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I've tried measuring acceleration due to gravity, using a photogate with free falling picket fence.

I've also tried photogate with picket fence on a low-friction cart and inclined plane.

So far, the latter has provided slightly less varied results, but in both cases I've found it very difficult to reduce the variation in my results.

What other experiment can be setup at home, to produce more consistent results?

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    $\begingroup$ Try using a simple pendulum or try a video of a falling object with a rule in the frames.. $\endgroup$
    – Farcher
    Jul 7, 2019 at 7:31
  • $\begingroup$ @Farcher I read the title and was about to post the same comment :) $\endgroup$
    – user207455
    Jul 7, 2019 at 7:33
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    $\begingroup$ It depends on how much time and money you would want to put into it. Kater's pendulum can give within 0.1% accuracy. $\endgroup$
    – user234190
    Jul 7, 2019 at 9:36
  • $\begingroup$ If repeat the experiment many times and average the results you will get a better estimate. $\endgroup$ Jul 7, 2019 at 13:14
  • $\begingroup$ The way graviometric surveys were done before precision, solid-state accelerometers was with pendulums. $\endgroup$ Jul 8, 2019 at 15:32

4 Answers 4

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enter image description here Using this pendulum and a tape measure plus cell phone stopwatch we calculated $g=9.7\pm 0.4 m/s^2$ using the formula for period length $T=2\pi \sqrt{ l/g}$ to calculate $g=l(2\pi/T)^2$.

With $l=1.38m$ and $T=(23.72/10)s$

Hints: period length is independent of starting point so just count several periods and stop the time for those cumulatively. Also this formula only holds for small angles so keep your starting angle below 10° or so.

A good thing about this pendulum is that the rock is pretty heavy which mitigates air resistance a bit (the wind does not help though...).

Improvement ideas: using a smaller starting angle both helps reduce air resistance, because it's proportional to velocity (to some power...), as well as improve accuracy because of the small angle approximation. Downside of smaller angle is more out of plain of swing motion though, so that is bad.

(Error estimate sponsored by @Paul T assuming bad reaction time of 0.3s on both ends of measuring $T$ and 2cm inaccuracy on measuring $l$)

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    $\begingroup$ Nice ans, lovely pic; here are ideas for improved precision. 1. Don't use a tree. 2. Maybe go indoors to avoid wind. 3. To determine length, use several different lengths and measure their difference accurately. Then by plotting all data you can deduce the unknown offset in the length. 4. Use of a video camera which sees both the pendulum and an accurate clock nearby would greatly improve timing. This could be regarded as "cheating" because it uses fancy equipment (i.e. a video camera) but I mention it anyway. This comment is not really for you; it is for anyone who wants to take it further. $\endgroup$ Aug 10 at 14:54
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Put up a 10m (measured precisely) length of cheap pipe. Fill it with distilled water. Measure the pressure at the bottom. (Amazon has sufficiently accurate gauges for about $13) For extra accuracy, measure the temperature and correct the water density.

$$p_{gauge} = (\rho_{water} - \rho_{air}) g h$$

Error estimate:

  • It's easy to measure ~10m to 1cm, and there might be a +/- 2mm meniscus to estimate. Call that 1.5cm total: 0.15%
  • Water density changes by 0.1% / 10C; call that .1%
  • Air density starts as a small effect, and small variations don't matter; negligible error
  • \$10 analog gauges have 2% divisions. Reading to a quarter of a division would be a 0.5% precision, but those don't have a quoted accuracy. Seems unlikely to be that good. \$13 gets you a digital gauge with quoted 1% accuracy. \$300 gets you a Bluetooth sensor that's quoted to 0.2% for this application.

The pressure measurement dominates by far, so 1% accuracy seems within reach for a reasonable cost.

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  • $\begingroup$ Pics or it did not happen. (And numbers!) Up voted though, as I would really love to know how accurate one can get that practically^^, $\endgroup$
    – Kuhlambo
    Aug 7 at 8:35
  • $\begingroup$ Added an error estimate. No pics: I've heard of this being done in undergrad physics labs, but I've never seen it done. $\endgroup$ Aug 7 at 11:48
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As suggested by @Kulambo's answer and @Farcher's comment a pendulum is a great method. There's a long history of pendulum gravimeters dating back to the 1600's.

The standard period of a pendulum $T = 2\pi \sqrt{L/g}$ depends on the small angle approximation, but is independent of the mass of the pendulum bob. The uncertainty in your determination of $g$ will depend on how well you can measure $L$ and $T$.

$$\delta g = g \sqrt{ \left(\frac{\delta L}{L}\right)^2 + 2\left(\frac{\delta T}{T}\right)^2}$$.

The small angle approximation $\sin\theta \approx \theta$ is good to about $2$% for $\theta = 20^\circ = 0.35$ rad. So keep your swing angle smaller than that.

Remember $L$ is the length from the pivot to the center of oscillation of the pendulum. With a light string and a small, heavy bob this will be very close to the center of mass of the bob. It is better to fix the string in place with a clamp, instead of tying it. This way it won't slide, so there won't be much frictional loss at the pivot. It's also easier to precisely locate the true pivot.

To better determine $T$, you can take the average of many swings. The biggest uncertainty will be your ability to press start and stop on your stopwatch at the correct times. If you time the pendulum making five complete swings then divide by five, that effectively cuts the start/stop error by a factor of five.

A $1$ m pendulum should have a period of about $2$ sec. Lets say you determine the pivot to center of oscillation length, $L$, to $1$ cm. Then you time five swings ($5T = 10$ sec) with $0.5$ sec of start/stop uncert. This will get you a determination of $g$ with better than $10$% precision.

$$\delta g = (9.8\,\mathrm{m/s}^2) \, \sqrt{\left( \frac{0.01\,\mathrm{m}}{1.00\,\mathrm{m}} \right)^2 + 2 \left( \frac{0.1\,\mathrm{s}}{2.0\,\mathrm{s}} \right)^2} \approx 0.7 \,\mathrm{m/s}^2$$

The limiting factor will likely be your ability to use the stop watch. You could do much better using your photogate. If you can get down to $0.01$ sec of start/stop precision on one swing, the measurement uncertainty will get close to $1$%. This would make the systematic error from the small angle approximation in the pendulum model take over as the dominate source of uncertainty.

The lack of friction and air resistance in the model may also be a factor if you are trying to get to the $1$% level. Historically, precision pendulums were operated in evacuated, low pressure chambers to reduce air resistance. Candela et al 2000 claim that one can measure $g$ to parts in $10^4$ using a Bessel pendulum. They also discuss how combining data from both pivots of a reversible pendulum causes the air resistance errors to cancel when calculating $g$. (The article is paywalled, but you can probably get it with the help of your local library.)

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  • $\begingroup$ You can use the AGM to calculate the true period from the small angle period. See physics.stackexchange.com/a/718837/123208 $\endgroup$
    – PM 2Ring
    Aug 7 at 0:50
  • $\begingroup$ @PM2Ring I was thinking about that. It would be cool to throw all the powers of theory at my stupid stone pendulum with a stopwatch combo^^, i would even try some estimate for friction loss, but alas, that is beyond my energy for the moment. $\endgroup$
    – Kuhlambo
    Aug 7 at 7:56
  • $\begingroup$ @Kuhlambo The damped pendulum equation is a bit trickier. The AGM is easy to calculate. ;) FWIW, amateurs have built pendulums with enough precision to detect the tidal variation, but your "stopwatch" needs to be an atomic clock. :) See the links in my answer here for details: physics.stackexchange.com/a/534306/123208 $\endgroup$
    – PM 2Ring
    Aug 7 at 8:18
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    $\begingroup$ My PhD advisor liked to say that the way to measure anything precisely was to figure out how to convert it into a frequency. Once you have an oscillator, you can measure its frequency more precisely just by observing for longer — which is also how you search for the class of systematic effects which cause the frequency to drift over time. $\endgroup$
    – rob
    Aug 7 at 10:27
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    $\begingroup$ "Remember L is the length from the pivot to the center of mass of the pendulum" - not quite right, only true for bobs of infinitesimal size. You want the "center of oscillation" which is below the center of mass for an extended size bob. It's easiest to see it when thinking of angular momentum: The I of the pendulum is mL^2 plus the I of the extended bob itself, so restoring torque takes longer to pull the bob back to center: Longer period, lower g. $\endgroup$ Aug 7 at 14:03
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One can use various smartphone sensors to measure $g$. See, e.g.,The Physics Teacher 59, 584 (2021); https://doi.org/10.1119/5.0056573

Abstract:

The use of sensors in smartphones to do physical experiments is a boom in the past decade, such as acceleration sensor,1,2 light sensor,3,4 magnetometer,5,6 camera,7,8 and gyroscope. However, few people study the application of proximity sensors in physical experiments, although, in our view, the employment of the proximity sensor is more accurate than other sensors (see Table I). For further research, this paper proposes a method to measure the oscillation period of a simple pendulum based on the proximity sensor integrated in the smartphone and determines the experimental value of the gravitational acceleration.

EDIT (8/6/2022): The cited article provides data on errors for $g$ as measured by 1. acceleration sensor; 2. light sensor; 3. magnetometer; 4. gyroscope: 1.40%; 0.71%; 0.20%; 0.94%, respectively ($g$= 9,66; 9.72; 9.77; 9.70 m/s^2, respectively).

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  • $\begingroup$ But, can you do better than me with that? probably, let's see some numbers. ^^ $\endgroup$
    – Kuhlambo
    Aug 6 at 18:23
  • $\begingroup$ @Kuhlambo : Please see the EDIT to my answer $\endgroup$
    – akhmeteli
    Aug 6 at 18:48
  • $\begingroup$ Yeah, I know but that's not as much fun, it's it? ^^ but don't worry you'll get the bounty if nobody else makes an effort. ;) $\endgroup$
    – Kuhlambo
    Aug 6 at 18:53
  • $\begingroup$ Isn't this like saying "You can measure $g$ using this magic $g$-measuring device"? $\endgroup$ Aug 10 at 14:45
  • $\begingroup$ @AndrewSteane : The OP asked: "What other experiment can be setup at home, to produce more consistent results?" I assume the OP has a smartphone at home, but I don't know what kind of a smartphone they have, and I cannot give detailed instructions. So I just hope that my answer may help the OP realize that their smartphone can have the capabilities they need. $\endgroup$
    – akhmeteli
    Aug 10 at 17:33

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