From a Calculated ωn to an Isolator You Can Order
Vibration isolation design begins with one square root. Idealise the machine and its mounts as a single mass on a spring, and the undamped natural frequency is ωn = √(k/m), with stiffness in newtons per metre and mass in kilograms. That expression hands back radians per second, because it comes straight out of the equation of motion. Nothing you can buy is specified that way: isolator catalogues, mount selection charts, floor-vibration criteria and shaker specifications all quote a natural frequency in hertz. Converting the calculated ωn is the first thing that happens between the sketch and the purchase order.
What the design hinges on
The Square Root Comes Out Angular
Catalogues Are Organised by Hertz
The √2 Threshold Decides Everything
Sizing a Mount from a Computed Natural Frequency
Stiffness values get iterated a dozen times before a mount is chosen, so keep the conversion open beside the spreadsheet rather than folding a 2π into each formula.
Enter the ωn your stiffness and mass produced
Work out √(k/m) for the supported mass and the total stiffness of all mounts in parallel, then type that figure into the rad/s field. The hertz side resolves as you type, and a comma typed instead of a decimal point is understood.
Check the result against the forcing frequency
Compare the hertz value with the lowest excitation the machine will produce. If the ratio is under 1.414 the mounts make matters worse, and softer stiffness or added mass has to go back into the calculation.
Turn a catalogue figure back into ωn
When a supplier answers with a rated natural frequency in hertz, press the swap button (↔) or type into the hertz field to recover ωn, which is what the transmissibility and static-deflection expressions want.
Copy the clean value into the design sheet
Each field has its own copy button that hands over the bare number, without unit symbols or thousands spacing, so a natural frequency drops straight into a spreadsheet cell or a mount-selection form.
Natural Frequency, Static Deflection and Where Isolation Begins
For a linear spring carrying its own load, the natural frequency depends only on how far that load pushes the spring down: ωn = √(g/δ). This makes static deflection an excellent first sanity check on a proposed mount, and it explains why genuinely low natural frequencies demand mounts that visibly sink under the machine. Because deflection appears under a square root, doubling the sag lowers the natural frequency by only a factor of √2.
| Static deflection under load | ωn (rad/s) | fn (Hz) | Attenuation begins above |
|---|---|---|---|
| 0.5 mm — steel shim / rigid pad | 140.07 rad/s | 22.29 Hz | 31.53 Hz |
| 1 mm — hard elastomer pad | 99.05 rad/s | 15.76 Hz | 22.29 Hz |
| 2.5 mm — bonded rubber mount | 62.64 rad/s | 9.97 Hz | 14.10 Hz |
| 5 mm — soft rubber mount | 44.29 rad/s | 7.05 Hz | 9.97 Hz |
| 10 mm — coil spring mount | 31.32 rad/s | 4.98 Hz | 7.05 Hz |
| 25 mm — soft spring / air mount | 19.81 rad/s | 3.15 Hz | 4.46 Hz |
| 50 mm — low-frequency air spring | 14.01 rad/s | 2.23 Hz | 3.15 Hz |
Notice how the rows chain together: doubling the deflection divides the natural frequency by √2, so each row's isolation threshold lands almost exactly on the previous row's natural frequency. That relationship is the quickest way to judge whether a proposed mount is a real improvement or merely a softer version of the same problem.
Stiffness Sweeps Without Retyping
Candidate stiffnesses are usually tried in a row. Each new ωn replaces the last and resolves immediately, so a shortlist of four mounts is compared in hertz in under a minute.
Deflection Targets in Buyable Units
A target sag translates through √(g/δ) into rad/s, and the hertz answer is what a mount supplier will confirm or reject. Both numbers stay visible while the load case is adjusted.
Frequency-Ratio Checks
Divide the lowest forcing frequency by the converted fn to get the ratio r that drives transmissibility. Anything under 1.414 is amplification, and the fix belongs back in the stiffness column.
Isolation Design Questions
How do I get ωn from stiffness and mass, and what is it in hertz?
Add the stiffness of every mount that carries the load in parallel, divide by the supported mass, and take the square root. A 100 kg unit on four mounts of 250 kN/m each has k = 1 MN/m, so ωn = √10 000 = 100 rad/s, which is 15.9155 Hz. Two cautions: use the mass actually resting on the mounts rather than the shipping weight, and take the dynamic stiffness from the datasheet, since elastomers are commonly 1.2 to 1.6 times stiffer dynamically than the static curve suggests.
Why does isolation only begin above √2 times the natural frequency?
Undamped transmissibility is T = 1/|1 − r²| with r the ratio of forcing frequency to natural frequency. Setting T = 1 gives r² = 2, so at r = 1.414 the mount passes the disturbance through unchanged and everything below that ratio is amplified. Real benefit needs distance: r = 2 gives T = 0.333, about 67% reduction; r = 3 gives 0.125, roughly 87.5%; r = 5 gives 0.042, close to 96%. The usual design aim is r of 3 or better at the lowest significant forcing frequency.
How far apart are the damped and undamped natural frequencies?
They differ by fd = fn√(1 − ζ²), and for the damping ratios that mounts actually deliver the gap is negligible. At ζ = 0.05 the damped value is 0.13% lower, at ζ = 0.10 it is 0.5% lower, and even at ζ = 0.15 it is only 1.1% lower. Damping earns its keep elsewhere: it caps the peak while the machine runs up through resonance, but it also degrades attenuation well above the corner, so heavily damped mounts trade high-frequency isolation for survivability at start-up.
Can static deflection alone predict a mount's natural frequency?
For a linear spring it can, because substituting k = mg/δ into √(k/m) cancels the mass and leaves ωn = √(g/δ). In millimetres that reduces to the shop-floor rule fn ≈ 15.76/√δ, so 5 mm of sag implies about 7.05 Hz and 25 mm implies about 3.15 Hz. It fails where the spring is not linear — air springs hold their height regardless of load, and elastomers stiffen as they compress — so treat it as a first estimate and confirm against the supplier's rated frequency at your actual load.
What happens if a forcing frequency lands exactly on fn?
At r = 1 nothing limits the response except damping, and transmissibility rises to roughly 1/(2ζ): about 10× at ζ = 0.05 and about 5× at ζ = 0.10. Motion an order of magnitude larger than the input tears mounts, fatigues brackets and destroys any precision the machine was supposed to hold. If the excitation is fixed, move the natural frequency instead — add mass to the base, soften the mounts, or split the difference with an inertia block — and re-run the conversion until the ratio is comfortably clear of that peak.
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