How to calculate viscous damping coefficient?

Question

The damping of a spring is calculated with:

$$[\zeta] = \frac{[c]}{\sqrt{[m][k]}}$$

Where c is the 'viscous damping coefficient' of the spring, according to Wikipedia. m is the mass, k is the spring constant, and zeta is the damping ratio.

How is the value of c calculated though? Is it a constant for the air through which the spring is moving or does it depend on the spring itself?

What data is required to calculate it and how can it be done?

I'm just looking at the oscillation of a spring vertically, and I have data for its decreasing amplitude, and the velocity of the spring at all points.

I have the value of the damping ratio, and I'm trying to find the value of 'c' in order to prove the above equation in an investigation.

There is almost no information about this online.

Answer 1

OK, I will assume you have the under-damped case.

If you continue reading the wikipedia article in question you'll find the solution for a underdamped oscillator writen as $$ x(t) = e^{- \zeta \omega_0 t} (A \cos(\omega_\mathrm{d}\,t) + B \sin(\omega_\mathrm{d}\,t )) $$ with $A$ and $B$ constant.

So, take you data, and plot all the maxima (or minima) as a function of time, fit an exponential {*} to that and $\zeta \omega_0$ pops right out.

If you also need to get $\omega_o$ from the data use $$ \omega_\mathrm{d} = \omega_0 \sqrt{1 - \zeta^2 } $$ where you get $\omega_\mathrm{d}$ by extracting the average period (i.e. time from peak to peak) in the data and noting that the period is $T_d = \frac{2 \pi}{\omega_\mathrm{d}}$.

Now you have two equations for two unknowns, so all you have left is a bit of algebra.

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{*} Or plot amplitude versus time on semi-log paper if you are doing this the old-school way. Or plot log(amplitude) versus time on linear--linear graph paper. Then extract the slope.

Answer 2

For the case of a real spring, damping (energy loss) comes from three areas:

1. Structural Damping. As the material flexes in cycles, there is internal losses that occur due to a hysteresis effect on the force-deflection relationship. This is a small but noticeable effect.

2. Contact friction. The spring is not floating in space by itself, but is in contact with other objects (like spring retainers and tappets). Where there is contact there is energy loss due to friction. If it is dry friction there is an equivalent damping coefficient calculated that depends in the frequency and amplitude of the oscillation (any vibrations book has it). If it is viscous friction then the damping coefficient depends on the laminar shearing of the fluid (any fluid dynamics book at some point relates viscous coefficient to damping based on geometry).

3. An finally as you mentioned there is aerodynamic drag that contributes to damping. This is the most difficult to calculate as you need to run a CFD simulation as the spring moves.

The combined effect can be measured (and tested) with a log decrement method. Hit the spring hard and measure the amplitude as a function of time. If you count the relative decrease and the # of cycles you can use that to calculate and overall effective damping coefficient.

What you want in the end is a relationship of the form $F_{\rm{damping}} = c\,\dot{x}$ and estimating the $c$ value experimentally or with a simulation. (This was first mentioned by other answers already)

PS. Be careful with real springs because not all of their mass is in motion and thus any $m$ value used in calculation is going to be inaccurate. Sometimes with vibrations the effective mass of a spring in one end of it is a low as 33% of the total mass.

Answer 3

The viscous damping coefficient is the coefficient $c$ in the formula

$$F=-cv$$

where $F$ is the damping force and $v$ is the velocity.

$c$ depends on what causes the damping. If it is a spring in air, then it is likely to be proportional both to the viscosity of the air and to the relevant area of the the spring leading to the damping. For low speeds in air you can probably use Stokes' approximation.

Answer 4

You can't "prove" the equation since the equation is actually the definition of zeta, which is not a physical parameter. M, c, and K are all physical parameters that could actually be measured for a simple harmonic oscillator with a lumped mass, a massless spring, and an ideal dashpot.

In some messy real world situations with springs that have mass and damping, it might be easier to use something like the aforementioned log decrement method, or some other method, to estimate the decay rate, and from that get a value directly for zeta. In this case, zeta is the "real" number and M, K, and or c may have vague meanings -- since it is no longer a "true" harmonic oscillator, it is a real system of which the harmonic oscillator is only an approximation. So in the case of a real physical system you can estimate zeta, and then if you have an approximation to M and k you can get an approximation for c using the definition.

As mentioned by the prior folks, you can try to measure all of these things (but "c" will be quite hard to measure for a spring) and even with "perfect" measurements you'd find some minor mismatch since the real system is not actually a perfect harmonic oscillator.

Hope this helps!

Answer 5

Consider the free fall of a body in a perfect vaccum, you know its fall acceleration/velocity and force acting on it. Now if the same body falls in a liquid you know its velocity of fall and the force which acted on it during the fall can be calculated comparing both situations mathematically hence you get the value of c, when F and velocity of fall is known. $C=F/v$