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Spring-Mass Oscillation Period Calculator

Calculate how long one full back-and-forth swing takes for a mass on an ideal spring, from its mass and the spring's stiffness.

Oscillation period
0.6283 s
Frequency
1.5915 Hz
Angular frequency
10 rad/s

How it works

A mass on an ideal spring, pulled aside and released, oscillates back and forth forever (ignoring friction and air resistance) at a rate set entirely by its own mass and the spring's stiffness — never by how far it was initially pulled. Hooke's law (restoring force F = -kx) combined with Newton's second law gives the equation of motion for simple harmonic motion, whose period works out to T = 2π√(m ÷ k), where m is the mass and k is the spring constant. Frequency is simply f = 1 ÷ T, and angular frequency is ω = √(k ÷ m) = 2π ÷ T.

Mass and spring constant are converted to kilograms and newtons-per-meter (the units this formula is published in) before it runs, the same convert-to-canonical-then-back pattern this site's Terminal Velocity and Body Fat Navy calculators use. The result — period in seconds, frequency in Hz, angular frequency in rad/s — is left as-is afterward, since time-based units don't need a metric/imperial toggle of their own, the same reasoning the Simple Pendulum Period calculator already established.

This is the same mass-spring system this site's Hooke's Law calculator models (force and stored energy from displacement) — that calculator answers 'how hard does it push back,' this one answers 'how fast does it oscillate.'

FAQ

Does a bigger displacement (pulling the spring further) change the period?

No — for an ideal spring obeying Hooke's law, the period is completely independent of amplitude; a gentle pull and a hard pull both complete one cycle in exactly the same time. This 'isochronism' is a defining feature of simple harmonic motion, the same reason a pendulum's period (at small angles) doesn't depend on how far it swings.

Why does a heavier mass oscillate slower?

More mass means more inertia resisting the spring's restoring force at every point in the swing, so it takes longer to accelerate back and forth across the same stiffness — the period grows with the square root of mass, so quadrupling the mass only doubles the period, not quadruples it.

Does this account for the spring's own mass or any damping?

No — it treats the spring as massless and the system as undamped (frictionless, no air resistance), the standard idealization for this formula. A spring with meaningful mass of its own oscillates slightly slower than this predicts (roughly as if one-third of the spring's mass were added to the attached mass), and any real damping causes the amplitude to decay over time even though the period itself stays essentially the same for light damping.

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