# Finding the damping constant and period from looking at a graph [closed]

I want to make sure there are no holes in my logic in solving this problem.

From the figure, I am instructed to determine the damping constant as accurately as possible.

Here's what my thought process is:

• "As accurately as possible" would, and must, given the information I'm supplied, mean I have to eye-ball the graph.

• I can tell the max displacement is $5 \ cm$.

• I like picking $t=2 \ s$, due to the fact that I do not know the argument in the cosine curve for the under-damped equation of motion, but I know the argument must equal $0$ here. If I picked $t=0$ to find the damping constant, I can't solve for $b$ in the argument of the exponential in the under-damped spring equation.

I keep mentioning the equation, so I owe it to just jot it down here so what I'm saying can be referenced. The general equation for an underdamped oscillator is:

$$x(t) = x_m \exp \left[-\frac{b}{2m}t \right] \cos\left(\omega_1t+\phi\right)$$

I'm going to eyeball that $x(t)$ at $t=2$ is roughly $2.5 \ cm$. I think that's not that bad an estimate.

Here, I have no way of knowing the underdamped resonant frequency, as I don't know the spring constant of this spring, but it doesn't matter at $t=2$, as I know cosine here must resolve to $1$ as it is at a peak, or where the argument is $0$ or an integer multiple of $\pi$.

I therefore have:

$$2.5\,\mathrm{cm} = 5 \,\mathrm{cm} \exp\left[\frac{-b}{2m}(2\,\mathrm{s})\right]$$

$$1/2 = \exp\left[-\frac{b}{m}\right]$$ $$\log(1/2) = -b/m$$

Which gives a damping constant $b$ of roughly $0.06$. Did I think this through properly?

## EDIT

I now would like to task myself on finding the period of oscillation. I think I've run into a problem, however. My aim for finding the period of oscillation would be too, naturally, find the angular frequency. So, I pick a time where the argument of cosine gives $0$ or an integer number of $\pi$, which I chose as $t=4$.

Thus, $$cos(\omega_1 t) = 1$$ $$\omega_1 t = n\pi$$

Where $n$ is an even integer (odd numbers will be $-1$).

Here's where I made my mistake at first. I figured any integer is valid, as it's the same result of $1$. I chose $2\pi$ arbitrarily.

$$\therefore \omega_1 4s=2\pi$$ $$\omega_1 = \pi/2 \implies T = 4s$$

This is clearly not true by looking at the graph. Then I figured, it may be due to the fact that I may have to consider the amount of cycles the graph has gone through. It looks like it has had $4$ cycles at $t=4$, as cosine has gone to maximum $4$ times. Thus, $\omega_1 4s = 8\pi$

This renders $T=4$, which looks like it makes sense.

I'm confused though as to why I couldn't arbitrarily choose an integer multiple of $\pi$ though, other than the fact that it was past $2\pi$ at $t=4$. Yes, it has, but why should that matter? It should render the same result - that being thre argument of cosine is $1$.

## closed as off-topic by Gert, Kyle Kanos, Jon Custer, peterh, John RennieNov 28 '17 at 8:25

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• I've applied some of what I allege are improvements to the math formatting. I did not fix one issue: the amplitude is a function of time and both the exponential and the sinusoid should have a $t$ in them. Oddly you got that right in the line when you explicitly insert the amplitudes. – dmckee Nov 26 '17 at 20:20
• Gah! That was a mistake on my part - I mean to include that. Yes, they emphatically are. I'll fix it now. – sangstar Nov 26 '17 at 20:21
• ::chuckles:: It happens. The accuracy of your calculation will depend on the accuracy of your reading of the graph. For that reason I would probably chose $t = 3$ as my working data. – dmckee Nov 26 '17 at 20:30
• I see why - it's a very reliable determination of $x(t)$! I've added another piece to this question, as well, if it enthuses you to tackle it. – sangstar Nov 26 '17 at 20:45