Is the drag coefficient the same thing as damping coefficient? Can I find the drag coefficient using the data of a damping oscillating sphere?

I am currently working on a lab experiment to find a relationship between the diameter of a sphere and its drag coefficient. I will be using a spring-mass system that oscillates vertically and then damps because of drag. How can I use the data obtained from that experiment in order to determine a drag coefficient for the oscillating object?

• Are you trying to find out how closely you can match existing steady-state drag data with that setup, or in detailing the drag changes at various points in the oscillations? Jul 20 '21 at 11:40
• I am trying to determine the drag coefficient of the ocilating object using that setup (spring-mass system). The drag coefficient is constant, it doesn't change at various points in the ocilations. I want to find how closely can I match the drag coefficient value of a sphere using that setup with the theoretical value which is 0.5. Jul 20 '21 at 12:26
• Why are you saying the drag coefficient is constant? Even without unsteady effects, it would still vary with Reynolds number. With the unsteady effects, it would be a very complicated variation. Jul 20 '21 at 13:45
• I meant that it will be constant for the same object that is oscillating. Jul 21 '21 at 9:29
• It will not be constant for the same object. It is a function of the speed and will vary continuously as the speed changes. Jul 21 '21 at 11:24

Assume the damping coefficient is a constant for the 1st order approximation, $$f_d = -b v$$. The equation of motion becomes: \begin{align*} F & = - b v - k x;\\ m \frac{d^2x}{dt^2} & = -b \frac{dx}{dt} - kx;\\ m \frac{d^2x}{dt^2}+b \frac{dx}{dt} + kx &= 0.\\ \frac{d^2x}{dt^2}+ 2\gamma \frac{dx}{dt} + \omega_0^2 x &= 0. \end{align*} where we define $$\gamma = \frac{b}{2m}$$ and $$\omega = \sqrt{k/m}$$.

The general solution for the above equation is: $$x(t) = e^{-\gamma t} \left\{ A \cos \sqrt{\omega_0^2 - \gamma^2} t + B \sin \sqrt{\omega_0^2 - \gamma^2} t\right\}$$ where $$A$$ and $$B$$ are two parameters to fit the initial conditions, .

The solution is depicted as the following figure. The oscillation is enveloped with the decaying function $$e^{-\gamma t}$$. Therefore, for each period $$T = \frac{2\pi}{\sqrt{\omega_0^2 - \gamma^2}}$$, the amplitude decreases by a factor $$e^{-\gamma T}$$. From measuring the decay rate of the amplitide, we may estimate $$\gamma$$, and the damping coefficeint $$b = 2 m \gamma$$.

You then may convert the damping coefficient to the speed-dependent dragging coefficient (for low speed): \begin{align*} F_d &= b v = \frac{1}{2} \rho C A v^2.\\ C&= \frac{b}{\rho A v} \end{align*} where $$\rho$$ is the density of the air, $$A$$ the cross-section area of the oscillator, and $$C$$ the dragging coefficient. The reference speed $$v$$ may use the average speed $$= \frac{1}{2} v_{max} = \frac{1}{2} \omega x_0$$. • Thanks a lot for your response. Is the damping coefficient the same thing as the drag coefficient? Jul 21 '21 at 9:38
• Yes. You may consider it as the first order approximation.
– ytlu
Jul 21 '21 at 15:12
• Edited. Adding the relation between the damping coefficient and the dragging coefficient.
– ytlu
Jul 21 '21 at 18:16
• The reference speed $v$ may use the average speed $= \frac{1}{2} v_{max} = \frac{1}{2} \omega x_0$ Where does the relationship for average speed come from? Oct 25 '21 at 22:39

The main issue here is that $$F_{\rm drag} = -\beta |v| v$$ where as $$F_{\rm damp} = -c v$$ and therefore you have a non-linear vibration problem that is not the same as the linear vibration problem of damping.

A secondary problem is that the spring itself has some damping associated with it and a fraction of the spring stores kinetic energy which as to be accounted for when fitting a model to the data.

• Will those issues give me a totally wrong value for drag coefficient or will the value be close to the actual one? Jul 21 '21 at 9:32
• @ytlu - You cannot 1st order approximate a quadratic function unless you specify an average speed to draw a tangent line from. In an oscillation the average speed is zero and the first order approximation is not going to yield anything useful, imho. Jul 21 '21 at 14:48
• @JAlex The dragging force $F_d = \frac{1}{2} \rho C_d v^2$. At lower speed, the dragging coefficient $C_d$ is propotional to $1/v$, make the dragging force proportional to $v$. If it is a simple quadratic function, then it will not able to stop. You may test it by a horizontal slipping. If $F_d \propto v^2$, the movement will be infinity.
– ytlu
Jul 21 '21 at 15:32
• @ytlu - Only with laminar flow you have $C_d = 24/R_e$ which you can approximate using linear drag . But this experiment might be in the turbulent region, which would have a much lower $C_d$ values than what would we calculated from the above approx. Jul 21 '21 at 15:44
• @ytlu - ok you have convinced me. I withdraw my objections. Jul 21 '21 at 20:40