Free and forced vibration of single-degree-of-freedom systems

Why it matters

Machine tools chatter, pumps and compressors shake their foundations, and shafts whirl when an exciting frequency comes close to a natural frequency. A single-degree-of-freedom (SDOF) spring–mass–damper model captures most of this behaviour and is the basis for choosing foundation springs, isolation mounts and dampers.

Key ideas

Model. A mass m (kg) on a spring of stiffness k (N/m) with a viscous damper of coefficient c (N·s/m), displaced x from its static equilibrium position. Measuring x from static equilibrium removes gravity from the equation of motion.

Free undamped vibration. After a disturbance the mass oscillates harmonically at the natural frequency ω_n = √(k/m), with constant amplitude. For a mass hanging on a spring, k = mg/δ_st, so ω_n = √(g/δ_st): the natural frequency follows from the static deflection alone. The energy method (d/dt(KE + PE) = 0) and Rayleigh's method give the same result and are handy for pendulums and rolling bodies.

Springs in combination. Springs in parallel share the deflection, so stiffnesses add: k = k₁ + k₂. Springs in series share the force, so compliances add: 1/k = 1/k₁ + 1/k₂.

Free damped vibration. The damping ratio ζ = c/c_c, with critical damping c_c = 2√(km) = 2mω_n, decides the response:

• ζ < 1, underdamped: decaying oscillation at ω_d = ω_n√(1 − ζ²);
• ζ = 1, critically damped: fastest return to rest without oscillation;
• ζ > 1, overdamped: slow, non-oscillatory return. The logarithmic decrement δ = ln(x_i/x_{i+1}) between successive peaks measures damping experimentally.

Forced vibration. Under F = F₀ sin ωt the steady-state response is at the forcing frequency ω, with amplitude given by the magnification factor and a phase lag φ behind the force. Near r = ω/ω_n = 1 the amplitude peaks (resonance), limited only by damping; at r = 1 the phase lag is 90°. The transient part decays because of damping.

Vibration isolation. The force transmitted to the foundation divided by the applied force is the transmissibility TR. TR < 1 only when r > √2, so the mounts must be soft enough that ω_n is well below the forcing frequency. In that region more damping increases TR, although some damping is needed to pass through resonance at start-up.

Rotating unbalance and whirling. A machine with unbalance m₀e excited at its running speed behaves the same way with F₀ = m₀·e·ω². A shaft carrying a disc whirls violently at its critical speed, which equals the natural frequency of transverse vibration ω_c = √(k/m) = √(g/δ_st).

Formulas

m·ẍ + c·ẋ + k·x = F₀ sin ωt — equation of motion.

ω_n = √(k/m) = √(g/δ_st) (rad/s); f_n = ω_n/(2π) (Hz); T = 1/f_n (s).

c_c = 2√(k·m) (N·s/m); ζ = c/c_c.

ω_d = ω_n·√(1 − ζ²) — ζ < 1.

δ = ln(x₁/x₂) = 2πζ/√(1 − ζ²); over n cycles δ = (1/n)·ln(x₀/x_n).

X = (F₀/k)/√[(1 − r²)² + (2ζr)²] — steady amplitude (m), r = ω/ω_n.

tan φ = 2ζr/(1 − r²) — phase lag.

TR = √[1 + (2ζr)²]/√[(1 − r²)² + (2ζr)²] — force transmissibility.

k_parallel = k₁ + k₂; 1/k_series = 1/k₁ + 1/k₂.

Worked examples

Example 1 (standard). A 20 kg mass is supported on a spring of 8 kN/m with a damper of 160 N·s/m. Find ω_n, ζ, ω_d, the logarithmic decrement and the ratio of two successive amplitudes.

1. ω_n = √(k/m) = √(8000/20) = 20 rad/s (f_n = 3.18 Hz).
2. c_c = 2√(km) = 2√(8000 × 20) = 800 N·s/m; ζ = 160/800 = 0.2.
3. ω_d = 20 × √(1 − 0.04) = 19.6 rad/s.
4. δ = 2π × 0.2/√(0.96) = 1.283.
5. x₁/x₂ = e^δ = 3.61.
6. ω_n = 20 rad/s, ζ = 0.2, ω_d ≈ 19.6 rad/s, δ ≈ 1.28, amplitude ratio ≈ 3.61

Example 2 (GATE level). A 100 kg machine stands on mounts of total stiffness 400 kN/m with ζ = 0.1. A harmonic force of amplitude 500 N acts at 900 rev/min. Find the steady amplitude, the transmissibility and the force transmitted to the floor.

1. ω_n = √(400 000/100) = 63.25 rad/s; ω = 2π × 900/60 = 94.25 rad/s; r = 94.25/63.25 = 1.490.
2. (1 − r²)² = (1 − 2.221)² = 1.490; (2ζr)² = (0.298)² = 0.0888; √(1.490 + 0.0888) = 1.2565.
3. Static deflection under F₀: F₀/k = 500/400 000 = 1.25 mm.
4. X = 1.25/1.2565 = 0.995 mm.
5. TR = √(1 + 0.0888)/1.2565 = 1.0435/1.2565 = 0.830.
6. Transmitted force = 0.830 × 500 = 415 N.
7. X ≈ 0.99 mm, TR ≈ 0.83, F_T ≈ 415 N. Because r is only just above √2 = 1.414, the isolation is poor; softer mounts (lower ω_n) would reduce TR further.

Common mistakes

• Mixing ω (rad/s) and f (Hz), or forgetting to convert rev/min.
• Adding series spring stiffnesses directly.
• Including gravity in the equation of motion when x is measured from static equilibrium.
• Thinking damping always helps isolation; above r = √2 it raises transmissibility.
• Writing ω_d = ω_n(1 − ζ²) without the square root.
• Using the log decrement over n cycles without dividing by n.

For GATE PI

Expect natural frequency of spring–mass systems (series/parallel springs, static deflection), damping ratio and log decrement from a given decay, steady amplitude or transmissibility at a given frequency ratio, and the condition for isolation. Practise working in rad/s throughout and quoting Hz only at the end.

Quick check

1. k = 400 N/m and m = 4 kg. What is f_n?
2. Two 1000 N/m springs act in series. What is the equivalent stiffness?
3. A spring deflects 10 mm statically under its mass. What is ω_n (g = 9.81 m/s²)?
4. For what frequency ratio does TR fall below 1?

Answers: 1. 1.59 Hz; 2. 500 N/m; 3. 31.3 rad/s; 4. r > √2.