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

Why it matters

Every elastic system has natural frequencies, and if an engine order, a road input or an unbalanced wheel excites one of them, the response can become large. The single-degree-of-freedom (SDOF) model, a mass on a spring, is how engineers first estimate a vehicle's ride frequency (about 1 to 1.5 Hz for comfort), an engine mount's natural frequency or a shaft's torsional frequency. Everything in the following vibration topics builds on it.

Key ideas

Degrees of freedom. The number of independent coordinates needed to describe the motion. A mass on a spring moving vertically, a pendulum swinging in a plane and a disc twisting on a shaft each need one coordinate (x or θ), so each is an SDOF system. Real structures have many degrees of freedom, but often one mode dominates and the SDOF model captures it.

Free undamped vibration. After an initial disturbance, the system oscillates with no external force and no energy loss. The motion is simple harmonic at the natural frequency, the amplitude stays constant, and kinetic and potential energy are exchanged with a constant total.

Equation of motion. For a mass m on a spring of stiffness k, measuring x from the static equilibrium position: m·ẍ + k·x = 0. Gravity only shifts the equilibrium position (static deflection δ = m·g/k); it does not change the frequency. The solution is x = X·sin(ω_n·t + φ), with X and φ fixed by the initial displacement and velocity.

Ways to find ω_n.

• Newton's second law (or d'Alembert) on a free-body diagram, giving m·ẍ + k·x = 0.
• Energy method: total energy T + U is constant, so d(T + U)/dt = 0. Useful for systems with several connected parts (levers, pulleys, rolling bodies).
• Rayleigh's method: maximum kinetic energy = maximum potential energy, with the velocity amplitude ω_n·X.
• Static deflection: for a system whose spring deflects under the weight of the mass, ω_n = √(g/δ). Larger static sag means lower frequency, which is why soft suspensions give a comfortable ride.

Equivalent springs.

• Parallel (both deflect equally): k_eq = k₁ + k₂.
• Series (both carry the same force): 1/k_eq = 1/k₁ + 1/k₂.
• Beams at the load point: cantilever with end load 3EI/L³; simply supported with central load 48EI/L³; fixed-fixed with central load 192EI/L³.
• Shaft in torsion: k_t = G·J_p / L, where J_p = π·d⁴/32.

Other SDOF systems.

• Simple pendulum (small angles): ω_n = √(g/L).
• Compound pendulum: ω_n = √(g·h / (k_G² + h²)), where h is the distance from pivot to centre of gravity.
• Torsional system (disc of mass moment of inertia J on a shaft): J·θ̈ + k_t·θ = 0, ω_n = √(k_t/J).
• Effect of spring mass: adding one-third of the spring's mass to the attached mass gives a good estimate (Rayleigh).

What changes the frequency. ω_n is proportional to √k and inversely proportional to √m: doubling the stiffness raises it by √2 (41%); doubling the mass lowers it to 1/√2 (71%).

Formulas

m·ẍ + k·x = 0

• m: mass (kg); k: stiffness (N/m); x: displacement from static equilibrium (m).

ω_n = √(k/m) and f_n = ω_n / (2π) and T = 1 / f_n

• ω_n: natural circular frequency (rad/s); f_n: natural frequency (Hz); T: period (s).

ω_n = √(g/δ)

• δ: static deflection under the weight of the mass (m); g = 9.81 m/s². Valid when the spring is loaded by the weight of the vibrating mass and the system is linear.

x(t) = x₀·cos(ω_n·t) + (v₀/ω_n)·sin(ω_n·t) and X = √(x₀² + (v₀/ω_n)²)

• x₀: initial displacement (m); v₀: initial velocity (m/s); X: amplitude (m).

k_eq = k₁ + k₂ (parallel) and 1/k_eq = 1/k₁ + 1/k₂ (series)

ω_n = √(k_t / J) and k_t = G·π·d⁴ / (32·L)

• k_t: torsional stiffness (N·m/rad); J: mass moment of inertia of the disc (kg·m²); G: shear modulus (Pa); d: shaft diameter (m); L: shaft length (m).

ω_n = √(k / (m + m_s/3))

• m_s: mass of the spring (kg).

Worked examples

Example 1 (standard: vehicle ride frequency). One corner of a car body carries 300 kg on a spring of stiffness 30 kN/m (tyre and damper neglected). Find the static deflection, the natural frequency and the period.

1. δ = m·g / k = 300 × 9.81 / 30 000 = 0.0981 m = 98.1 mm.
2. ω_n = √(k/m) = √(30 000 / 300) = √100 = 10.0 rad/s.
3. Check: ω_n = √(g/δ) = √(9.81 / 0.0981) = 10.0 rad/s.
4. f_n = 10.0 / (2π) = 1.59 Hz; period T = 1 / 1.59 = 0.628 s.

Example 2 (GATE level: rigid bar with a spring). A uniform rigid bar of mass 3 kg and length 1 m is pinned at one end and held horizontal by a vertical spring of stiffness 1000 N/m at its free end. Find the natural frequency of small oscillations. What is it if the spring is moved to the middle of the bar?

1. Measure θ from the static equilibrium position, so the weight and the static spring force cancel.
2. Inertia about the pivot: J_O = m·L²/3 = 3 × 1² / 3 = 1.0 kg·m².
3. Spring at the free end: a rotation θ stretches it by L·θ, giving a restoring moment k·L²·θ.
4. Equation: J_O·θ̈ + k·L²·θ = 0, so ω_n = √(k·L² / J_O) = √(1000 × 1 / 1.0) = 31.62 rad/s, f_n = 5.03 Hz.
5. Spring at mid-span: restoring moment k·(L/2)²·θ = 250·θ. ω_n = √(250 / 1.0) = 15.81 rad/s, f_n = 2.52 Hz: halving the lever arm halves the frequency.

Example 3 (springs in series). A 10 kg mass hangs from two springs in series of 4 kN/m and 6 kN/m. k_eq = 4000 × 6000 / 10 000 = 2400 N/m. ω_n = √(2400/10) = 15.49 rad/s, f_n = 2.47 Hz. In parallel the same springs would give k = 10 000 N/m and f_n = 5.03 Hz.

Common mistakes

• Giving ω_n in rad/s when the question asks for f_n in Hz, or the reverse. f_n = ω_n / 2π.
• Including gravity in the equation of motion when x is measured from static equilibrium; it cancels.
• Adding stiffnesses in series. For series springs add the flexibilities (1/k).
• Using the mass moment of inertia about the centre of mass instead of about the pivot for a pinned bar.
• Saying ω_n is proportional to k or inversely proportional to m. It goes with √(k/m).
• Using δ in millimetres in √(g/δ).

For GATE ME

Expect natural frequency of mass-spring systems with springs in series or parallel, of levers and bars with springs (energy method), of pendulums and torsional discs, and from static deflection. Many questions hide the trick in the geometry (spring at a different point from the mass, a pulley, rolling). Practise writing T and U for a system and differentiating, which handles any of these quickly.

Quick check

1. f_n for m = 5 kg, k = 500 N/m?
2. Static deflection 25 mm. f_n?
3. Springs of 2 kN/m and 3 kN/m in series: k_eq?
4. If the mass is made four times larger, what happens to f_n?
5. Does the amplitude of free undamped vibration change with time?

Answers: 1. √100 / 2π = 1.59 Hz. 2. √(9.81/0.025) / 2π = 19.81 / 6.283 = 3.15 Hz. 3. 1.2 kN/m. 4. It halves. 5. No, it stays constant.