# Damping Coefficient of SHM

In lab for my physics of digital systems class, we were told to find the damping coefficient of a spring experiencing simple harmonic oscillation. We were given the formula

$$x = A e^{\left( -\frac{b}{2m}t \right)} \cos \omega t \, .$$

I recorded the following information in lab:

• $t = 113 \, \text{s}$
• $\text{oscillations} = 96$
• $\text{amplitude} = .06 \, \text{m}$
• $k = 4.1432 \, \text{N/m}$
• $m = 0.15 \, \text{kg}$

I know that I can calculate $\omega$ as the $\sqrt{k/m}$ and everything else seems like it's just a matter of plugging things in, which isn't the issue at all. However, I have no idea what the difference between $A$ and $x$ are. In my notes, I have $A$ as the max amplitude and $x$ as amplitude. Does this mean that .06 m is the value of $A$ or $x$? If it's one or the other, what would the value of the other one be? If it helps, the spring started at .6625 m (with .15 kg attached) and we stretched it to .6925 m before releasing it and beginning the counting of oscillations and time elapsed. We stopped counting and timing once the spring's bottom reached half of what the amplitude was, so in this case, .6625 m. I don't know if I explained this problem sufficiently enough, so feel free to ask me questions in case I've missed something.

In a nutshell, I want to understand the difference between $x$ and $A$ in the above formula.

• Welcome to Physics Stack Exchange. Great job writing a clear question. Note that this site supports mathjax which allows you to write nicely formatted math in your posts. I edited that into this one, and if you look at the edit history or the link I gave you can see how it works. – DanielSank Feb 3 '16 at 3:58
• Thank you for the warm welcome and for making my question more readable. I'm having a hard time getting the exponent in the equation to display properly. It should be e raised to -bt/2m. Thanks! – Pat Mulvihill Feb 3 '16 at 4:14
• I think the superscript notation is actually hard to read here because of the fraction. I'd inline either the exponential itself (as I had in my edit) or inline the fraction. Anyway, to your puzzle about $A$ versus $x$: $x$ is the time dependent position of the oscillator, while $A$ is the maximum position attained over each cycle. Does that answer it? – DanielSank Feb 3 '16 at 4:40
• Thanks, it looks great the way it is now. Regarding the question, in my particular case, .06m would continue to be maximum amplitude then (A). But how would I find x? Is it somewhere in front of me and I'm missing? I don't have any data for amplitude over time. – Pat Mulvihill Feb 3 '16 at 5:04
• Ok, when you say $\text{amplitude} = 0.06 \, \text{m}$, what do you mean by "amplitude"? – DanielSank Feb 3 '16 at 5:29

## 1 Answer

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 :)

• This helped. Setting x to 1/2A allowed me to cancel A on both sides eliminating A from the equation completely. Thank you! – Pat Mulvihill Feb 3 '16 at 18:05
• Great! In your calculation it would be nice if you also think about the uncertainty on determining the value of b using your method and the sources of experimental error, and how to improve your methodology. – cvb0rg Feb 3 '16 at 21:35