Calculators
Mass–Spring Period Calculator
Calculate period, frequency, and angular frequency for an ideal mass–spring oscillator using mass and spring stiffness.
Find the natural timing of a mass on a linear spring.
For an ideal undamped oscillator, angular frequency ω = √(k/m), period T = 2π√(m/k), and frequency f = 1/T. Enter mass m in kilograms and spring stiffness k in N/m. For 1 kg and 100 N/m, ω = 10 rad/s, T ≈ 0.628318530718 s, and f ≈ 1.59154943092 Hz. Larger mass increases the period; larger stiffness decreases it. The model assumes a massless linear spring, motion along one axis, and no damping or friction. Amplitude does not change the ideal period while the spring stays in its linear elastic range. For a vertical spring, gravity shifts equilibrium but does not change this period. Real spring mass, damping, and nonlinear stiffness need a different model. Positive inputs from 1e-12 to 1e12 are supported; results use browser floating-point arithmetic rounded to 12 significant digits. Inputs remain on your device.
Frequently Asked Questions
Everything you need to know about this tool, how it works, and privacy.
How is the oscillation period calculated?
T = 2π√(m/k), where m is mass in kg and k is stiffness in N/m. Frequency f = 1/T and angular frequency ω = √(k/m).
What is an example?
A 1 kg mass on a 100 N/m spring has a period of approximately 0.628318530718 seconds, frequency 1.59154943092 Hz, and angular frequency 10 rad/s.
Does gravity change the period?
For an ideal vertical linear spring, gravity changes the equilibrium position but not the period about equilibrium.
Does amplitude affect the result?
Amplitude does not affect the ideal linear oscillator period. Large motion that makes the spring nonlinear or exceeds its elastic limit is outside this model.
Can I use a real spring with damping?
This calculation gives the ideal undamped natural period. It ignores spring mass, friction, damping, and nonlinear stiffness. Enter positive mass and stiffness from 1e-12 to 1e12 in the stated SI units.