Hang a mass on a spring, nudge it, and it returns at one particular rate — its natural frequency. $f_n = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$, read aloud f-n equals one over two pi, root k over m. The subscript n means natural. $k$ is the spring rate in newtons per metre — how many newtons each metre of squash costs. $m$ is the supported mass in kilograms: what the springs actually carry, which on a real machine means the equipment plus its base frame plus any liquid in it, not the shipping weight off the datasheet. $f_n$ comes out in hertz — cycles per second. Stiffer rings faster; heavier rings slower.

The $2\pi$ in the denominator is the single most expensive character on this page. The bare root $\sqrt{k/m}$ is $\omega_n$, the natural frequency in radians per second; dividing by $2\pi$ converts it to hertz. Textbook analysis is written in $\omega_n$; nameplates and vibration analysers read in hertz. Read one as the other and you are wrong by a factor of 6.283 — and when a vibration number looks wrong “by about six”, that is almost always exactly what happened.

There is a shortcut that needs neither $k$ nor $m$. Set the machine down and measure how far the mounts sink: the static deflection $\delta_{st}$, in metres, subscript st for static. Then $f_n = \dfrac{1}{2\pi}\sqrt{\dfrac{g}{\delta_{st}}}$, with $g = 9.81\ \mathrm{m/s^2}$. The mass cancelled — a heavier machine on the same mount sinks further, and the extra sag exactly pays for the extra mass. A tape measure has replaced a datasheet. Convert millimetres to metres first: inside a square root, a factor of 1000 becomes a factor of about 32, which is just wrong enough to look plausible.