Natural Frequency
Learn the natural frequency formula for spring-mass systems and vibrating structures, with worked examples and applications.
The Formula
ωn = √(k / m)
The natural frequency is the rate at which a system oscillates when displaced from its equilibrium position and released without any external driving force or damping. Every physical structure has one or more natural frequencies determined by its mass and stiffness properties. Understanding natural frequency is critical for engineers because when an external force matches a system's natural frequency, resonance occurs and vibration amplitudes can grow dangerously large.
The formula fn = (1/2π)√(k/m) gives the natural frequency in hertz (cycles per second) for a simple spring-mass system. The related quantity ωn = √(k/m) gives the angular natural frequency in radians per second. These two are related by ωn = 2πfn.
The formula reveals an intuitive relationship: stiffer systems (larger k) vibrate faster, while heavier systems (larger m) vibrate slower. Doubling the stiffness increases the frequency by a factor of √2 (about 41%), while doubling the mass decreases it by the same factor.
For more complex structures like beams, plates, and shafts, the natural frequency depends on geometry, material properties, and boundary conditions. A simply supported beam of length L, for example, has natural frequencies fn = (n²π/2L²)√(EI/(ρA)), where n is the mode number, E is Young's modulus, I is the moment of inertia, ρ is density, and A is the cross-sectional area.
One of the most famous examples of resonance failure is the Tacoma Narrows Bridge collapse in 1940 in the United States. Wind-induced vibrations matched the bridge's natural frequency, causing oscillations that grew until the structure failed. Modern engineering practice requires careful analysis of natural frequencies to avoid such catastrophic resonance.
Variables
| Symbol | Meaning |
|---|---|
| fn | Natural frequency (Hz, cycles per second) |
| ωn | Angular natural frequency (rad/s) |
| k | Spring constant or stiffness (N/m or lb/in) |
| m | Mass of the vibrating object (kg or slugs) |
| π | Pi, approximately 3.14159 |
Example 1
A 5 kg mass is attached to a spring with stiffness k = 2000 N/m. Find the natural frequency.
ωn = √(k/m) = √(2000/5) = √400 = 20 rad/s
fn = ωn / (2π) = 20 / (2 × 3.14159) = 20 / 6.2832
fn ≈ 3.18 Hz (the system oscillates about 3.18 times per second)
Example 2
A car suspension has a natural frequency of 1.5 Hz. If the sprung mass is 400 kg, what is the spring rate?
From fn = (1/2π)√(k/m), solve for k: k = m(2πfn)²
k = 400 × (2 × 3.14159 × 1.5)²
k = 400 × (9.4248)² = 400 × 88.83
k ≈ 35,530 N/m (about 203 lb/in). This is a typical value for passenger car springs.
When to Use It
Natural frequency analysis is essential in many branches of engineering and physics.
- Structural engineering: ensuring buildings and bridges avoid resonance from wind or earthquakes
- Automotive engineering: tuning suspension systems for ride comfort
- Aerospace: preventing flutter in aircraft wings and control surfaces
- Machinery design: avoiding resonance in rotating equipment and engine mounts
- Musical instrument design: tuning strings, membranes, and air columns
- Electronics: designing oscillator circuits and MEMS sensors