Spring Natural Frequency Calculator
Enter spring constant k and mass m to compute the natural frequency f=(1/2π)√(k/m), period T=2π√(m/k) and angular frequency ω=√(k/m). k=100, m=1 → f≈1.592 Hz, T≈0.628 s.
Input Data
Results
At a glance:Natural (resonant) frequency is the free-vibration characteristic frequency of a mass–spring system with no damping and no external force. Once m and k are fixed, the system has a unique set of f, T, ω: f=(1/2π)√(k/m) (Hz), T=2π√(m/k) (s), ω=√(k/m) (rad/s), where k is the spring constant (N/m, from Hooke's law F=−kx) and m the oscillator mass (kg). Physical meaning: (1) a stiffer spring (larger k) gives a higher f (faster oscillation); (2) a larger mass gives a lower f (more inertia, slower); (3) f is independent of amplitude (isochronism, discovered by Galileo). History: Hooke stated the spring law in 1676; Galileo observed the pendulum's isochronism; the mass–spring system is the canonical simple-harmonic-motion model. Applications: (1) mechanical-clock balance wheels; (2) vehicle suspension (spring + damper); (3) building tuned-mass dampers (Taipei 101); (4) spring scales; (5) tuning forks and piezoelectric oscillators.
Formula
Natural frequency: f = (1/2π)·√(k/m) (Hz)
Period: T = 1/f = 2π·√(m/k) (s)
Angular frequency: ω = 2πf = √(k/m) (rad/s)
k spring constant (N/m), m mass (kg)
$$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}, \quad T = 2\pi\sqrt{\frac{m}{k}}, \quad \omega = \sqrt{\frac{k}{m}}$$How to Use
- Enter spring constant k (N/m) and mass m (kg).
- The tool outputs f=(1/2π)√(k/m), period T and angular frequency ω.
Case Studies
Vehicle suspension and tuned damper
Car suspension: k≈20000 N/m, body m≈400 kg (shared by 4 wheels) → T=2π√(400/20000)=0.889 s.
Too soft/slow → wallowing; too stiff → harsh. A damper (γ≈0.3) settles it in 2–3 cycles.
Taipei 101 TMD: m=660 t, k≈1700 N/m (equivalent) → T≈124 s, countering the building's sway.
Content review: Calculatorism Editorial Team. Results are for reference only; please refer to the relevant authorities for the official figures.