Single-DOF vibration calculator
Compute the characteristics of a spring-mass-damper system with one degree of freedom: natural angular frequency and natural frequency, Lehr damping ratio and damped frequency, plus the magnification factor and resonance position under harmonic force excitation. Enter mass, stiffness and damping, optionally excitation frequency and force amplitude – the calculator returns every value live as you type.
Single-DOF vibration calculator
Model: linear single-DOF oscillator with constant mass, linear spring and velocity-proportional (viscous) damping, small displacements. Nonlinear characteristics, dry friction, multiple degrees of freedom and absorbers are not covered. Reference calculator based on linear vibration theory (Gross/Hauger/Schröder/Wall, Technische Mechanik 3, Chapter 5).
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Formulas and fundamentals
The single-DOF oscillator is governed by the equation of motion m·x″ + d·x′ + c·x = F(t). Without damping and excitation it oscillates at the natural angular frequency ω₀ = √(c/m); the corresponding natural frequency is f₀ = ω₀/(2·π). The natural frequency depends only on the ratio of stiffness to mass: a stiffer spring raises it, a larger mass lowers it. The static deflection under a constant force F₀ is x_stat = F₀/c.
Damping is expressed through the Lehr damping ratio D = d/(2·√(c·m)) = δ/ω₀ with the decay coefficient δ = d/(2·m). The denominator d_crit = 2·√(c·m) is the critical (aperiodic) damping: for D < 1 the system oscillates with decaying amplitude (underdamped) at the damped angular frequency ω_d = ω₀·√(1−D²) and f_d = ω_d/(2·π); at D = 1 it is critically damped, and for D > 1 it creeps back to rest without oscillating. Since D² is usually small, the damped frequency lies only slightly below the undamped one.
Under harmonic force excitation F₀·cos(Ω·t) a forced vibration develops at the excitation angular frequency Ω. With the frequency ratio η = Ω/ω₀ the amplitude is x₀ = x_stat·V(η) with the magnification factor V(η) = 1/√((1−η²)² + (2·D·η)²). For D < 1/√2 ≈ 0.707, V has a maximum at η_res = √(1−2·D²) with the peak value V_max = 1/(2·D·√(1−D²)); the resonance peak therefore lies just below η = 1 rather than exactly at the natural frequency. The phase shift between excitation and response is tan φ = 2·D·η/(1−η²) and always equals 90° at η = 1.
Worked example
A 10 kg mass hangs from a spring with stiffness c = 10,000 N/m and is damped by a damper with d = 40 Ns/m. The natural angular frequency is ω₀ = √(10000/10) = 31.62 rad/s, so f₀ = 5.03 Hz. The critical damping is d_crit = 2·√(10000·10) = 632.5 Ns/m, giving the damping ratio D = 40/632.5 = 0.0632 – a lightly damped, underdamped case.
The damped natural frequency f_d = 5.03·√(1−0.0632²) = 5.02 Hz lies only marginally below it. If the system is excited exactly at its natural frequency with F₀ = 100 N (η = 1), the magnification is V = 1/(2·D) = 7.91: the static deflection x_stat = 100/10000 = 10 mm grows to about 79 mm. The resonance peak itself is at η_res = √(1−2·0.0632²) = 0.996 with V_max ≈ 7.92. Without sufficient damping the amplitude near resonance would become dangerously large.
Frequently asked questions
What is the difference between undamped and damped natural frequency?
The undamped natural frequency f₀ = ω₀/(2π) with ω₀ = √(c/m) follows from stiffness and mass alone. The damped natural frequency f_d = f₀·√(1−D²) additionally accounts for damping and is always smaller. For light damping (small D) the two are practically equal; only strong damping makes the difference noticeable.
What does the Lehr damping ratio D mean?
D = d/(2·√(c·m)) relates the actual damping to the critical damping d_crit = 2·√(c·m). D < 1 is the underdamped case (decaying oscillation), D = 1 the critically damped case (fastest return without overshoot, e.g. in measuring instruments), D > 1 the overdamped creep case (slow return without oscillation). The formulas for D, ω_d and the critical damping d = 2·√(m·c) are derived in Gross, Hauger, Schröder, Wall: Technische Mechanik 3 (Kinetik), Chapter 5 Schwingungen, Section 5.2.3, Equations (5.31), (5.36) and (5.40).
Where is resonance – at η = 1 or beside it?
For the force-excitation magnification factor the maximum is not exactly at η = 1 but at η_res = √(1−2·D²) and thus just below it. The peak value is V_max = 1/(2·D·√(1−D²)). A maximum exists only for D < 1/√2 ≈ 0.707; for stronger damping the curve falls monotonically from V(0) = 1. These relations and the magnification factor V = 1/√((1−η²)²+(2Dη)²) are given in Gross, Hauger, Schröder, Wall: Technische Mechanik 3, Section 5.3.2, Equations (5.61) and (5.62) and Figure 5.23a.
Why is the magnification exactly 1/(2D) at η = 1?
Substituting η = 1 into V(η) = 1/√((1−η²)²+(2·D·η)²) makes the first term vanish, leaving V(1) = 1/(2·D). For light damping this value becomes very large – at D = 0.05 it is ten times the static deflection. This is why the excitation frequency should keep enough distance from the natural frequency or be sufficiently damped.
How are stiffness, mass and natural frequency related?
The natural frequency rises with the square root of stiffness and falls with the square root of mass: f₀ ∝ √(c/m). To double the natural frequency you must quadruple the stiffness or quarter the mass. This is the key lever in vibration tuning, for instance to avoid resonance with known excitation frequencies.
Which systems does this calculation apply to?
Linear oscillators with one degree of freedom and constant parameters m, c and d under small displacement. Nonlinear spring characteristics, dry (Coulomb) friction, several degrees of freedom or absorbers are not covered. For two-degree-of-freedom systems or vibration absorbers the corresponding extended models are required.
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