Stretch. Release. Repeat.
Spring oscillator, 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
F = −kx · T = 2π√(m/k)
A worked example
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 the interactive experiment ↗A little question. A clearer picture.
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.