Springs and simple harmonic motion, explained.
The spring force points toward equilibrium and grows with displacement. Spring potential energy is largest at the turning points, where speed is zero. At equilibrium, potential energy is smallest and speed is greatest.
The relationship to remember
Key formulas
- Hooke’s law
- F = −kx
- Force points back toward equilibrium
- Period
- T = 2π√(m/k)
- Independent of amplitude
- Angular frequency
- ω = √(k/m)
- In rad/s; f = ω/2π
- Spring potential energy
- PE = ½kx²
- Largest at the turning points
- Total energy
- E = ½kA² = ½m·vmax²
- vmax = Aω at equilibrium
Worked example
Problem. A 0.5 kg mass on a spring with k = 50 N/m is pulled 0.1 m from equilibrium and released. Find the period, the maximum speed and the total energy.
- Angular frequency: ω = √(k/m) = √(50 / 0.5) = 10 rad/s.
- Period: T = 2π/ω = 2π/10 = 0.628 s.
- Maximum speed: vmax = Aω = 0.1 × 10 = 1.0 m/s.
- Total energy: E = ½kA² = ½ × 50 × 0.1² = 0.25 J. Check: ½m·vmax² = ½ × 0.5 × 1.0² = 0.25 J.
Answer. The period is 0.628 s, the maximum speed is 1.0 m/s and the total energy is 0.25 J.
A quick check
A 1 kg mass on a 20 N/m spring has a period of about 1.40 s. At amplitude 0.5 m, total energy is 2.5 J.
Inside the default experiment
- Angular frequency = √(k/m) = 4.47 rad/s. Period = 2π/ω = 1.4 s.
- Spring energy = ½kx² = 2.5 J. Kinetic energy = ½mv² = 0 J.
- Their sum remains ½kA² = 2.5 J.
When this model applies
Horizontal ideal Hooke’s-law spring, no friction or damping. Displacement is measured from equilibrium.
Make a prediction. Then test it.
At which point is the object fastest? Watch the energy bars.
Try it: Spring oscillator simulator ↗Frequently asked questions
Wondering about the why? Start here.
Does a larger amplitude change the period?
Not for an ideal linear spring. It increases maximum speed and stored energy, while the period stays the same.
How does doubling the mass affect the period?
The period is multiplied by √2. To double the period, multiply mass by four while keeping stiffness constant.