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My physics textbook states that in measuring the time period of a pendulum it is advised to measure the time between consecutive passage though the mean position in the same direction. This results in better accuracy than measuring time between consecutive passage through an extreme position.

Why is one method of finding the time period better than the other? How can this affect the accuracy of the final result? I think it shouldn't make any difference, as the time taken for one full oscillation is independent of the choice of the start/end point. Is the statement in the book really correct?

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    $\begingroup$ The best way to eliminate error is to time a large number cycles (not trying to time one cycle accurately). I was taught to do this as follows: time 10 cycles (say 1s each); calculate roughly how long 50 cycles will take (50s); come back after 50s and time exactly when a cycle completes. Use the first timing to work out how many cycles there have been and improve the accuracy of timing of a single cycle by x5. Repeat with 250 oscillations, and you so on. Our lecturer had actually used this technique when his research team discovered the first pulsars. $\endgroup$
    – Martin Bonner supports Monica
    Nov 28, 2019 at 13:08
  • $\begingroup$ @MartinBonnersupportsMonica, Thanks. Actually, my question was not about the number of oscillations taken into account, but where to start/end the oscillation - whether at the mean or at the extremes. If you could add this, then if you wish you may type your comment as answer so that users can vote. $\endgroup$
    – Vishnu
    Nov 28, 2019 at 13:30
  • $\begingroup$ Yes, I realize what your question was about. My comment was basically "it doesn't matter". $\endgroup$
    – Martin Bonner supports Monica
    Nov 28, 2019 at 13:38

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The issue with measuring at the ends is that the pendulum “dwells” at the end point as it turns around, so that there is a greater spread of time for which it looks “at the extreme point” than for which it looks “at the midpoint”.

This spread of time introduces error, whether you are triggering a stopwatch by hand or having some kind of sensor make the decision.

If you tried recording all of the points and determining the period, you’d get the same result — either you take the time between zero crossings of the position, or you take the time between peaks. To get the time between peaks, you need to take a derivative of the position measurement and find its zero crossings, and taking the derivative of a set of data points introduces error.

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  • $\begingroup$ Is this true, that the pendulum "dwells" at the top of the swing, much like the cartoon Wile E. Coyote who has a moment to think before he falls off a cliff? Although the top of the swing would be a parabola, I think there would be a definite top point without any dwell. $\endgroup$
    – foolishmuse
    Nov 27, 2019 at 17:31
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    $\begingroup$ It’s true to the extent that there is a period of time for which the velocity is close enough to zero that the pendulum’s changes in displacement are below the threshold of whatever sensor you’re using to measure position. $\endgroup$
    – RLH
    Nov 27, 2019 at 17:38
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    $\begingroup$ @foolishmuse There will indeed be a definite top point. However, if you have any position uncertainty in your sensor, the period of time where the pendulum is within this region of uncertainty is much longer as the velocity approaches zero at the ends. It transitions quickly through it in the center. $\endgroup$
    – Cort Ammon
    Nov 28, 2019 at 5:24
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    $\begingroup$ @foolishmuse: There is a definite top point, but it's harder to measure or perceive. The "dwell" is a matter of imprecise observation or measurement; rather than an actual physical fact. $\endgroup$
    – Flater
    Nov 28, 2019 at 12:10
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    $\begingroup$ For a good illustration, try the same problem in space instead of in time. Plot the graph of a sine function (with x and y scaled equally and coordinate axes not drawn at 0) but using a line slightly thicker than you'd prefer. Try to mark the points where the function takes the values 0 and 1 by hand using only a ruler. The zeros will be easy, the ones will be surprisingly unclear. $\endgroup$
    – mlk
    Nov 28, 2019 at 14:18
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To measure the time period it is useful to use a fiducial (reference) mark which in this case could be a vertical line drawn on a piece of card and placed “behind” the pendulum bob/string.

It is assumed that the time for a number of complete oscillations will be measured to enable one to find a more accurate value of the period than just measuring the time of just one complete oscillation.

If the fiducial mark is placed at an extreme of the motion of the bob then one can estimate when the bob reaches that mark to start and stop the timer.
However because the amplitude of motion of the bob will decrease with time the estimation of exactly when the bob stops will become progressively more difficult as the position at which this happens can only be guessed.
Even a small error in the location of the position of the bob will result in a relatively large error in the timing because the bob/string would be moving slowly.

Putting the fiducial mark at or near the centre of an oscillation does not require any estimate of when the bob (or string) passes across the fiducial mark as the bob/string will always pass the fiducial mark. Also because the speed of the bob/string is a maximum at this position the error in taking a reading when the bob/string is not quite passing the fiducial mark is going to be relatively small.

An order of magnitude calculation to estimate the possible error in the measurement of a time interval when the bob has not quite reached the fiducial mark by $1\,\rm mm$ at the centre and extreme of a swing..

A simple pendulum of length of $1$ metre has a period of approximately $2$ seconds so $\omega \approx 3 \,\rm s^{-1}$.
For an angular swing of about $5^\circ$ the amplitude of motion is approximately $160$ mm.
Approximating the motion of the bob to a straight line gives an equation for the displacement in millimetres $x = 100\sin (3\,t)$ where the displacement is zero when time $t=0$.
For the bob to move from $x=0\,\rm mm$ to $x=1\,\rm mm$ takes approximately $0.003\,\rm s$ and to move from $x=99\,\rm mm$ to $x=100\,\rm mm$ takes approximately $0.05\,\rm s$.

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  • $\begingroup$ You could put the fiducial mark anywhere on the swing, but near the center is the maximum velocity. The extrema will lower as energy is removed (friction ...) so placing the fiducial mark at the equilibrium position lets you average over many swings, say 20 passes of the equilibrium. $\endgroup$
    – TazAstroSpacial
    Nov 28, 2019 at 6:30
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The reason is that it is (supposedly) marginally easier to judge when the pendulum sweeps past a point when it is moving quickly than when it is moving slowly. In think the point is debatable if you are judging it by eye and manually triggering a stop watch, as your own reaction time comes into play. What's much more important is to measure the elapsed time taken for a very large number of oscillations together, rather than individually, and take the average--that way you can greatly reduce the error associated with judging the exact start and end points.

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    $\begingroup$ Reaction time shouldn’t really matter. Assuming you have the same reaction time for each pass, you would be offset by the same amount. Although you may start the timer late, you would end it an equally late amount, so it shouldn’t actually effect the elapsed time you’re trying to measure. $\endgroup$
    – JMac
    Nov 27, 2019 at 17:20
  • $\begingroup$ Reaction time error can be minimized by measuring the time for several (small) swings. Better yet, use an electronic timer with a light beam sensor. $\endgroup$
    – R.W. Bird
    Nov 27, 2019 at 18:26
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    $\begingroup$ Reaction time isn't an issue for manually timing a regular periodic motion like a pendulum. A typical human reaction time is about 30 ms, and it limits timing unpredictable events. However, most students can time a sequence of pendulum oscillations with a variance of just a few ms after a bit of practice. $\endgroup$
    – Stephen C. Steel
    Nov 28, 2019 at 16:24
  • $\begingroup$ "Assuming you have the same reaction time for each pass" Hmm. $\endgroup$
    – Lightness Races in Orbit
    Nov 29, 2019 at 17:28

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