I'm currently doing a small project in college for structures being damped. We're adding weights to a structure and measuring the damping affect. We've calculated the damping ratio and coeffecients, however, we currently have damping ratio increasing with mass, which I do not believe is correct following the below equation:

$$\zeta=\frac{c}{2\sqrt{mk}} $$

The above equation, assuming $k$ and $c$ are constant, suggests that $\zeta$ decreases as mass increases.

However, the results we are getting are entirely different, and we do not have a constant value for the damping coefficient:

enter image description here

We got the initial value of damping ratio by interpreting the graphs by using the following equation and process:

enter image description here

$${\sqrt{\frac{\left(\ln \left(\frac{x_n}{x_0}\right)\right)}{4(\pi n)^2+ \left(\ln \left(\frac{x_n}{x_0}\right)\right)}}} = \zeta$$

Where the above was rearranged from:

$$\Large{\frac{x_n}{x_0}= e^{\frac{-2\pi n \zeta }{\sqrt{1-\zeta^2}}}} $$

where $n=$ the number of full wave cycles between $x_0$ and $x_n$

Ultimately, I'm curious if the damping ratio should decrease with increase in mass, and should the damping coefficient stay constant?

enter image description here

Edit 2: Graphs for Unweighted vs 4.004 damping

enter image description here

enter image description here

  • $\begingroup$ Is the following the protocol you followed? : (1) Find $\zeta$ using the formula that relates the ratio of two amplitudes to the damping ratio : To be more specific about your calculation, as an example, the $n$ for the two amplitudes marked in red circles in your plot is $1$, right? ; (2) Find the critical damping coefficient using the already-known value of the mass and spring constant ; (3) Use $\zeta$ and $c_{\text{critical}}$ to find the damping coefficient $c$. $\endgroup$
    – Ajay Mohan
    Commented Dec 10, 2019 at 12:48
  • $\begingroup$ Yeah that's what we done exactly, n is one. $\endgroup$ Commented Dec 10, 2019 at 14:23

1 Answer 1


I assume that your structure is a simple single spring mass damper system? For such a system the assumption of decreasing dampingratio $\zeta$ for increasing mass $m$ is correct. You need more dampingforce to stop a heavier system, therefore resulting in a smaller $\zeta$.

In reality though it could be that the additional mass affects the damping behaviour of your damper. There could be more reasons why your results don't fit the expectations. If you can supply some more information about your system it could be easier to pinpoint to the problem.

Edit1: Added image that shows the expected plot for different $\zeta$ values and different masses. The given data is flipped. My assumption are $c=2.5 kg/s$ and $k=162~kg/s^2$.

enter image description here

  • $\begingroup$ You're correct that is a simple spring-mass damper. I've edited my original post with a mass vs damping ratio graph. That's all the info I can provide at the moment. $\endgroup$ Commented Dec 10, 2019 at 0:49
  • $\begingroup$ From what I am seeing it looks like you may have accidentally switched the order of your experiments or masses. I just took your data and tried to guess the values for c and k and calculated the theoretical graph. $\endgroup$
    – scheepan
    Commented Dec 10, 2019 at 16:02
  • $\begingroup$ I don't understand how that has happened. But I will take your word for it. I've double checked my experiment. It could have just been a mix up, but highly doubt it. $\endgroup$ Commented Dec 11, 2019 at 0:20
  • $\begingroup$ I've added two new spreadsheets, you can see for the first image, where the system is not damped by additional weights, that the amplitude is a lot higher. Compare this to the second image, where the amplitude is a lot less with higher damping masses. The damping ratios are still the same. So, I'm not sure what I've done wrong here. (notice on the top right of both images the 'final value', which is the damping ratio. $\endgroup$ Commented Dec 11, 2019 at 0:26
  • $\begingroup$ Somehow I can't reproduce your graphs with the solution for a simple spring mass damper system. It seems that your decay is not exponential but linear. Also you are talking about damping masses, are you using a tuned mass system? Somehow all those things you said and your data dont fit together for me. $\endgroup$
    – scheepan
    Commented Dec 11, 2019 at 12:05

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