The following is a 2018 F=ma exam question. I know that this isn't a homework site, but I think that my question is conceptually relevant. Here's the problem:

A group of students wish to measure the acceleration of gravity with a simple pendulum. They take one length measurement of the pendulum to be $l = 1.00 \pm 0.05\ \rm m$. They then measure the period of a single swing to be $T = 2.00 \pm 0.10\ \rm s$. Assume that all uncertainties are Gaussian. The computed acceleration of gravity from this experiment illustrating the range of possible values should be recorded as

A) $9.87 \pm 0.10\ \rm{m/s^2}$

B) $9.87 \pm 0.15\ \rm{m/s^2}$

C) $9.9 \pm 0.25\ \rm{m/s^2}$

D) $9.9 \pm 1.1\ \rm{m/s^2}$

E) $9.9 \pm 1.5\ \rm{m/s^2}$

I know that the pendulum period formula is $T = 2\pi\sqrt{l/g}$, and therefore, that $g = \frac{4\pi^2l}{T^2}$

I don't know much about this error stuff, so I went about it in a fairly straightforward way and I still can't seem to find the flaw:

Obviously the biggest value for $g$ will be when $l$ is biggest and $T$ is smallest. The biggest $l$ is $1.00 + 0.05 = 1.05$. The smallest $T$ is $2.00 - 0.10 = 1.90$. $4\pi^2\times 1.05/(1.90)^2 = 11.48$.

The smallest value for $g$ is when $l$ is smallest and $T$ is biggest. Going through the same process, you get $g = 8.50$. The difference between those two values is $2.97$ which is about $3$. Therefore, the error has to be something $\pm 1.5$, which would make the answer E. It is not E, but D.

The solution on the F=ma website said complicated things about adding squares of error and things I haven't learned.

So my question is what is wrong with my logic here?

  • $\begingroup$ Acvording to what you have provided, the answer is indeed E and not D $\endgroup$ Jan 11, 2019 at 0:43

3 Answers 3


The problem with your method is that it implicitly assumes the errors are (anti)correlated. Well, what does that mean?

It means you assume that when the length fluctuates high (low), then the period measurement must fluctuate low (high). Since a larger (smaller) length and a shorter (longer) time both increase (decrease) the value of $g$, the fluctuations in $g$ are overestimated.

The question then is, how do you add uncorrelated errors? Answer: in quadrature. The next question is, why in quadrature? You could write books on that, but in short, it's the Pythagorean theorem:

$$ a^2 + b^2 = c^2 $$

which is exactly how you add uncorrelated lengths.

A somewhat loose explanation is, suppose:

$$ C = AB $$

and you measure:

$$ A \pm a $$


$$ B \pm b $$

What is the uncertainty in $C$?

Well, gaussian errors are distributed as:

$$ P_a(a) \propto e^{-a^2} $$


$$ P_b(b) \propto e^{-b^2}$$

so that:

$$ P_c(a, b) \propto P_a(a)P_b(b) $$ $$ P_c(a, b) = e^{-a^2}e^{-b^2} = e^{-(a^2+b^2)}\equiv e^{-c^2} =P_c(c) $$

where the fluctuations in $C$ given by those in $A$ and $B$ added in quadrature:

$$ c = \sqrt{a^2 + b^2}$$

  • $\begingroup$ But when you want to describe the range of possibility, don't you want to consider the worst case? $\endgroup$ Jan 11, 2019 at 15:22
  • $\begingroup$ @QFTUNIverse When you quote a 1 sigma error bar made from 1 sigma errors, you want exactly that: a 1 sigma variance--not the worst case. Now I have worked projects were we have included 99-percentile cases (1% chance), but then you're doing quantiles, and it's a bit more complicated. BTW: on 2nd thought, my "loose" explanation is too loose (unless $A=B$). $\endgroup$
    – JEB
    Jan 11, 2019 at 17:12

When you give a measurement result with an extended standard uncertainty, you set a probability that the "true" value is in your range. For example, the probability that the true value of $l$ is between 0.95 and 1.05 is 95%. Or, the probability that the value is not in this range is 5%. Same idea for $T$.

As a result, your calculation is too pessimistic. You put yourself in the situation for which the errors would be maximum on both $l$ and on $T$ which is "unlikely". Without accurate calculation, you can understand that the probability that the true value of $g$ is outside your range is less than 5%. This is the reason why, most often, it is the squares of uncertainties that are added.


An error equation is obtained from the differential of the equation. Take the differential operator of both sides and you get:

dg = 4 pi^2 (dI/T^2 - 2IdT/T^3)

dg/g = dI/I - 2dT/T

As we are after the magnitude of errors, the negative sign gets switched for a plus. There's a bit more to it but with uncorrelated guassian errors it works out smoothly. The equation is then one of relative errors.

  • $\begingroup$ As I said in my question, I know basically nothing about error propagation. My question was less about the proper solution and more about what is wrong with the method I was using. $\endgroup$ Jan 11, 2019 at 1:08
  • $\begingroup$ Methodology wise you were looking at maximum and minimum errors. The key is that the errors are guassian so don't have a min/max but a variance. $\endgroup$ Jan 11, 2019 at 2:05

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