.
5. Running amplitude: X = (0.01/50) × 11/10.006 = 0.22 mm.
6. Passing through resonance: X ≈ (m₀·e/M)/(2ζ) = 0.0002/0.1 = 2.0 mm, ten times the running amplitude, which is why some damping is kept. Answer: k ≈ 112 kN/m (δ_st ≈ 4.4 mm), TR ≈ 0.105, running amplitude ≈ 0.22 mm.

Common mistakes

• Expecting the steady response at ω_n; it is at the forcing frequency.
• Adding damping to improve isolation when r > √2; it makes TR worse there.
• Making mounts stiffer to "hold the machine still" when it runs above resonance.
• Using rpm for ω, or Hz where rad/s is needed.
• Using the force-excitation amplitude formula for rotating unbalance, where the force grows with ω².
• Forgetting that TR = 1 at r = √2 regardless of damping.

For GATE ME

Frequent questions: steady-state amplitude and phase; magnification factor at resonance; transmissibility and the r > √2 condition; spring stiffness for a required isolation; rotating unbalance amplitude; absorber tuning. Practise sketching MF and TR against r for several ζ, and solving r² = 1 + 1/TR.

Quick check

1. ζ = 0.05. Magnification factor at resonance?
2. Above what frequency ratio does isolation start?
3. Undamped mounts, r = 3. Transmissibility?
4. Does more damping help or hurt isolation at r = 4?
5. An absorber must suppress vibration at 50 Hz with m₂ = 2 kg. Absorber stiffness?

Answers: 1. 10. 2. r = √2. 3. 1/(9 − 1) = 0.125. 4. Hurts. 5. k₂ = 2 × (2π × 50)² = 197 kN/m.