Forced vibration, resonance and vibration isolation

Why it matters

Engines, unbalanced wheels, road roughness and gear meshing all apply periodic forces or motions, and a structure responds at the forcing frequency. If that frequency is near a natural frequency the response can be many times the static deflection (resonance). Engine mounts, suspension design and machine foundations rely on the same two curves: magnification factor and transmissibility.

Key ideas

Forced vibration. A harmonic force F₀·sin ωt acts on a mass-spring-damper. The response has a transient part (free vibration at ω_d, which dies out because of damping) and a steady-state part at the forcing frequency ω. Steady-state is what matters for design.

Frequency ratio and magnification factor. With r = ω/ω_n, the steady-state amplitude X divided by the static deflection F₀/k is the magnification factor (dynamic amplification). It is about 1 for r much less than 1 (stiffness-controlled), peaks near r = 1 (damping-controlled), and falls towards 0 for r much greater than 1 (mass-controlled).

Resonance. At r = 1 the amplitude is limited only by damping: X / (F₀/k) = 1/(2ζ). With ζ = 0.05 the response is 10 times the static deflection. The exact peak occurs at r = √(1 − 2ζ²), slightly below 1. At resonance the response lags the force by 90°.

Phase. The displacement lags the force by φ, where tan φ = 2ζr/(1 − r²): nearly 0° well below resonance, 90° at resonance, approaching 180° well above it.

Rotating unbalance. An unbalanced mass m₀ at eccentricity e in a machine of total mass M gives a force m₀·e·ω², which grows with speed. At high speed the amplitude tends to m₀·e/M.

Transmissibility. The ratio of the force transmitted to the foundation to the applied force amplitude (or, for base excitation, the ratio of mass motion to base motion). It is the same expression in both cases.

• For r < √2, TR > 1: the mounts amplify the force.
• For r > √2, TR < 1: isolation. The larger r, the better the isolation.
• In the isolation region, more damping makes TR larger (worse); damping is only useful to limit the resonance peak while the machine runs up through it.

Designing an isolator. Choose the mount stiffness so that the mounted natural frequency is well below the forcing frequency, typically r = 2.5 to 5. Soft mounts mean large static deflection (δ = g/ω_n²), which is the practical limit. Adding mass (an inertia block) to the machine also lowers ω_n and reduces the amplitude of motion.

Vehicle examples. Engine mounts are tuned so that the mount natural frequency (about 8 to 15 Hz) is well below the firing frequency at idle. Suspension ride frequency (about 1 to 1.5 Hz) isolates the body from road inputs above about 2 Hz, while the damper controls the body resonance. A tuned vibration absorber (a small spring-mass tuned to the excitation frequency) can cancel the motion of the main mass at that one frequency.

Formulas

m·ẍ + c·ẋ + k·x = F₀·sin ωt

• F₀: force amplitude (N); ω: forcing frequency (rad/s).

X = (F₀/k) / √[(1 − r²)² + (2ζr)²]

• X: steady-state amplitude (m); r = ω/ω_n: frequency ratio; ζ: damping ratio.

tan φ = 2ζr / (1 − r²)

• φ: phase lag of displacement behind force (0° to 180°).

X_res = F₀ / (2ζ·k) = F₀ / (c·ω_n)

• Amplitude at r = 1 (m).

X = (m₀·e/M)·r² / √[(1 − r²)² + (2ζr)²]

• Rotating unbalance: m₀: unbalanced mass (kg); e: eccentricity (m); M: total machine mass (kg).

TR = √[1 + (2ζr)²] / √[(1 − r²)² + (2ζr)²]

• TR: force transmissibility (transmitted force / applied force), also motion transmissibility for base excitation.

TR = 1 / (r² − 1) (undamped, r > √2)

F_T = TR·F₀

• F_T: amplitude of force transmitted to the foundation (N).

Worked examples

Example 1 (standard: amplitude and transmitted force). A 100 kg machine stands on mounts of total stiffness 400 kN/m with ζ = 0.1. An unbalance produces a harmonic force of amplitude 1000 N at 1200 rpm. Find the amplitude, the phase and the force transmitted to the floor.

1. ω_n = √(400 000 / 100) = 63.25 rad/s; ω = 2π × 1200 / 60 = 125.66 rad/s; r = 125.66 / 63.25 = 1.987.
2. 1 − r² = 1 − 3.948 = −2.948; 2ζr = 0.397.
3. Denominator = √(2.948² + 0.397²) = √(8.690 + 0.158) = 2.975.
4. X = (1000 / 400 000) / 2.975 = 0.0025 / 2.975 = 8.40 × 10⁻⁴ m = 0.84 mm.
5. Phase: tan φ = 0.397 / (−2.948), so φ = 172° (displacement nearly opposite to force, as expected above resonance).
6. TR = √(1 + 0.158) / 2.975 = 1.076 / 2.975 = 0.362; F_T = 0.362 × 1000 = 362 N.

Example 2 (GATE level: isolator design). A 200 kg compressor runs at 1500 rpm. Only 10% of its unbalanced force may reach the floor. Neglecting damping, find the required mount stiffness and the static deflection.

1. Undamped: TR = 1/(r² − 1) = 0.1, so r² = 11 and r = 3.317.
2. ω = 2π × 1500 / 60 = 157.08 rad/s; ω_n = ω / r = 157.08 / 3.317 = 47.36 rad/s.
3. k = m·ω_n² = 200 × 47.36² = 448 600 N/m = 449 kN/m (total for all mounts).
4. Static deflection δ = g / ω_n² = 9.81 / 2243 = 0.00437 m = 4.4 mm.
5. During start-up the machine passes through ω_n (452 rpm); some damping is needed to keep that transient resonance in check.

Example 3 (resonance amplitude). A force of 500 N acts at the natural frequency on a system with k = 200 kN/m and ζ = 0.1. X_res = 500 / (2 × 0.1 × 200 000) = 12.5 mm, five times the static deflection of 2.5 mm.

Common mistakes

• Writing TR = 1/√(1 − r²); the undamped result is 1/|1 − r²|, with no square root.
• Thinking damping always helps isolation. Above r = √2 it increases the transmitted force.
• Mixing Hz and rad/s inside r. Both ω and ω_n must be in the same units.
• Using X_res = F₀/(2ζω_n); the correct form is F₀/(2ζk) or F₀/(c·ω_n).
• Choosing a mount with r slightly above 1 and expecting isolation. You need r > √2, and in practice 2.5 or more.
• Forgetting that the unbalance force m₀·e·ω² itself changes with speed.

For GATE ME

Expect amplitude and phase of a forced damped SDOF system, resonance amplitude, rotating unbalance amplitude, force transmitted to a foundation, and isolator stiffness for a target transmissibility. Remember the landmarks: r = 1 (resonance, φ = 90°), r = √2 (TR = 1 for any damping), r > √2 (isolation, damping hurts). Practise each formula with r below and above 1.

Quick check

1. At what frequency ratio is TR equal to 1 for every damping ratio?
2. Peak magnification factor for ζ = 0.05?
3. Undamped TR at r = 3?
4. Phase lag at resonance?
5. Force at 10 Hz on a system with f_n = 8 Hz, undamped. TR?

Answers: 1. r = √2. 2. 1/(2 × 0.05) = 10. 3. 1/(9 − 1) = 0.125. 4. 90°. 5. r = 1.25, TR = 1/|1 − 1.5625| = 1.78.