How does position change if the damping force depends on velocity squared instead of linearly?

Question

How would a graph of the motion of a spring over time change if the initial damping force depended linearly on velocity was changed to depend solely on velocity squared? and why would it change in that way?

Answer 1

The mass will oscillate similarly to a linearly-damped system in which damping is gradually lowered over time.

For linear damping (physically approximated with magnets and copper blocks) there is a dimensionless number that determines whether they are overdamped, critically damped, or underdamped, and thus the shape (decaying sinusoid vs exponential-like) of the motion which has the same shape regardless of it's amplitude.

For quadratic damping (physically approximated with large objects moving though low-viscosity fluids), small amplitudes are more under-damped and large amplitudes are more over-damped.

Suppose a mass is released far from equilibrium, it will initially approach the equilibrium, slowing down in doing so, like an over-damped system. However, it will settle into a decaying oscillation pattern that will take longer to decay than the linear case.

There is no exact algebric mathematical formula for this curve, unfortunately, but it is easy to solve numerically.