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

Free undamped vibration of a single-degree-of-freedom system: equation of motion, natural frequency by Newton, energy and static-deflection methods, equivalent stiffness of springs and beams, and the effect of spring mass.

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Why it matters

Every machine, mount, shaft, robot arm and sensor bracket has natural frequencies. If an excitation (an unbalanced rotor, gear mesh, motor ripple or road input) lands near one, the response becomes large. The single-degree-of-freedom (SDOF) model, one mass on one spring, is how engineers estimate a natural frequency on the back of an envelope, and it underlies the damped, forced and isolation topics that follow.

Key ideas

Degree of freedom. The number of independent coordinates needed to describe the motion. An SDOF system is described by a single coordinate, x (translation) or θ (rotation): a mass on a spring, a pendulum, a disc on a torsion shaft, a beam with a dominant end mass.

Free undamped vibration. After an initial disturbance (displacement or velocity), the system oscillates with no external force and no energy loss. The motion is simple harmonic at the natural frequency, and the amplitude stays constant. It is an idealisation: real systems have some damping, but for light damping the free-vibration frequency is almost the same.

Equation of motion. For a mass m on a spring of stiffness k, measuring x from the static equilibrium position, Newton's law gives m·ẍ + k·x = 0. Gravity is balanced by the static spring force (k·δ_st = m·g) and drops out, so a vertical spring–mass system has the same natural frequency as a horizontal one.

Ways to get ω_n.

  • Newton / D'Alembert: write the equation and read off ω_n² = (coefficient of x)/(coefficient of ẍ).
  • Energy method: with no losses, T + U = constant, so d(T + U)/dt = 0 gives the same equation. Convenient for systems with levers, pulleys and rolling bodies.
  • Rayleigh's method: T_max = U_max; useful for including the mass of a spring or beam.
  • Static deflection: ω_n = √(g/δ_st), handy when the deflection under the weight is known or measured.

Equivalent stiffness.

  • Springs in parallel (sharing the same displacement): stiffnesses add.
  • Springs in series (sharing the same force): compliances add.
  • Beams as springs (load P at a point, k = P/δ): cantilever with end load 3EI/L³; simply supported with central load 48EI/L³; fixed–fixed with central load 192EI/L³.
  • Shafts in torsion: k_t = G·J/L.

Effect of spring or beam mass. By Rayleigh's method a uniform helical spring adds one third of its mass to the end mass (m + m_s/3); a uniform cantilever adds 33/140 of its mass to an end mass; a simply supported beam adds 17/35 of its mass to a central mass.

Pendulums. Simple pendulum (small angles) ω_n = √(g/L). Compound pendulum pivoted at a distance h from its centre of mass ω_n = √(m·g·h / I_O) with I_O = m(k_G² + h²).

Formulas

m·ẍ + k·x = 0 (equation of motion) x(t) = X·sin(ω_n·t + φ), with X = √(x₀² + (v₀/ω_n)²) and tan φ = x₀·ω_n / v₀

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

  • k = stiffness (N/m), m = mass (kg)

ω_n = √(g/δ_st) ; f_n ≈ 15.76 / √(δ_st in mm) Hz (or 0.4985 / √(δ_st in m) Hz)

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

k_cantilever = 3EI/L³ ; k_SS,centre = 48EI/L³ ; k_fixed,centre = 192EI/L³ (E in Pa, I in m⁴, L in m)

ω_n = √(k_t / J) (torsional; k_t = GJ_p/L in N·m/rad, J = mass moment of inertia in kg·m²)

ω_n = √(k / (m + m_s/3)) (spring of mass m_s)

ω_n = √(g/L) (simple pendulum) ; ω_n = √(m·g·h / I_O) (compound pendulum)

Worked examples

Example 1 (standard): machine on springs. Given: a 200 kg machine on four identical springs, each 50 kN/m, sharing the load. It is pushed down 5 mm and released with a downward velocity of 0.1 m/s.

  1. Parallel springs: k = 4 × 50 = 200 kN/m.
  2. ω_n = √(200 000 / 200) = 31.62 rad/s → f_n = 31.62 / 2π = 5.03 Hz.
  3. Static deflection δ_st = m·g/k = 200 × 9.81 / 200 000 = 9.81 mm (check: √(9.81 / 0.00981) = 31.6 rad/s).
  4. Amplitude X = √(0.005² + (0.1 / 31.62)²) = √(25 × 10⁻⁶ + 10.0 × 10⁻⁶) = 5.92 mm. Answer: f_n ≈ 5.03 Hz, amplitude ≈ 5.9 mm.

Example 2 (GATE level): cantilever with an end mass, including beam mass. Given: steel cantilever, L = 0.5 m, rectangular section 40 mm wide × 10 mm deep, E = 200 GPa, density 7850 kg/m³; a 10 kg mass at the free end.

  1. I = b·h³/12 = 0.04 × 0.01³ / 12 = 3.333 × 10⁻⁹ m⁴
  2. k = 3EI/L³ = 3 × 200 × 10⁹ × 3.333 × 10⁻⁹ / 0.125 = 16 000 N/m
  3. Ignoring beam mass: ω_n = √(16 000 / 10) = 40.0 rad/s → 6.37 Hz.
  4. Beam mass = 7850 × 0.04 × 0.01 × 0.5 = 1.57 kg; effective share = (33/140) × 1.57 = 0.370 kg.
  5. ω_n = √(16 000 / 10.37) = 39.28 rad/s → f_n = 6.25 Hz. Answer: f_n ≈ 6.25 Hz (6.37 Hz if the beam mass is ignored, about 2% high).

Common mistakes

  • Reporting ω_n in rad/s when the question asks for f_n in Hz, or the reverse.
  • Adding stiffnesses of springs in series.
  • Using total deflection from the free length instead of the static deflection under the weight.
  • Thinking gravity changes the natural frequency of a vertical spring–mass system.
  • Using a beam's deflection formula with load in kg instead of N, or I in mm⁴ with E in Pa.
  • Forgetting the 1/3 (spring) or 33/140 (cantilever) mass correction when the question says to include it.

For GATE ME

Typical items: natural frequency from k and m or from static deflection; equivalent stiffness of combined springs, levers and beams; energy-method problems with pulleys, rolling cylinders or rigid bars on springs; compound pendulums; effect of spring mass. Practise writing T and U for a system and differentiating, and the standard beam stiffness results.

Quick check

  1. m = 4 kg, k = 400 N/m. Find f_n.
  2. Two 1000 N/m springs in series carry a 1 kg mass. ω_n?
  3. A machine deflects its mounts by 1 mm. Approximate f_n?
  4. Does a vertical spring–mass system have a different ω_n from a horizontal one?
  5. A spring of mass 0.6 kg carries 2 kg. Effective mass for ω_n?

Answers: 1. ω_n = 10 rad/s, f_n = 1.59 Hz. 2. k = 500 N/m, ω_n = 22.4 rad/s. 3. 15.76 Hz. 4. No. 5. 2.2 kg.

Interview questions

All Theory of Machines and Vibrations interview questions

Try answering each one aloud before you open it.

  1. 1.What is a single-degree-of-freedom system in the context of vibrations?Concept

    A single-degree-of-freedom system is one whose configuration is fixed by one independent coordinate, a displacement x or a rotation θ. Examples are a mass on a spring, a disc on a torsion shaft, a pendulum, or a beam whose dominant end mass moves in one direction. The general model has a mass (or inertia), a spring and a damper; in free undamped vibration the damper and any external force are absent, and the system has exactly one natural frequency.

  2. 2.Explain the concept of free undamped vibration in a single-degree-of-freedom system.Concept

    Free undamped vibration occurs when a system oscillates without any external force acting on it and without any energy loss due to damping. In a single-degree-of-freedom system, this means the system will continue to oscillate indefinitely at its natural frequency once it is set into motion. The motion is purely sinusoidal and is determined by the system's mass and stiffness.

  3. 3.What is the natural frequency of a single-degree-of-freedom system, and how is it calculated?Concept

    The natural frequency of a single-degree-of-freedom system is the frequency at which the system naturally oscillates when disturbed from its equilibrium position and then left to vibrate freely. It is calculated using the formula ω_n = √(k/m), where ω_n is the natural frequency in radians per second, k is the stiffness of the system in N/m, and m is the mass in kg.

  4. 4.Why is it important to know the natural frequency of a mechanical system?Application

    Knowing the natural frequency of a mechanical system is crucial because it helps in avoiding resonance, which can lead to excessive vibrations and potential failure. If a system is subjected to external forces at its natural frequency, it can experience large amplitude oscillations, leading to damage or failure. Designing systems to operate away from their natural frequencies ensures stability and longevity.

  5. 5.What happens if a single-degree-of-freedom system is subjected to a force at its natural frequency?Application

    If a single-degree-of-freedom system is subjected to a force at its natural frequency, it will experience resonance. This means the amplitude of the system's oscillations will increase significantly, potentially leading to structural damage or failure. This is why it is critical to design systems to avoid operating at or near their natural frequencies.

  6. 6.How does the mass of a system affect its natural frequency?Application

    The mass of a system is inversely related to its natural frequency. As the mass increases, the natural frequency decreases, assuming the stiffness remains constant. This is because a larger mass requires more force to achieve the same acceleration, resulting in slower oscillations.

  7. 7.Describe how the stiffness of a system influences its natural frequency.Application

    Natural frequency rises with stiffness, but as its square root: ω_n = √(k/m), so doubling k raises ω_n by √2 (about 41%), and quadrupling k is needed to double it. A stiffer spring gives a larger restoring force per unit displacement, so the mass is accelerated back faster. This is why stiffening a bracket is an effective way to push a resonance above the operating frequency, while adding mass pulls it down.

  8. 8.Calculate the natural frequency of a system with a mass of 5 kg and a stiffness of 200 N/m.Numerical

    To calculate the natural frequency, use the formula ω_n = √(k/m). Here, k = 200 N/m and m = 5 kg.

    1. ω_n = √(200/5)
    2. ω_n = √40
    3. ω_n ≈ 6.32 rad/s
  9. 9.A system with a natural frequency of 10 rad/s has a mass of 2 kg. What is the stiffness of the system?Numerical

    To find the stiffness, use the formula ω_n = √(k/m). Rearrange to find k: k = ω_n² * m. Here, ω_n = 10 rad/s and m = 2 kg.

    1. k = 10² * 2
    2. k = 100 * 2
    3. k = 200 N/m
  10. 10.Why is damping often neglected when calculating natural frequency?Concept

    In most structures and machines damping is light, with a damping ratio of a few per cent, and the damped natural frequency ω_d = ω_n√(1 − ζ²) differs from ω_n by well under 1%. Neglecting damping gives a simple equation, m·ẍ + k·x = 0, whose solution is pure simple harmonic motion at ω_n = √(k/m). Damping must be included when you need the decay of free vibration or the amplitude at resonance, because there it controls the answer.

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