Damping of horizontal mass-spring system at high velocities [closed]

Question

I am trying to study the damping of a horizontal mass-spring system at high velocities. The equation for damping due to a viscous fluid at low velocities is:

$$ma+cv+kx=0$$

I changed the equation to :

$$ma + kx +\frac{\rho A v^2 C}{2} = 0.$$ However, when I graph displacement vs time, I get a weird graph.

Why does this not look like a normal damped sine/cosine wave. How can I make it look so? Have I made a mistake in the formula?

code snippet:

for run in range(runs):

    times.append(times[run] + dt)

    x = positions[run]
    v = velocities[run]
    dxdt = v
    v2 = (v**3)/abs(v)
    dvdt = -(k/m) - (l30*v2)/m

    positions.append(x+dxdt*dt)
    velocities.append(v+dvdt*dt)

positions, times and velocities are all arrays.

Answer 1

You forget the direction of the damping force.

In the linear equation the $-v$ term always has the opposite sign to $v$, which is correct.

But in your quadratic damping equation $-v^2$ is always negative. If $v$ is negative, that force is increasing the speed of the particle, not decreasing it, and the particle shot off to infinity as in your graph.

Change $-v^2$ to something like $-v^3/|v|$, which has the same magnitude as $-v^2$ but the sign is always opposite to $v$. Note, in a computer program you probably have to handle $v=0$ as a special case, to avoid calculating $0/0$. (The force is $0$ when $v=0$, of course)

Answer 2

Change your equation to: $$ ma + kx + \frac{\rho Av^2C}{2} = 0 $$ and it should work.