Natural Frequency Calculator
Calculate natural frequency for three classic oscillating systems: spring-mass (f = (1/2π)·√(k/m)), pendulum (f = (1/2π)·√(g/L)), and cantilever beam. A live SVG animation oscillates the chosen system at the calculated frequency with adjustable playback speed.
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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. 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.
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). Resonance disasters like Tacoma Narrows happen when a periodic input (wind vortices) happens to match a structure's natural frequency and damping is insufficient.
Frequently Asked Questions
Does the mass of a pendulum bob affect its frequency?
Why does doubling the spring constant only multiply the frequency by √2?
√(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?
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?
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.