Free & Forced Vibrations
Natural Frequency, Damping, Resonance & Transmissibility — Complete Guide
Last Updated: March 2026
Key Takeaways 📌
- Natural frequency: ωn = √(k/m) rad/s, fn = (1/2π)√(k/m) Hz — the frequency at which a system vibrates freely.
- Damping ratio: ξ = c/(2√(km)) — determines how quickly vibrations decay.
- Underdamped (ξ < 1): oscillates and decays. Critically damped (ξ = 1): fastest return without oscillation. Overdamped (ξ > 1): slow return, no oscillation.
- Resonance: When forcing frequency ≈ natural frequency → amplitude becomes very large (dangerous).
- Forced vibration: Steady-state amplitude depends on frequency ratio (ω/ωn) and damping ratio.
- Vibrations are the second-most tested TOM topic in GATE ME (2–4 marks).
1. Vibration Basics
A vibration is a repetitive back-and-forth motion of a body about an equilibrium position. Every vibrating system has three essential elements:
| Element | Role | Parameter |
|---|---|---|
| Mass (m) | Stores kinetic energy | kg |
| Spring (k) | Stores potential energy, provides restoring force | N/m (stiffness) |
| Damper (c) | Dissipates energy (friction) | N·s/m (damping coefficient) |
| Classification | Description |
|---|---|
| Free vibration | System vibrates after initial disturbance — no external force during motion |
| Forced vibration | System vibrates under a continuous external periodic force |
| Undamped | No energy dissipation — vibration continues forever (theoretical) |
| Damped | Energy dissipated by friction — vibration decays over time |
2. Free Undamped Vibrations
The simplest vibrating system: a mass attached to a spring with no friction. The equation of motion is:
mẍ + kx = 0
Solution: x(t) = A sin(ωnt + φ)
Natural Frequency
ωn = √(k/m) (rad/s)
fn = ωn/(2π) = (1/2π)√(k/m) (Hz)
Tn = 1/fn = 2π√(m/k) (seconds — time period)
The natural frequency depends only on the system’s stiffness and mass — not on the initial displacement or velocity. Stiffer springs and lighter masses vibrate faster.
Static Deflection Method
If the static deflection under the weight is δst = mg/k, then:
ωn = √(g/δst)
fn = (1/2π)√(g/δst)
This is useful when k is unknown but static deflection can be measured.
3. Free Damped Vibrations
Adding a viscous damper (dashpot) to the spring-mass system:
mẍ + cẋ + kx = 0
Damping Ratio
ξ = c / cc = c / (2√(km)) = c / (2mωn)
Where cc = 2√(km) = critical damping coefficient
| Condition | ξ Value | Response | Example |
|---|---|---|---|
| Underdamped | 0 < ξ < 1 | Oscillates with exponentially decaying amplitude | Car suspension, guitar string |
| Critically damped | ξ = 1 | Returns to equilibrium fastest without oscillation | Door closer (ideal setting) |
| Overdamped | ξ > 1 | Returns to equilibrium slowly without oscillation | Heavy-duty shock absorber |
Damped Natural Frequency (Underdamped)
ωd = ωn√(1 − ξ²)
The damped frequency is always lower than the undamped natural frequency.
Logarithmic Decrement
δ = ln(xn/xn+1) = 2πξ/√(1 − ξ²)
Measures the rate of decay — ratio of successive peak amplitudes.
4. Forced Vibrations & Resonance
When a periodic external force F₀sinωt acts on a damped spring-mass system:
mẍ + cẋ + kx = F₀sinωt
The steady-state response is:
Amplitude of Forced Vibration
X = (F₀/k) / √[(1 − r²)² + (2ξr)²]
Where r = ω/ωn = frequency ratio
Magnification factor (MF) = X/(F₀/k) = 1/√[(1−r²)² + (2ξr)²]
Phase Lag
φ = tan⁻¹[2ξr / (1 − r²)]
Resonance occurs when r ≈ 1 (ω ≈ ωn). The magnification factor becomes very large (limited only by damping). At exact resonance with no damping, amplitude → ∞ (theoretical). In practice, damping limits the peak amplitude to MFmax = 1/(2ξ).
| Frequency Ratio r | Behaviour |
|---|---|
| r << 1 (ω << ωn) | MF ≈ 1 — system follows the force quasi-statically |
| r ≈ 1 (ω ≈ ωn) | MF >> 1 — resonance, large amplitudes, dangerous |
| r >> 1 (ω >> ωn) | MF ≈ 0 — system cannot follow high-frequency force |
Transmissibility (Force Transmitted to Foundation)
TR = √[1 + (2ξr)²] / √[(1−r²)² + (2ξr)²]
For vibration isolation: TR < 1 when r > √2 (operating speed > √2 × natural frequency).
5. Springs — Series & Parallel
Springs in Parallel
keq = k₁ + k₂ + k₃ + …
Same displacement, forces add. Stiffer overall.
Springs in Series
1/keq = 1/k₁ + 1/k₂ + 1/k₃ + …
Same force, displacements add. More flexible overall.
6. Worked Numerical Examples
Example 1: Natural Frequency
Problem: A 10 kg mass is attached to a spring of stiffness 4000 N/m. Find the natural frequency.
ωn = √(k/m) = √(4000/10) = √400 = 20 rad/s
fn = 20/(2π) = 3.18 Hz
Tn = 1/3.18 = 0.314 s
Example 2: Damping Ratio
Problem: m = 5 kg, k = 2000 N/m, c = 40 N·s/m. Find the damping ratio and state the type of damping.
cc = 2√(km) = 2√(2000 × 5) = 2√10,000 = 2 × 100 = 200 N·s/m
ξ = c/cc = 40/200 = 0.2
ξ < 1 → Underdamped
ωd = ωn√(1−ξ²) = 20 × √(1−0.04) = 20 × 0.98 = 19.6 rad/s
Example 3: Resonance — Magnification Factor
Problem: A system with ωn = 50 rad/s and ξ = 0.1 is subjected to a forcing frequency ω = 48 rad/s. Find the magnification factor.
r = ω/ωn = 48/50 = 0.96
MF = 1/√[(1−0.96²)² + (2×0.1×0.96)²]
= 1/√[(1−0.9216)² + (0.192)²] = 1/√[(0.0784)² + 0.0369]
= 1/√[0.00615 + 0.0369] = 1/√0.04305 = 1/0.2075
MF = 4.82
Near resonance — amplitude is 4.82 times the static deflection.
7. Common Mistakes Students Make
- Confusing ωn (rad/s) with fn (Hz): ωn = 2πfn. Many formulas use ωn. If a problem gives frequency in Hz, convert before substituting.
- Using damped frequency in undamped formulas: ωd = ωn√(1−ξ²) is the actual oscillation frequency of a damped system. ωn is used in MF and transmissibility formulas.
- Forgetting that resonance peak shifts with damping: The peak MF does not occur exactly at r = 1 for damped systems — it shifts slightly below. For small ξ, the difference is negligible.
- Mixing up series and parallel spring formulas: Parallel springs add stiffnesses directly. Series springs add reciprocals. This is the OPPOSITE of resistors in circuits, which trips students who have studied electrical engineering.
8. Frequently Asked Questions
What is natural frequency?
Natural frequency is the frequency at which a system vibrates when disturbed and allowed to oscillate freely without any external force. It depends only on the system’s stiffness (k) and mass (m): ωn = √(k/m). Every structure, machine, and component has a natural frequency, and operating near it causes resonance — dangerous amplification of vibrations.
What is resonance and why is it dangerous?
Resonance occurs when the frequency of an external periodic force matches the natural frequency of a system (r = ω/ωn ≈ 1). At resonance, even a small force produces very large amplitude vibrations because energy is continuously added in phase with the system’s motion. This can cause catastrophic failure — the Tacoma Narrows Bridge collapse (1940) is the most famous example.