How this instrument works
Natural frequency is the rate a system oscillates at when disturbed and then left alone — no motor, no push, just stiffness and inertia trading energy back and forth. For a single mass on a spring, Newton's second law and Hooke's law combine into one differential equation whose solution oscillates at angular frequency ω = √(k ⁄ m) radians per second. Dividing by 2π converts that into hertz, cycles per second, which is the figure most drawings and datasheets actually list.
The square root is not decoration; it decides how sensitive the result is to each input. Quadruple the spring constant k and the natural frequency only doubles, since doubling a square root needs a fourfold increase underneath it. Quadruple the mass instead and that same figure is cut exactly in half — heavier parts settle into slower cycles, which is one reason a car body sways at a couple of hertz while a small tuning fork rings in the hundreds.
The formula describes an idealized single-degree-of-freedom system: one lumped mass, one massless spring, zero friction. Real hardware has damping, which pulls the actual oscillation rate slightly below this undamped value, and most structures carry many possible mode shapes rather than one. Engineers still reach for this formula constantly because it gives the lowest, usually most troublesome, mode first — the one a designer most needs to keep clear of whatever else is driving the structure nearby.
- Enter the stiffness in the Spring constant k, N/m field — newtons of restoring force per metre of stretch or compression.
- Enter the Mass, in kilograms or grams; the instrument converts automatically.
- Read the result in the Natural frequency field, reported in Hz.
- Compare that figure against any nearby driving frequency — a motor's RPM divided by 60, or a footstep cadence — converted to Hz first.
- Change either input and watch which one moves the frequency more: stiffness by a square root, mass by an inverse square root.
Worked example — a 2 kg mass on a 1,000 N/m spring
Take a 2 kg instrument enclosure mounted on an isolator spring rated at 1,000 N/m, an ordinary vibration-isolation setup for benchtop equipment. First the angular frequency: ω = √(k ⁄ m) = √(1000 ⁄ 2) = √500 = 22.3607 rad/s — how fast the phase angle advances once the mass is tapped and left to ring on its own.
Convert to hertz by dividing by 2π: f = 22.3607 ⁄ 6.28319 = 3.55881 Hz, about 3.56 cycles every second. A designer would now check that nothing nearby drives at close to 3.56 Hz — a fan, a compressor, footfall on an adjacent floor — because forcing a mount at its own natural frequency is the resonance condition that widens vibration amplitude far beyond what the same force causes off-resonance, the textbook mechanism blamed for the Tacoma Narrows Bridge's dramatic 1940 collapse.
Questions
What is the difference between natural frequency and resonant frequency?
For an undamped system the two are identical, both equal to √(k ⁄ m) ⁄ 2π. Add damping and they split slightly: natural frequency stays fixed by k and m alone, while the resonant frequency — the drive rate that produces the largest response — drops a little below it, by an amount set by the damping ratio. Lightly damped hardware, most metal structures included, keeps the two close enough that engineers often use the terms loosely.
Does a stronger push change the natural frequency?
No. Natural frequency depends only on stiffness k and mass m, never on how hard the system is struck, how far it is displaced, or how long it keeps ringing. A gently plucked guitar string and a hard-plucked one settle on the same pitch — only the loudness and how long the note lasts differ, not the underlying rate, as long as the displacement stays small enough for the spring to behave linearly.
Why does doubling the mass not simply halve the natural frequency?
Because the formula runs through a square root: f is proportional to 1 ⁄ √m, not to 1 ⁄ m directly. Doubling the mass divides frequency by √2, about 0.707 times, not by 2 — for the 2 kg, 1,000 N/m example here that is a drop from 3.559 Hz to 2.516 Hz at 4 kg. A full halving needs four times the mass, not two.
Can this formula be used for a real building or bridge?
Only as a first estimate, and only for the lowest mode. Real structures are distributed-mass systems with many natural frequencies, not one lumped mass on one spring, and engineers turn to finite-element models for precise values. The single-degree-of-freedom formula still earns its keep for quick early estimates and for explaining why heavy, soft structures such as suspension bridges vibrate slowly while light, stiff parts such as circuit boards vibrate fast.
What units does this calculator expect?
Spring constant k in newtons per metre and mass in kilograms, with grams also accepted and converted automatically. The result comes back in hertz, cycles per second — multiply by 2π afterward if a downstream calculation needs angular frequency in radians per second instead.