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Pretend we have a mass which is attached to a vertical spring. If we were to extend the spring by dragging it down and then releasing it, the weight would start to oscillate. Because of friction from surrounding air, the amplitude will decrease and eventually reach zero. I know that the damping force can be calculate using the formula $F = -b*v$

Now onto my question. Is it possible to calculate the constant $b$? And does this suggest that air always has the same damping constant?

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If the mass doesn't move too quickly, the resistive force acting on it will be due to the air's viscosity, and will be proportional to the body's velocity and in the opposite direction to it. As you say, $\mathbf F_{\text{res}}=-b\mathbf v$, in which $b$ is a constant for a particular size and shape of body and for air at a particular temperature. [For example, according to the nineteenth century Anglo-Irish physicist George Gabriel Stokes, if the body is a sphere of radius $r$, then $b=6\pi\eta r$ in which $\eta$ is the viscosity of air at the relevant temperature.] So the equation of motion of the mass is $$m\frac{d_2x}{dt^2}=-b\frac{dx}{dt}-ax$$ in which $-ax$ is the 'restoring force' from the spring. The solution to this equation is $$x=Ae^{-bt/2m}\sin{(\omega t+\phi)}\ \ \ \ \ \ \text{in which}\ \ \ \ \ \ \omega=\sqrt{\frac km-\frac{b^2}{4m^2}}$$ The easiest way to determine the value of $b$ is from the damped oscillations themselves. Their initial amplitude is $A$ and their amplitude at time $t$ is $Ae^{-bt/2m}.$ Therefore the time, $t_{1/2}$, for their amplitude to fall to half its initial value is given by $$Ae^{-(bt_{1/2}/2m)}=\tfrac A2\ \ \ \ \ \ \ \ \ \text{giving}\ \ \ \ \ \ \ \ \ b=\frac{2m\ \ln 2}{t_{1/2}}$$ So, knowing the value of the mass, $m$, we can determine $b$ by measuring the time taken for the amplitude to halve.

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  • $\begingroup$ So If I were to calculate the damping constant for one object in its oscilliation, i cannot assume that the constant is going to be the same for a different object which is oscillating in the same place. The reason being that the objects differ in size? $\endgroup$
    – Gabriel
    Commented Feb 25, 2023 at 0:25
  • $\begingroup$ In practice the oscillation part makes it hard to establish when the amplitude is exactly halved… $\endgroup$
    – ZeroTheHero
    Commented Feb 25, 2023 at 0:47
  • $\begingroup$ @Gabriel That's exactly right. In general $b$ is different for different objects. $\endgroup$
    – Philip Wood
    Commented Feb 25, 2023 at 8:52
  • $\begingroup$ @ZeroTheHero. Yes indeed. You'd place markers in advance close to the line of oscillation, one for the initial point of release and the other at the half-amplitude point. If the damping is light, so it takes several oscillations for the amplitude to halve, you could time between when the body passes the two marks at the top of its oscillation. You could no doubt work out a procedure to correct for errors of a fraction of a cycle! Alternatively note the amplitude, $A_n$, at the top of the $n$th oscillation and use $$A_n=Ae^{-bnT/2m}$$ in which $T$ is the periodic time. $\endgroup$
    – Philip Wood
    Commented Feb 25, 2023 at 9:38
  • $\begingroup$ From the formula for the angular frequency, we can solve for b. Then we see that the damping constant b is only dependent on m, k and T (where T is the period). But if b is also dependent on the size and shape of the oscillating object and the temperature in the room, why are these factors not included in the formula? $\endgroup$
    – Gabriel
    Commented Mar 1, 2023 at 2:02
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Yes you can do so, it's actually a Standard undergrad experiment to measure this dampening explicitly and determine the dampening constant. (Essentially the amplitude of the pendulum will be bounded by an exponential with exponent proportional to this constant)

However this does not mean that the air has one universal constant for this dampening. The constant is heavily dependent on the shape of the object you are looking at. Further the dampening constant depends on the density of the air and as such on external conditions such as pressure and temperature of the lab you are working in.

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