Spring-Mass Energy Calculator
Split a spring-mass oscillator’s energy into potential and kinetic at any displacement, and find its speed, period and frequency.
How it works
total E = ½k·A² · PE = ½k·x² · KE = E − PE · v = √(k(A² − x²) ÷ m) · ω = √(k ÷ m) · T = 2π ÷ ω
A mass on an ideal spring holds a constant total energy ½kA², swapping between elastic potential energy ½kx² and kinetic energy as it oscillates. At the extremes of the motion (x = A) it is momentarily still and all the energy is potential; passing through equilibrium (x = 0) all of it is kinetic, giving the top speed. The angular frequency ω = √(k/m) fixes the period 2π/ω, which depends on the stiffness and mass but not on the amplitude.
Worked example
A 0.5 kg mass on a 200 N/m spring with 0.1 m amplitude has total energy ½ × 200 × 0.1² = 1.0 J. At equilibrium (x = 0) all of it is kinetic, so v = √(200 × 0.1² ÷ 0.5) = 2.0 m/s. Its angular frequency is √(200 ÷ 0.5) = 20 rad/s, a period of 0.314 s (about 3.18 Hz).
Sources: Simple harmonic motion — Wikipedia
Frequently asked questions
Where is the mass moving fastest?
At the equilibrium point, x = 0, where the spring is neither stretched nor compressed. All the energy is kinetic there, giving the maximum speed v = A·√(k/m).
Does the amplitude change the period?
No. For an ideal spring the period 2π√(m/k) depends only on the mass and the spring constant. A larger swing simply means higher speeds, not a longer time per cycle — a defining feature of simple harmonic motion.
How does a stiffer spring or heavier mass affect the motion?
A stiffer spring (larger k) raises ω and shortens the period, so it oscillates faster; a heavier mass (larger m) lowers ω and lengthens the period, so it oscillates slower.
Related calculators
This calculator is for educational purposes. Double-check important results before acting on them.