I need the value for the damping coefficient of air for a mass-spring system simple harmonic motion experiment. I can't seem to find a value for this anywhere else.
[Physics] the damping coefficient of air? How about steel
dragfrictionnewtonian-mechanicsoscillatorsspring
Related Solutions
Yes, one can calculate an equivalent mass. Take the following coordinates, that describe the position of the particles' centers with respect to the center of mass of the set, $G$. In the sketch that follows I have arbitrarily chosen $m_1 < m_2$, and so $G$ falls closer to $m_2$:
Since $G$ will remain fixed in the absence of external forces, it makes sense to use the coordinates above. The equations of motion of the two particles in these coordinates read $$m_1\frac{\mathrm{d}^2 x_1}{\mathrm{d} t^2} + c \frac{\mathrm{d}l}{\mathrm{d} t} + kl = 0 \\ m_2\frac{\mathrm{d}^2 x_2}{\mathrm{d} t^2} + c \frac{\mathrm{d}l}{\mathrm{d} t} + kl = 0$$ where $l = x_1 + x_2$ (the distance between centers). Adding these two equations and deviding through by $2$ one obtains $$\frac{m_1 + m_2}{2}\frac{\mathrm{d}^2 l}{\mathrm{d} t^2} + c \frac{\mathrm{d}l}{\mathrm{d} t} + kl = 0$$ which is of the form of the original equation (the one from Wikipedia) if one takes $$m = \frac{m_1 + m_2}{2}$$
Thus, the equivalent mass is just the arithmetic average of the two masses.
Let's start by plotting the damped harmonic oscillation equation:
$$x = Ae^{-\gamma t} \cos (\omega t)$$
In the above example $\gamma = 0.8$ and $\omega = 2.5\times\pi$. It should now be more clear that when $t = 0$, $x=A$ i.e. the initial displacement of the oscillator from it's equilibrium position. Here $A = 1~\mathrm{m}$. Let's assume down is the positive $x$-axis for simplicity.
Now the way one would conventionally go about finding the damping coefficient is by finding $\gamma$ form the envelope function $Ae^{-\gamma t}$ (plotted in green). You would ideally use a data logger of some fashion to sample the position of the oscillator as a function of time (hopefully with enough resolution) and then you find $x$ at the exact same point in the cycle (constant phase) for each oscillation. It is usually the easiest to locate the peaks (plotted in red) where $$\cos(\omega t) = 1 \quad \therefore \quad \omega t = 2n\pi\; (n=1,2,3,\ldots) $$
You can either fit an exponential curve to these data points or alternatively fit a linear curve to the log-log of the data.
Now in your case, it seems you only have 2 data points. The position at $t =0$, and the time $t_{1/2}$ when $x= A/2$. It's not ideal, but you have everything you need to calculate the damping coefficient now.
Hope this helps :)
Best Answer
If the Reynolds number is confortably larger than one ($\mathit{Re} = \frac{U L}{\nu} \gg 1$, where $U$ is the velocity of the mass, $L$ its diameter and $\nu$ the kinematic viscosity of air), which I think is the most likely scenario, you can use an empirical drag coefficient to calculate the air resistance, which is proportional to the velocity squared.
In fact, I believe Newton himself was able to give the first estimation of the drag coefficient from assuming the resistance to be proportional to the momentum flux $\propto \rho_{\text{air}}U^2$ and using a pendulum to observe the progressive decrease in amplitude due to this drag force.
You can use the following expression, which will give you a rough, but possibly sufficiently accurate value for your purposes (which I ignore):
$$ F_{\text{damp}} = 1/2 C_D \rho_{\text{air}} A U(t)^2 $$ where A is the crossectional area of your mass, $U$ is the velocity and $C_D$ is the drag coefficient, which depends on the particular geometry of your mass. For tabulated values of $C_D$ you can look in the Wikipedia article. In any case, always bear in mind that this force is only an average, since for high Reynolds numbers, the flow of air around the mass will not be steady, but highly oscillatory due to turbulence.
Only For the case of very small Reynolds number do you have a damping that is steady and proportional to the velocity (and not the velocity squared). This is the Stokes drag, which for a sphere, can be calculated as
$$F_{\text{damp}} = 6\pi\rho_{\text{air}}\nu R U(t)$$
where $R$ is the radius of the sphere.
I hope this helps. Note that in all this, I have neglected effect of air on the spring itself, which is probably reasonable given the uncertainties involved.