Spring and Mass

How the model works

A mass m on a spring of stiffness k with a viscous damper b obeys Newton's second law:

m·x'' = −k·x − b·x'

Without damping it oscillates at angular frequency ω₀ = √(k/m), with period T = 2π√(m/k). The damping ratio ζ = b / (2√(km)) sets the character of the motion:

• ζ < 1, under-damped: the mass oscillates while the amplitude decays.
• ζ = 1, critically damped: the fastest return to rest with no overshoot.
• ζ > 1, over-damped: a slow creep back with no oscillation.

Energy: kinetic ½mv², elastic ½kx², and the energy the damper has turned into heat. Their sum stays constant, which is a useful check on the numerical method (RK4).

Ideal massless spring obeying Hooke's law, horizontal motion, and no friction other than the linear damper.

Guided experiments

1. Heavier is slower. Change the mass from 1 kg to 4 kg with k fixed. Prediction: the period doubles, from 1.26 s to 2.51 s at k = 25 N/m. Try m = 4 kg
2. Energy sloshing. With no damping, watch the energy bars. Prediction: kinetic energy peaks at the centre and elastic energy at the ends. They swap twice per period and the total stays flat. Undamped energy view
3. Critical damping. Set ζ = 1, then 0.3 and 3. Prediction: ζ = 1 settles fastest without crossing zero. ζ = 3 is slower because the damper resists the return. Set ζ = 1

Challenge

A car suspension has m = 300 kg per wheel and k = 30 000 N/m. Which damping coefficient b makes it critically damped? Would you choose slightly less or slightly more in practice?

References

• OpenStax, University Physics Volume 1, §15.1 and §15.5 “Damped Oscillations”.
• MIT OpenCourseWare 18.03, “Damped harmonic oscillators”.