Why is the damping force proportional to $v$ and not $v^2$?

Question

We use (source)

$$ \Sigma F=ma=F_{spring}+F_{damp} $$

for the forces acting on an object undergoing damped harmonic oscillation. We define

$$ F_{damp}=\beta v $$

where $\beta$ is the drag constant and $v$ is the velocity.

My question is why $F_{damp}$ is proportional to $v$ and not $v^2$ because if we have a damped pendulum, the damping force should relate to the air resistance (drag force) which is proportional to $v^2$:

$$ F_{drag}=\frac{1}{2} \rho C_D A v^2 $$

If we let $\beta = \frac{1}{2} \rho C_D A$, we get $F_{drag} = \beta v^2$

So why is the damping force proportional to $v$ and not $v^2$?

Answer 1

At low velocity $v$ the flow of the fluid around the object is mostly laminar and the drag force a viscous response, which is proportional to $v$.

But at higher velocity, flow becomes turbulent and inertial forces acting on the flowing fluid have to be taken into account. In those conditions the drag force becomes proportional to the square of $v$.

Answer 2

Damped harmonic oscillations are an extremely broad paradigm, and there are many physically dissimilar examples for which the force behaves in completely different ways as a function of velocity.

• In the standard Amontons-Coulomb model of friction, we have $F\propto v^0 \operatorname{sign}(v)$.

• In the case of viscous drag, we have $F\propto v^1$.

• For high velocities, we typically have, approximately, $F\propto v^2\operatorname{sign}(v)$.

The reason that people like to talk about $F\propto v^1$ is not physics, it's simply that the resulting solutions come out to have a simple analytic form. One way to see why the exponent 1 is mathematically special is that in this case, the equations of motion can be put in the form $x''+ax'+bx=0$ (the homogeneous case, i.e., free oscilations). Then we can take $x=e^{rt}$, where $r$ is a complex number, and the solutions correspond to values of $r$ that are roots of a quadratic.

Answer 3

As pointed out by Gert and Kyle Kanos, for low velocities (as determined by the Reynolds number) the flow of air will be laminar. At high velocities the quadratic dependence comes about due to turbulence making the flow far away from the object independent of the flow in the immediate neighborhood of the object. One can then consider the change in momentum of the air flow that is intercepted by the object. Clearly the change in momentum if a given quantity of intercepted air is proportional to the velocity and the amount of intercepted air per unit time is also proportional to the velocity, therefore the friction force must be quadratic in the velocity.

At low velocities this reasoning becomes invalid, the flow of the air perturbed by the moving object is no longer local. One can show using the Navier Stokes equations that this leads to a linear dependence of the friction velocity on the velocity for small velocities (allowing one to ignore the $\vec{v}\cdot\vec{\nabla}\vec{v}$ term). However, this is only true for an object moving at a uniform velocity; precisely the long range effect of the moving object on the fluid will cause the friction force to depend on the entire history of the object's trajectory. The general formula for the friction force in the low velocity limit of a spherical object of radius $R$ moving at velocity $\vec{v}(t)$ is:

$$\vec{F}(t) = -2\pi\rho R^3\left[\frac{1}{3}\vec{a}(t) + \frac{3\nu \vec{v}(t)}{R^2} + \frac{3}{R}\sqrt{\frac{\nu}{\pi}}\int_{-\infty}^{t}\frac{\vec{a}(t')dt'}{\sqrt{t-t'}}\right]$$

where $\vec{a}(t)$ is the acceleration of the object, $\nu$ is the kinematic viscosity $\frac{\eta}{\rho}$ where $\eta$ is the dynamic viscosity and $\rho$ is the density of the fluid. The second term in the brackets yields the familiar Stokes formula for the friction force. The first term is the effect of the inertia of fluid, if the object accelerates then part of the fluid will accelerate with it due to the no-slip boundary conditions. The last term yields the effect of the history of the object's motion on the friction force.

Answer 4

This is discussed on the Wikipedia entry on Drag (emphasis theirs):

The equation for viscous resistance or linear drag is appropriate for objects or particles moving through a fluid at relatively slow speeds where there is no turbulence (i.e. low Reynolds number, $\displaystyle R_{e}<1$). Note that purely laminar flow only exists up to Re = 0.1 under this definition. In this case, the force of drag is approximately proportional to velocity, but opposite in direction. The equation for viscous resistance is: $$ \mathbf F=-b\mathbf v $$

So because the author is assuming a laminar flow for the air around the oscillating mass, it uses the linear form (Stokes' limit) for drag.

Note also that the quadratic form requires a Reynolds number of $\gtrsim$1000 before it is valid (also depends on shape of moving object).