# Homework Help: Uncertainty of an experiment

1. Feb 14, 2014

### ajcoelho

1. The problem statement, all variables and given/known data

Hi. I don't know if this is the right place to talk about my problem. Anyway, here it is:

I've made a classic experiment to find g. For that I used a gravitational pendulum and measured several periods of oscillation for differentes values of L.

After that, i made a linear regression and using the slope i got g.

Now I have to estimate a value for the uncertainty of the value of g that i obtained.

The thing is. I have to consider several different uncertainties!: uncertainty of L, uncertainty of T, reaction time and the uncertainty on linear regression (that was made "by eye").

How do I get all this together to get a single value of uncertainty?? How do i even get the value for the uncertainty on linear regression?

I would be very appreciate if someone could help me!!

Thanks guys!

2. Relevant equations

No idea!!

2. Feb 14, 2014

### haruspex

L: The main problem with L is that there are really two Ls: the distance from pivot to mass centre (important for the force in the equation of motion) and the radius of inertia. Try to estimate bounds on these from the physical shape and materials of the pendulum. Look at the equation of motion to see how these two lengths affect the period.

T: Reaction time might not matter if it is constant. You react to the start event and to the end event. You could (if you still have access to equipment) test yourself. Set one stopwatch going. When it hits ten seconds start a second watch. When it hits 20 s, stop the second watch. Start again at 30s, and so on. See what accumulated error you get over a couple of minutes. Repeat this to measure the variance.

Regression: This is tricky. It is not independent of the first two errors. If you had no experimental inaccuracies and you're plotting the right two variables you will now have a dead straight line (according to theory), so not much room for regression line error. Do you have to do it by eye?

There are two more sources of error I can think of: air resistance, and the fact that a simple pendulum only approximates SHM. The period is not really independent of the amplitude.

3. Feb 14, 2014

### ajcoelho

I really have to do the linear regression by eye :(

So, I have an uncertainty of +/- 0.0005m for L and +/-0.01s for T (uncertainties of the tape measure and stopwatch)

How can i relate these values and then estimate a GLOBAL uncertainty? I will not add all these uncertainties, right?

And now im already assuming no error in linear regression (r=0.99). And what if pearson coefficient was way less smaller, how could i calculate uncertainty in linear regression?

4. Feb 14, 2014

### BvU

You measure T for a number of lengths L, I suppose. Then you plot T2 versus L and find a straight line. With a supposed T2 = 4 $\pi^2$L/g you get a slope 4 $\pi^2$/g and an intercept 0.
Using numerical linear regression (Kirchner 2001) you can calculate errors in slope and intercept, but if you want to do it correct, you also need to weigh your T2 points properly. Tedious.

Linear regression by eye is not so bad. The thing to do is mark the center of gravity of your points on the graph (<x>, <y>). The regressed line goes through that point. Now take your ruler and let it go from the intercept to this center. Wiggle -- while holding on to the center -- to find the maximum slope. Where you don't believe it any more, draw a line. Wiggle same way to find the minimum slope. Where you don't believe it any more, draw a line. Usually where you don't believe it any more is close to 3σ from the mean (sope). So you have lines with slope +3σ and -3σ. Make a good guess.

If the center of gravity is pretty far from x=0, the error in the intercept is mainly determined by the slope. Otherwise you can get an estimate by hanging on to the slope and shifting the ruler up and down. Same 3σ idea. The error in the intercept is generally square root of the sum of both contributions squared. The further you get from x=0, the higher the correlation between the errors in slope and intercept.

If there is a good argument to force the intercept to be zero, with no error at all, all you can do is wiggle the ruler while hanging on to the origin. Much smaller errors result, but you really need a good argument...

By now you have (either by calculating or by "wiggling") found a slope +/- uncertainty where all the scatter from measurement uncertainties is processed as best you can.

From your earlier posts I estimate this is not a post-graduate experimental physics lab exercise, but an introduction in mechanics. Hence the wiggling.

The error in the slope is a number, say you find 4 +/- 0.01. That is 0.5%. Divide by 4 $\pi^2$: still 0.5%. Inverting the quotient gets you something that also has 0.5% uncertainty.

I greyed out the stuff above. If you find something you can use, fine. But I suspect that the sraight line is so straight you will have difficulty extracting an uncertainty by eye. And you put it to us that that's what can be done.

In that case you might concentrate on systematic errors: errors that don't go away when you do a lot of measurements because they reappear in every measurement the same way. Calibration errors are often systematic. E.g: what if your tape measure is 1 mm off at 1 m length and what if your stopwatch is 0.002 seconds per second too slow. The first would introduce 0.1% error, so 0.05% error in the square root and thus in g. The second 0.2 % in T and 0.4% in T2.

Last edited: Feb 14, 2014
5. Feb 14, 2014

### haruspex

I find that hard to believe, for the reasons I gave. You can measure a distance that accurately, sure, but how closely does it match the distance from pivot to mass centre, and what is the radius of inertia?
Let's go back to first principles for a moment. The component of the weight perpendicular to the pendulum rod is mg sin(θ). That exerts a torque mgLsin(θ), where L is from pivot to mass centre. If the radius of inertia is k, we have $k^2\ddot \theta = - gL\sin(\theta)$. E.g. for a bob with mass 30 times that of the rod, even assuming you measure the position of the mass centre correctly, the k2:L ratio will be off by one part in 180.
Not in this case. The usual here would be to apply a root-sum-square method, as with standard deviations. It's not quite straightforward, though. The error in measuring on a linear scale, as with a perfect tape measure, has a uniform distribution (+/- half a mm you say). The error due to stretch in the tape will have perhaps a normal distribution. Ideally, you would posit a distribution for each source of error, express each fractional error as one standard deviation, add the squares of these, then take the square root. You would declare the result to be one standard deviation in the answer.
Up till here I've only been discussing errors in an individual measurement.
In your regression analysis, there is one point that know perfectly. You should ensure your line passes through this. For further reading I refer you to http://en.wikipedia.org/wiki/Simple_linear_regression#Normality_assumption.

6. Feb 15, 2014

### BvU

I think the further reading is a bit too much of the good stuff. And my wiggling may not be applicable because the line is so straight.

Dear AJ, could you show some of your work? Describe the instructions for the exercise, show the plot or the measurements? What level/context are we dealing with ?

From my undergraduate practical work I remember the reversible physical pendulum where the length measurement was done with a bloody expensive invar rod + thermomenter + micrometer and still was the biggest uncertainty source. Here Haru's arguments go the same way even stronger.

There is also the finite amplitude issue (the formula comes out when $\sin \theta = \theta$ is valid).

7. Feb 15, 2014

### haruspex

Thanks for the reminder - I did mention it in post #2 but intended to say a bit more in post #5.
You can get an idea of the contribution from this error by expanding sin(θ) to the next term:
$k^2\ddot\theta=−gL\sin(\theta)≈-gL\theta(1-\theta^2/6)$.
However, it's not clear how to turn that into an error term on the period. You can get an upper bound by using the amplitude of the swing: $\theta^2 ≤\theta_{max}^2$. You can then treat it like a fractional error in L.

8. Feb 16, 2014

### BvU

Experimentalists let the pendulum swing for hours (at night in an otherwise empty building) and thus set up an extrapolation graph to correct T (twice: for damping and for amplitude...)

AJ should reveal some more of his experiment: was there a wire or a rod, how was it attached, etc. Where does the 0.0005 for the tape measure come from ? And if you measure L, do you know where to start, too (not at the hook on the tape but at e.g. 0.1000 +/- 0.0005 m; where is L = 0 on the rod/wire pivot point)

9. Feb 16, 2014

### ajcoelho

I'm not in an advanced level. so what it was asked me to do was just an estimative of the value of the uncertainty.

L +/- 0.0005 m came from the tape measure (the uncertainty is half the value of the lowest divisor scale...). I'm assuming the sphere i used in pendulum is a single point so we may not treating it as a rigid object as you were doing.´
I think that the only concern on the estimative that i have to do is related with L +/- 0.0005 m and T +/- 0.001 s (in this last case it is the lowest divisor scale because it's a digital stopwatch).

I'm I thinking right? how will I do an estimative only with these numbers?

10. Feb 16, 2014

### haruspex

If we assume those are the only two sources of error:
Each of those errors has a uniform distribution.
Express each as a fractional error, e.g. if the measured time was t and the error is +/-Δt then the fractional error is +/-Δt/t.
If w = f(x, y), and there is a small change (i.e. error) Δx in x, the change in w, Δw = Δx∂f/∂x.
If the function f looks like Axayb then ∂f/∂x = Aaxa-1yb = aw/x, so Δw = Δxaw/x or Δw/w = aΔx/x. So we take the fractional error in x, multiply by the power that x has in f(x, y), and that gives the fractional error in w. Similarly y. If we knew Δx and Δy we could add these add up: Δw/w = aΔx/x + bΔy/y. But here we only know ranges for the errors, and we presume them to be independent, so they might some times partly cancel and other times reinforce.
So, long story short, if sx is the standard deviation in Δx/x and sy the s.d. of Δy/y etc., we would have $s_w^2 = a^2s_x^2+b^2s_y^2$.
It remains to calculate the standard deviations of the fractional errors in t and L and to determine a and b.
What is the formula for g in terms of t and L? What does that give you for a and b?
Do you know what the standard deviation is for a uniform distribution (look it up on the net if you don't)?

11. Feb 16, 2014

### BvU

Poor AJ. How complicated does it have to be ? You found a slope in the T2 plot, which we expect to be ${4 \pi^2 \over g}$. If your slope has 1% statistical error, so does your g (from ${1\over 1+\epsilon } \approx 1-\epsilon$).

If your tape measure has 1% systematic error (miscalibration), your slope has 1% systematic error (same expression; the slope would be something like $\Delta T^2 / \Delta L$). Statistical errors are somewhat accounted for in the slope uncertainty.

If your stopwatch has 1% systematic error, slope has 2% ( (from $(1+\epsilon)^2 \approx 1+2\epsilon$). Statistical errors are accounted for in the slope uncertainty.

Reaction times and fluctuations therein are a multiple of the lowest divisor scale (try it out with a friend and two stopwatches!), but we could claim they are in the statistical error in the slope.

Similarly, measurement uncertainties of 0.5 mm are pretty hard to achieve! You should realize that in fact you measure the difference of two readings (0 and L) (see earlier posts).

All in all, if you learn from this that experimental physics isn't all that exact a science, this relatively uncomplicated experiment has taught you something valuable. There will be those that cry out "but it is!" and their noses can be rubbed in zillions of counter-examples
Of course we all try to estimate statistical and sytematic errors to the best of our 'current knowledge'. Most of the times that works out reasonably well.

12. Feb 16, 2014

### haruspex

Most of the complication is in combining the two sources to get a single uncertainty estimate. Your last post did not cover that.

13. Feb 18, 2014

### ajcoelho

Exactly. I'm still not getting on how to do it...

14. Feb 18, 2014

### BvU

Show us the plot, or the raw data. Perhaps we can comment on that.
Point is, statistical errors are reduced by doing a fair number of measurements. They end up in the statistical error in the slope. How big was that error ? (r=0.99 tells me it is > 0).

Systematical errors stay around until the final result and end up in the uncertainty thereof.

How big was the 'sphere' ? Did you use a steel wire, a plastic rod, something else ? How was it suspended ?

Difficult to say something that helps you without more information, sorry. You obviously want to do the right thing.

15. Feb 19, 2014

### haruspex

Maybe a worked example will help.
You measure a distance L with a supposedly perfect tape, so the only error, δL, is in rounding to the nearest gradation, giving a uniformly distributed error of +/-ΔL. Likewise, you measure a time interval T with a uniformly distributed error δT in +/-ΔT. From these you wish to compute a quantity A=2L/T2.
The actual distribution of the error δA in A is a bit messy. So we first consider the errors as fractional errors, δL/L, δT/T, then turn these two uniform distributions into Gaussian approximations with zero mean and the right variances, namely, (ΔL/L)2/3, (ΔT/T)2/3.
Now A+δA = 2(L+δL)/(T+δT)2, whence δA/A ≈ δL/L+2δT/T. Those last two terms are independent and we've made them Gaussian random variables, so var(δA/A) = var(δL/L)+2var(δT/T) = (ΔL/L)2/3 + 2(ΔT/T)2/3. One standard deviation for δA/A is the square root of that.
Going back to the original problem, you plotted a straight line for L2 against T, making g the slope, yes? Most tools for linear regression suppose X is measured exactly and Y is where the uncertainty lies. That doesn't apply here: you have uncertainty in both. There are methods for that, but I'd need to look them up.

But as BvU and I have indicated, the errors discussed above are likely to be trivial compared with systematic errors arising from failing to take into account that the mass centre and radius of inertia of the pendulum are other than you suppose.