3
$\begingroup$

We know and have been taught that due to friction on the surface of maybe a spring mass system, the body faces damping. But what is bothering me, is the fact that force due to damping is proportional to the velocity of movement of the body, i.e., $F = -bv.$ But we already know that kinetic friction on solids is independent of velocity and dependent on the normal force exerted by the surface on the body.

This is where the confusion arises. What really causes the damping of the mass?

$\endgroup$
2
  • 1
    $\begingroup$ Learning physics starts with very simple approximations, moving to slightly less simple approximations, and so forth. There are many different types of damping with many different regions of applicability. Start simple (no damping), add a constant damping term, add another damping term... $\endgroup$
    – Jon Custer
    Commented Sep 19 at 16:51
  • 1
    $\begingroup$ Usually, air or liquid around the oscillating body. If flow of fluid around the body is laminar, the additional force on the body due to the fluid has a component proportional to its velocity. $\endgroup$ Commented Sep 19 at 23:10

3 Answers 3

4
$\begingroup$

A purely "ideal" friction damped oscillator would be a mess.

$$ma = F$$

would have

$$ F = F_{spring} + F_{kinetic} + F_{static} $$

Ofc:

$$ F_{spring} = -kx $$

The kinetic friction would be:

$$ F_{kinetic} = -mg\mu_k \cdot {\rm sgn}(\dot x) $$

and the static term..idk. It only exists at $\dot x= 0$, and it opposes the direction of acceleration, which I can write as:

$$ F_{static} = -mg\mu_s \cdot {\rm sgn}(\ddot x) \cdot (1-{\rm sgn}^2 \dot x) $$

for:

$$ m\ddot x + mg\big( \mu_k\cdot{\rm sgn}(\dot x) + \mu_s\cdot {\rm sgn}(\ddot x) \cdot (1-{\rm sgn}^2 \dot x) \big) + kx = 0 $$

Yuck.

$\endgroup$
3
$\begingroup$

Anything involving friction is a 'rough' (pun not intended but celebrated anyway) approximation. For example, your statement that "kinetic friction on solids is independent of velocity" isn't completely true. If it WERE true, then a block with initial sliding velocity on a rough surface would not come to rest in the way it does. But it is the typical approximation that is made in physics classes. Friction laws are mostly approximations, not on the same level as Newton's laws or constitutive laws. See my answer here for further comments.

Why are my heavier objects sliding on a smaller incline than lighter objects? Coefficient of Static Friction

The assumption of damping proportional to velocity probably developed from historical approaches to differential equations in that 1) it's sort-of accurate for everyday use, and 2) it gives you exponential decay superimposed on a sine wave so its a mathematically solvable ODE - appealing to mathematicians, and useful for pedagogical reasons. A third possible reason is the mathematical analogies with electronic circuits and PID control, although I guess maybe its a chicken-and-egg question of which came first.

The actual source of mechanical damping is often fluid like oil being forced through valves and orifices, as in vehicle shock absorbers, and the mechanical energy ultimately ends up being dissipated into heat.

$\endgroup$
0
$\begingroup$

the standard model of SHM consists of a vertical spring holding up a weight against gravity and a small shock absorber. The example you cite is more complicated and requires a different constituitive equation for the damping term.

so, the simplest SHM example is taught first. the more complex examples come later in the course.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.