I'm a high school science teacher, my primary degree is in Physics, so I have a solid grasp of the background. I'm running into a strange issue with a pendulum lab I had my students complete.

We're using iOLab devices (http://www.iolab.science/) as our pendulum and I'm having the students measure the time by reading the $\Delta t$ from the graph generated by the software. We set the software to record the accelerometer data, and the y-axis data makes a nice sine wave that corresponds to the motion of the pendulum. I've tested it with the force sensor as well, and the same wave appears. To reduce errors, I instructed the students to find the $\Delta t$ for three complete waves, and then we divide by three, to average them.

I tested it myself trying to eliminate errors, and I've encountered the same following issue as the students. When we plug the length (in $m$) and the period (in $s$) into the equation for the period of a pendulum (solved algebraically for the acceleration due to gravity), we get a value approximately 4 times higher than the known value. I've checked and double-checked the algebra, and verified that I'm entering the numbers correctly into the calculation. When I plugged the length and the known value for the acceleration due to gravity into the equation to find the period, I got a calculated period that was about 3 times what we measured.

I cannot figure out anything that could be causing that much of a difference, unless the acceleration due to gravity at our school is different than the rest of the planet.

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    $\begingroup$ Please provide some actual data. $\endgroup$
    – Farcher
    Mar 1, 2021 at 23:43
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    $\begingroup$ I’ll bet that if you tied a fishing weight to a string and used a clock to measure the period, it would work fine. $\endgroup$
    – G. Smith
    Mar 1, 2021 at 23:46
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    $\begingroup$ The link says the ioLab can measure force. If you are measuring the centrifugal force the pendulum exerts on the support, it reaches a mex twice per pendulum period. Likewise, velocity reaches a max twice per period. Are you making a mistake like that? $\endgroup$
    – mmesser314
    Mar 2, 2021 at 0:10
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    $\begingroup$ using a simple weight on a string and a clock won't work Yes, it certainly will. Don’t pretend that a low-tech approach won’t work just because you prefer to have your students use a high-tech approach. $\endgroup$
    – G. Smith
    Mar 3, 2021 at 1:58
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    $\begingroup$ @G.Smith, when I taught high school physics, I liked the low tech approach at least as much as the high tech approach because the students didn't have to learn the technology AND the physics at the same time. And believe me, "stone age" equipment can get fairly creative and fairly accurate depending on how it is used. $\endgroup$ Mar 4, 2021 at 1:47

1 Answer 1


One of my mentors likes to say, “Never make a measurement unless you already know the answer.” By which he meant that, any time you are building a new measurement instrument, which is what most experimental apparatus is, you’re going to discover several times that what it’s measuring isn’t quite the same as what you thought it was measuring. Here you have a new data-acquisition-and-analysis pipeline that’s giving you unexpected results; you are reasonably assuming that some things in the pipeline are different than you believed when you built them.

Here you seem to have a pendulum of length $\ell$, operated on Earth where the gravitational acceleration is $g$, which you predict should have period

$$ \tau = 2\pi \sqrt{\ell/g} $$

but you are observing $\tau_\text{calc} \approx \tau_\text{measured} \cdot 3$. You’ve also solved this for the gravitational acceleration,

\begin{align} g &= \left(\frac{2\pi}{\tau}\right)^2 \ell, & \text{but }g_\text{measured} &\approx g_\text{reference} \cdot 4. \end{align}

Those two statements taken together are a little surprising: you would predict $(\tau_\text{calc} / \tau_\text{measured})^2 = (g_\text{measured}/g_\text{reference})$ with exactly the same data involved, but one doesn’t ordinarily get $3^2 \approx 4$ with just one algebra error. I can imagine a few reasons why that might happen, but I don’t think it’s productive for me to speculate.

A common problem for intro-level students is “frequency doubling”: for some people it seems like the “period” of a signal should be the interval between zero-crossings, rather than the interval between peaks or between crests. In a comment, mmesser314 suggests

The link says the ioLab can measure force. If you are measuring the centrifugal force the pendulum exerts on the support, it reaches a max twice per pendulum period. Likewise, velocity reaches a max twice per period. Are you making a mistake like that?

to which you respond with interest. Note that halving your period would give $(\tau_\text{calc} / \tau_\text{measured}) = 2$ and $(g_\text{measured}/g_\text{reference}) = 4$.

You write in a comment that you are reluctant to do the experiment with a stopwatch because your digital timer is more sensitive than human reflexes and a stopwatch. This is a statement that you want to be able to quantify. I like to have my college students measure their own reaction times, by having a proper clock generate some noise at a known interval and letting them measure it. I find typical stopwatch reaction times of about a quarter-second, and jitter of about a tenth of a second. (The athletic kids tend to be better at stopwatches; apparently it, too, is a skill.). If you are getting timing discrepancies that are a factor of three, you don’t need a high-tech DAQ system: you should be able to tell whether the pendulum is really oscillating at 3 Hz, 1 Hz, or 0.3 Hz by looking at it and counting “mississippi-one, mississippi-two.”


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