Vibration of a single-degree system
Many structures can be pictured, at a first approximation, as a single mass on a spring with a damper: a water tank on a slender staging, a machine on its foundation, a footbridge swaying sideways, or one storey of a building. Displaced and released, the mass swings back and forth at its natural frequency, which depends only on its mass and the stiffness of its support. Heavier means slower; stiffer means faster.
Damping removes energy each cycle, so free vibration dies away. Structural damping is small, typically a few percent of critical for concrete and steel buildings, and it barely changes the natural frequency. It is measured from a free-vibration record: the ratio of successive peaks gives the logarithmic decrement, which for light damping is about two pi times the damping ratio.
When a harmonic force drives the system, the response depends on the frequency ratio, driving frequency over natural frequency. Well below resonance the structure follows the force almost statically. Near a ratio of one the dynamic amplification can be very large, limited only by damping; at 2 percent damping the peak is about 25 times the static deflection. Well above resonance the response falls below the static value, which is why machine foundations are often tuned to be either much stiffer or much softer than the running speed.
The same model underlies earthquake engineering: a response spectrum is simply the peak response of many single-degree systems with different periods to the same ground motion. Footbridge comfort checks, machine foundation design and wind-induced vibration of chimneys all start here before moving to multi-degree models.
Always convert mass to kilograms and stiffness to N/m; then ω comes out directly in rad/s.
Natural circular frequency ω = √(k / m); period T = 2π / ω; frequency f = 1 / TDamped frequency ω_d = ω √(1 − ζ²)Log decrement δ = ln(u_n / u_(n+1)) ≈ 2π ζ (light damping)Dynamic amplification D = 1 / √[(1 − r²)² + (2 ζ r)²], r = forcing / natural frequency
Using weight in newtons instead of mass in kilograms in ω = √(k/m).
Problem. A machine and its base have a mass of 20 000 kg on supports with total stiffness 8 × 10⁶ N/m. Find ω, T and f. A free-vibration test shows successive peaks of 10.0 mm and 8.8 mm; estimate the damping ratio.
- ω = √(8 × 10⁶ / 20 000) = √400 = 20.0 rad/s.
- T = 2π / 20 = 0.314 s; f = 1 / 0.314 = 3.18 Hz.
- δ = ln(10.0 / 8.8) = ln(1.136) = 0.128.
- ζ ≈ δ / (2π) = 0.128 / 6.283 = 0.020, about 2 percent of critical.
Answer. ω = 20 rad/s, T = 0.314 s, f = 3.18 Hz, damping about 2 percent
Quick quiz
1Doubling the mass changes the period by what factor?
√2, about 1.41 times longer.
2Where is dynamic amplification largest?
Near resonance, when forcing frequency equals natural frequency.
3Does light damping change the natural frequency much?
No. At 5 percent damping the damped frequency is only about 0.13 percent below the undamped one.
References and further reading
Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.
- F. P. Beer, E. R. Johnston, Vector Mechanics for Engineers: Statics and Dynamics, McGraw Hill, latest edition.
- R. C. Hibbeler, Engineering Mechanics: Statics, Pearson, latest edition.
- S. Timoshenko, D. H. Young, Engineering Mechanics, McGraw Hill, 4th edition, 1956.
Free and official sources
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