Period of a Spring-Mass Oscillator

The remarkable claim in $T = 2\pi\sqrt{m/k}$ is what it leaves out: the amplitude. Pull the mass twice as far and it takes exactly as long to come back. The reason is the shape of the spring's restoring force, $F = -kx$ — doubling the displacement doubles the distance to be covered, but it also doubles the force and therefore the acceleration, and the two effects cancel exactly. That cancellation is the definition of simple harmonic motion, and it is why springs and pendulums became clocks: a mechanism whose rate does not drift as its drive weakens is worth a great deal. The rest of the formula is intuition made quantitative. More mass means more inertia and a slower bounce; more stiffness means a firmer push back and a faster one. The square root softens both: quadrupling the mass only halves the frequency.

One kilogram on a 100 N/m spring gives $T = 2\pi\sqrt{1/100} = 0.628$ s, a frequency of 1.59 Hz. Run in reverse, the relation becomes a measurement. A 250 kg machine sitting on vibration isolators is observed to bounce at 3 Hz, so $T = 0.333$ s and $k = 4\pi^2 m/T^2 = 4\pi^2(250)/0.111 \approx 89$ kN/m — the effective stiffness of the whole set of mounts, which is exactly the number you need before deciding whether the machine will resonate with anything the floor is doing.

Notice that $g$ does not appear, which surprises people who have hung the mass vertically rather than laid it on a table. Gravity does change something: it shifts the equilibrium position down by the static deflection $\delta = mg/k$. But it shifts the whole motion, not its shape, so the oscillation about the new equilibrium runs at the same period as before. That gives a genuinely useful field trick — substitute $k = mg/\delta$ and the formula becomes $T = 2\pi\sqrt{\delta/g}$, so the natural frequency of a mounted machine can be read straight off how far it settled onto its mounts. A millimetre of static deflection means about 15.8 Hz; ten millimetres means 5 Hz. The same $2\pi\sqrt{\text{inertia}/\text{stiffness}}$ skeleton carries the pendulum, where $L/g$ plays the part of $m/k$.

Three things to watch. The $m$ is the oscillating mass, and a heavy spring oscillates too — adding about a third of the spring's own mass to $m$ is the standard correction, and it matters on soft suspension springs. The $k$ must be the effective stiffness of the entire arrangement, and springs combine the opposite way to resistors: parallel springs add directly, $k = k_1 + k_2$, while springs in series add reciprocally. Getting that backwards is common and the error is large. Finally, real springs are linear only over part of their travel; push into coil bind or use a progressive rate and the period becomes amplitude-dependent again, which means the answer this page gives quietly stops applying at exactly the point you were probably interested in.