# Finding the effect of air resistance on period of oscilation without differential equation

I'm currently working on the "cavendish" experiment and wish to use/develop a method separate from the casus we've been provided. Now I've nicely calculated and derived everything I need to know, including all the corrections that have to be made for the mass of the rod, the finite size of the masses, etc. However: I'm running in to a brick wall when it comes to finding a way to bring air resistance in to the picture.

For those of you don't know: The cavendish experiment is a torsion balance that is used to measure gravity. The way it works is by suspending a rod with some small masses on either end by a wire with low torsion spring coefficient. These small masses will start to move towards an equilibrium where the force supplied by the torsion in the wire and the force of gravity supplied by two larger masses brought in close proximity of the smaller masses cancel out.

The problem is that these small masses have a certain moment of inertia, so they'll end up oscillating about the position of equilibrium. This oscillation is of course dampened by air resistance. The most common way to go about finding the effect of air resistance is by setting up and solving a differential equation, which I prefer not to do. Because of this, I'm now looking in to alternative methods of (numerically) finding the way air resistance affects the period of the oscillation.

My work so far So far, I've determined that air resistance is irrelevant when determining the equilibrium position of the bar (which is used to determine $G$). Where air resistance does become relevant is in finding the value for $\kappa$ (torsion spring coefficient). Most of you will know that one can describe kappa with $\kappa=2m(\frac{\pi L}{T})^2$ where m is the point mass's mass and L the length of the rod. Because $T$ gets changed by friction, $\kappa$ can not be accurately determined without finding a way to go from $T_{measured}$ to $T_{frictionless}$. Herein lies my struggle, as I haven't been able to do so.

The reason I'm exploring this question of doing it without differential equations is in order to explore the science. I've got no problem actually working through the differential equation.

• Do you have some idea how the friction torque caused by air drag depends on angular velocity of the beam? – Gert Jan 18 '17 at 21:44
• If you can find a way to perform the oscillation in a vacuum jar, you can directly get the data that you need. – David White Sep 6 '18 at 17:08
• Upon re-reading this, it occurs to me that determining the period of oscillation for various amplitudes may be the way to go. You will find that, due to various non-linearities (mostly drag), the period will increase with amplitude. If you can fit a line (not sure if it will be linear or parabolic) and extrapolate to “zero amplitude”, you will have your answer. – Floris Sep 9 '18 at 12:05

The Cavendish balance (CB) is very slow moving, so the air flow is almost laminar.

In laminar flow, a sphere accelerating through a fluid experiences a force because the fluid has to be accelerated. It turns out that the effect of this force is to make it appear that the sphere has inertia increased by half the inertia of the displaced fluid. In the case of the CB, this will add a few grams at most to the apparent mass - and this will decrease the response frequency in the same way that a slightly lower $\kappa$ would do. The equation of motion for simple harmonic motion is easily extended to include linear drag, and the expression for the shift in resonant frequency is well known (see for example http://hyperphysics.phy-astr.gsu.edu/hbase/oscdr.html#c1)

As the velocity increases, drag will become proportional to square of velocity (and thus, square of amplitude). That means the force will be largest when the torsion is smallest (namely, at the zero crossing) - and it will be a bit harder to explore the relationship of resonant frequency. The basic relationship - resonant frequency is lower when there is drag - remains the same.

So if you want to determine the torsion constant from the resonant frequency, I recommend observing the oscillation for a large number of cycles - you can then estimate the damping, and should be able to use the correction for the lightly damped simple harmonic oscillator to obtain the correct value for $\kappa$.

It would be interesting to compare that with the result you would get from the inertial argument in my second paragraph.

• Yeah, this is the well known method for doing this. You eventually add up with the equation $I\phi''+B\phi'+\kappa (\phi-\phi_0)$ which has the solution $\phi(t)=A exp(-\frac{B}{2I}t) \sin{\omega t}+\phi_0$. But I'm looking for alternative ways to approach the problem (possibly numerically). Do you know of any? – Mitchell Faas Jan 18 '17 at 22:11
• For small amounts of damping, as in this experiment, it doesn't really matter if the damping force is proportional to the velocity or not. The important quantity is the amount of energy taken out of the system per cycle of oscillation. You can measure that from the rate of decay of the amplitude and convert that to an equivalent linearized damping constant, and estimate the shift in resonant frequency. In practice, you will only measure a significant amplitude decay over several days, and the frequency correction will be most likely be negligible compared with other sources of error....... – alephzero Jan 18 '17 at 23:15
• ..... Google "logarithmic decrement" or "log dec" for the way to convert the amplitude decay into a damping parameter. – alephzero Jan 18 '17 at 23:16
• In "real world" situations like this, trying to calculate damping from first principles is very difficult, and in any case you would need to validate your calculations by doing experiments. So the practical solution is just do one experiment to measure the damping of your actual device, and forget about the calculations! – alephzero Jan 18 '17 at 23:19
• I agree that the experimental determination is most likely the way to go here. And I also agree that the impact of damping will be very small. – Floris Jan 18 '17 at 23:24