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Natural Frequency Calculator

This natural frequency calculator covers three classic oscillating systems: spring-mass (f = (1/2π)·√(k/m)), pendulum (f = (1/2π)·√(g/L)), and cantilever beam. A live oscillation animation renders the chosen system moving at the calculated frequency with adjustable playback speed.

Input

N/m
kg
Presets

Result

Natural frequency f
Period T
Angular frequency ω
Formulas
Spring-mass: f = (1 / 2π) · √(k / m)
Pendulum: f = (1 / 2π) · √(g / L) (small-angle approximation)
Cantilever: f₁ = (β₁L)² · √(EI / (ρA·L⁴)) / (2π), β₁·L = 1.8751
Period T = 1 / f, Angular ω = 2π · f
Oscillation animation

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About Natural Frequency

Every elastic system has at least one frequency it "wants" to oscillate at when displaced and released — the natural frequency (also called the resonant frequency or eigenfrequency in structural analysis). It depends only on the system's geometry, mass distribution, and elasticity — not on how hard you push it. Push a swing once: it returns at its own frequency regardless of how big the initial push was. This calculator covers the three textbook cases that appear in every intro physics and engineering vibrations course. The same principles underlie modal analysis, vibration isolation design, and structural dynamics in civil and mechanical engineering.

Spring-mass: f = (1 / 2π) · √(k / m)

A mass m on a spring of stiffness k oscillates at f = √(k/m) / (2π). Stiffer spring → higher frequency. Heavier mass → lower frequency. To double the frequency you must either quadruple k or quarter m — frequency depends on the square root. A car suspension at k = 30 kN/m and m = 350 kg per wheel resonates at ~1.5 Hz, comfortably below the 4–8 Hz vertical-motion sensitivity of the human body.

Pendulum: f = (1 / 2π) · √(g / L)

For small swing angles, a simple pendulum's frequency depends only on its length and the local gravity — not on the bob's mass or the swing amplitude. A 1 m pendulum on Earth (g = 9.81 m/s²) has T ≈ 2.01 s, which is why grandfather clocks have a 1 m pendulum that "ticks" each second (each tick is half a period). The same 1 m pendulum on the Moon (g = 1.625) would have T ≈ 4.93 s — visibly slower. Above large amplitudes (> ~15°), the small-angle approximation breaks down and the actual period is slightly longer.

Cantilever beam: f₁ = (β₁L)² · √(EI / (ρA·L⁴)) / (2π)

A beam fixed at one end and free at the other. The fundamental coefficient β₁·L = 1.8751 comes from solving the transcendental equation cos(βL)·cosh(βL) = −1. Doubling the length drops the frequency by a factor of 4 (L⁴ in the denominator). Stiffer material (higher E) or larger cross-section moment (I) raises it; denser material (higher ρ) lowers it.

Why "natural" frequency?

"Natural" because it's intrinsic to the system. A different external force at the natural frequency produces resonance — amplified vibration. The amplitude growth is limited only by damping (covered in the Mechanical Resonance Frequency Tool, which adds the damping ratio ζ and Q-factor to the calculations here). Resonance disasters like Tacoma Narrows happen when a periodic input (wind vortices) happens to match a structure's natural frequency and damping is insufficient. In acoustics, the same concept explains why hollow cavities amplify certain pitches — explored further in the Acoustic Resonance Frequency Calculator.

Calculating natural frequency from static deflection

A useful shortcut for spring-mass systems: if you know the static deflection δst (how far a mass sags under gravity), then f = (1/2π)·√(g/δst). This is equivalent to the spring formula because δst = mg/k, so k/m = g/δst. Engineers often measure static deflection with a ruler to estimate natural frequency without knowing k directly — a handy field technique for isolator pads and machinery mounts.

Frequently Asked Questions

Does the mass of a pendulum bob affect its frequency?
No — surprisingly, the mass cancels out of the equation. A heavy bob and a light bob on the same string oscillate at the same frequency (for small angles). This is because gravity scales the restoring force in exact proportion to the mass that resists motion — heavier mass = stronger restoring force, exactly canceling. Galileo noticed this watching chandeliers swing in church and tested it formally around 1602.
Why does doubling the spring constant only multiply the frequency by √2?
Because frequency depends on √(k/m) — the square root. To double f, you'd need k to quadruple. This is why "stiffer" springs raise pitch slowly: piano string tension might need to quadruple to raise a note by an octave (or you could shorten the string by half, which is much easier — hence why low piano notes use longer strings, not heavier ones).
What's the difference between f, ω, and T?
f = frequency in Hz (cycles per second). T = period in seconds (time per cycle). They're reciprocals: T = 1/f. ω = angular frequency in rad/s, useful for the math because differentiating sin(ωt) is clean. ω = 2πf. Physicists often use ω; engineers and musicians prefer f.
When does the small-angle pendulum approximation break down?
The exact pendulum equation is θ̈ + (g/L)·sin(θ) = 0, but sin(θ) ≈ θ is only accurate within ~1% up to about 15°. At 30° the period is 1.7% longer than predicted; at 60° it's 7% longer. For amplitudes beyond a few degrees, you need elliptic integrals or numerical integration. Real grandfather clocks swing only ~3° to stay accurate.
Why is the cantilever fundamental coefficient 1.8751 instead of something nicer?
The boundary conditions (fixed at one end, free at the other) lead to a transcendental equation 1 + cos(βL)·cosh(βL) = 0. The first root is at βL ≈ 1.8751… — a messy real number with no closed form. Higher modes are at 4.6941, 7.8548, etc. Simply-supported beams have the much nicer βL = nπ exactly, because their boundary conditions match perfectly with sine waves.
Why does the animation slow down at high frequencies?
For systems where f is high (say 20+ Hz), the visual oscillation becomes too fast for the eye to follow — it would just look like a blur. The Speed dropdown lets you select 0.1× or 0.05× slow-motion so the motion stays visible. The displayed frequency value is always the actual natural frequency, not the slowed-down playback rate.
How does damping affect the natural frequency?
Damping slightly lowers the frequency at which a system actually oscillates after being released — the damped natural frequency is fd = fn·√(1 − ζ²), where ζ is the damping ratio. For lightly damped structures (ζ < 0.1, common in metals and concrete), the difference is less than 0.5% and is usually negligible. For heavily damped systems like rubber isolators (ζ ≈ 0.5), fd is about 13% lower than fn. This tool calculates undamped natural frequency fn; use the Mechanical Resonance Frequency Tool to add damping effects.
Can I calculate natural frequency from static deflection instead of spring constant?
Yes — if you know the static deflection δst (how much a mass compresses a spring or isolator under its own weight), use f = (1/2π)·√(g/δst). For example, a machine mount that deflects 10 mm under load has fn ≈ (1/2π)·√(9.81/0.01) ≈ 5 Hz. This is mathematically identical to the spring-mass formula because k/m = g/δst, so you don't need to know k or m separately — just the deflection. Enter k = g·m and m = 1 in the Spring-Mass panel, adjusting the ratio to match your measured δst.
What happens to a structure when it is driven at its natural frequency?
Resonance occurs — the driving force adds energy each cycle in phase with the motion, causing amplitude to grow. Without sufficient damping, oscillations can become destructive. Real-world examples include the Tacoma Narrows Bridge collapse (1940), vibration fatigue in aircraft components, and washing machines "walking" across a floor during a spin cycle. Engineers deliberately design resonant frequencies to be far from expected excitation frequencies — a technique called frequency separation. The Harmonic Series Calculator can help identify which harmonics of a driving force might coincide with a structure's natural frequency.