Free vibration of undamped single-degree-of-freedom systems

Why it matters

Every aeroelastic and vibration problem in an aircraft, from wing bending under gusts to flutter of a control surface, starts from the natural frequencies of the structure. The undamped single-degree-of-freedom (SDOF) oscillator is the building block: a multi-degree-of-freedom structure, once split into its normal modes, behaves like a set of independent SDOF oscillators. Knowing how stiffness and mass set the natural frequency lets you predict, and shift, resonances before a part is built.

Key ideas

Degree of freedom. The number of independent coordinates needed to describe the configuration of a system. An SDOF system needs one coordinate: the displacement x of a mass on a spring, the angle θ of a pendulum or the twist of a disc on a shaft.

Free vibration. Motion that follows an initial disturbance (an initial displacement x₀, an initial velocity v₀, or both) with no external force acting afterwards. "Undamped" means no energy is dissipated, so the motion never decays. This is an idealisation: real structures always have some damping, but light damping changes the natural frequency very little, so the undamped model gives the frequency accurately.

Equation of motion. Measuring x from the static equilibrium position, Newton's second law for a mass m on a linear spring of stiffness k gives

m·ẍ + k·x = 0

Gravity does not appear: the static spring force k·δ_st exactly balances the weight m·g, so both cancel when x is measured from equilibrium. The same equation follows from energy: kinetic energy T = ½·m·ẋ² plus potential energy U = ½·k·x² is constant, and d(T + U)/dt = 0 gives the same result.

Solution. The general solution is simple harmonic motion

x(t) = A·sin(ω_n·t + φ) = x₀·cos(ω_n·t) + (v₀/ω_n)·sin(ω_n·t)

The amplitude A and phase φ depend on the initial conditions. The natural frequency ω_n does not: it is a property of the system alone (k and m). Velocity leads displacement by 90° and acceleration is 180° out of phase with displacement, with peak values A·ω_n and A·ω_n².

Equivalent stiffness and mass. Real components are reduced to an equivalent k and m.

• Springs in parallel: k_eq = k₁ + k₂. Springs in series: 1/k_eq = 1/k₁ + 1/k₂.
• A cantilever of length L with a tip mass: k = 3·E·I/L³ for a tip load.
• A simply supported beam with a central mass: k = 48·E·I/L³.
• A shaft in torsion: k_t = G·J/L, with a disc of polar mass moment of inertia J₀, giving ω_n = √(k_t/J₀).
• If the spring's own mass m_s is not negligible, add an effective share of it: m_s/3 for a coil spring and about 0.236·m_beam for a cantilever with tip mass (Rayleigh's method, a later topic).

Static deflection method. Because k·δ_st = m·g, the natural frequency can be found from the static deflection alone: ω_n = √(g/δ_st). A stiffer support deflects less and vibrates faster.

Limits of validity. The linear model needs small displacements (spring force proportional to x; for a pendulum, sin θ ≈ θ), a rigid lumped mass and no damping. It connects directly to damped vibration (next topic), where a damping term c·ẋ is added, and to multi-DOF modal analysis, where each mode is an SDOF oscillator with modal mass and modal stiffness.

Formulas

m·ẍ + k·x = 0 — equation of motion; m mass (kg), k stiffness (N/m), x displacement from static equilibrium (m).

ω_n = √(k/m) — natural circular frequency (rad/s).

f_n = ω_n/(2π) — natural frequency (Hz). T = 1/f_n = 2π/ω_n — period (s).

x(t) = x₀·cos(ω_n·t) + (v₀/ω_n)·sin(ω_n·t) — response to initial displacement x₀ (m) and initial velocity v₀ (m/s).

A = √(x₀² + (v₀/ω_n)²) — amplitude (m); tan φ = x₀·ω_n/v₀ for the form A·sin(ω_n·t + φ).

ω_n = √(g/δ_st) — from static deflection δ_st (m), g = 9.81 m/s²; applies to a mass on a linear spring under gravity.

k = 3·E·I/L³ (cantilever, tip load), k = 48·E·I/L³ (simply supported, central load) — E Young's modulus (Pa), I second moment of area (m⁴), L length (m).

ω_n = √(k_t/J₀), k_t = G·J/L — torsional system; G shear modulus (Pa), J polar second moment of area of the shaft (m⁴), J₀ mass moment of inertia of the disc (kg·m²).

ω_n = √(g/L) — simple pendulum of length L (m), small angles only.

Worked examples

Example 1 (standard). A mass m = 2 kg on a spring k = 200 N/m is displaced x₀ = 0.05 m and released with velocity v₀ = 0.5 m/s. Find ω_n, T, the amplitude and the peak acceleration.

1. ω_n = √(k/m) = √(200/2) = √100 = 10 rad/s.
2. T = 2π/ω_n = 2π/10 = 0.628 s (f_n = 1.59 Hz).
3. A = √(x₀² + (v₀/ω_n)²) = √(0.05² + (0.5/10)²) = √(0.0025 + 0.0025) = 0.0707 m.
4. Peak acceleration = A·ω_n² = 0.0707 × 100 = 7.07 m/s².

Answer: ω_n = 10 rad/s, T = 0.628 s, A = 70.7 mm, peak acceleration 7.07 m/s².

Example 2 (GATE level). An aluminium cantilever strip (E = 70 GPa), L = 0.5 m long, 30 mm wide and 5 mm thick (bending about the thin direction), carries an instrument of mass 1.5 kg at its free end. Neglecting the strip's mass, find the natural frequency in Hz.

1. Second moment of area: I = b·h³/12 = 0.030 × 0.005³/12 = 3.125 × 10⁻¹⁰ m⁴.
2. Tip stiffness: k = 3·E·I/L³ = 3 × 70 × 10⁹ × 3.125 × 10⁻¹⁰ / 0.5³ = 65.625/0.125 = 525 N/m.
3. ω_n = √(k/m) = √(525/1.5) = √350 = 18.71 rad/s.
4. f_n = ω_n/(2π) = 18.71/6.283 = 2.98 Hz.

Answer: f_n ≈ 2.98 Hz. Check on the neglected mass: the strip weighs 2700 × 0.03 × 0.005 × 0.5 = 0.2025 kg; adding 0.236 × 0.2025 = 0.048 kg to the tip mass lowers f_n to about 2.93 Hz, a 1.6 % change.

Common mistakes

• Giving ω_n (rad/s) when the question asks for f_n (Hz), or the reverse. They differ by 2π.
• Including gravity in the equation of motion when x is measured from static equilibrium; weight only shifts the equilibrium position.
• Thinking a larger initial displacement changes the frequency of a linear system. It changes only the amplitude.
• Combining springs the wrong way: springs that share the same deflection are in parallel (stiffnesses add); springs that carry the same force are in series (compliances add).
• Using mm and N/mm mixed with kg: keep SI (m, N/m, kg) throughout, and take I in m⁴.
• Using the beam stiffness formula for the wrong boundary condition or load position.

For GATE AE

Expect direct numericals: natural frequency from k and m, from static deflection, or from a beam stiffness (cantilever or simply supported) with a lumped mass; springs in series and parallel; torsional pendulums; and how f_n scales when mass, length or stiffness change (for a cantilever f_n ∝ 1/L^1.5 at fixed tip mass). Practise writing the equation of motion by energy for systems with levers and pulleys, and reading amplitude and phase from initial conditions.

Quick check

1. What is ω_n for m = 5 kg and k = 125 N/m?
2. A mass causes a static deflection of 4 mm on its spring. Find f_n.
3. If the mass is doubled at constant stiffness, by what factor does ω_n change?
4. Two equal springs of stiffness k are placed in series. What is the equivalent stiffness?
5. Does doubling the initial displacement change the period of a linear undamped oscillator?

Answers: 1. 5 rad/s; 2. ω_n = √(9.81/0.004) = 49.5 rad/s, f_n = 7.88 Hz; 3. it falls by 1/√2 (≈ 0.707); 4. k/2; 5. No, only the amplitude doubles.