How to calculate the errors in the measured period of a pendulum?

[Lotto]

Homework Statement

I measured 20 periods of a pendulum and did it 10 times, so from that I calculated a statistical error $\sigma_A$. Let's say that my reaction time is 0,3 s. What is the total error of the measured period $T$?

Relevant Equations

$\sigma_T=\frac{1}{20}\sqrt{\sigma^2_A+\sigma^2_B}$

So let's say that my reaction time is 0,3 s, so is then a systematic error $\sigma_B$ for 20 periods 0,6 s? Beacause first I had to start a stopwatch and then stop it. So then $\sigma_T=\frac{1}{20}\sqrt{\sigma^2_A+\sigma^2_B}$?

 

[kuruman]

How did you calculate $\sigma_A$ and what experimental quantity is it due to?

The relevant equation that you quoted assumes statistically independent uncertainties. Let's say that your reaction time is zero. How would that affect $\sigma_A$?

BTW, I hope that when you timed the 20 periods, you started and stopped your timer as the string was passing through the vertical. Do you see why?

 

[Lotto]

How did you calculate $\sigma_A$ and what experimental quantity is it due to?

The relevant equation that you quoted assumes statistically independent uncertainties. Let's say that your reaction time is zero. How would that affect $\sigma_A$?

BTW, I hope that when you timed the 20 periods, you started and stopped your timer as the string was passing through the vertical. Do you see why?

Well, $\sigma_A$ is just a standard deviation for $20T$.

I am not sure how to deal with the reaction time when calculation the final error. From a statistical point of view, is my equation above correct? Because the mean value of $T$ in my case is 0,94 s, so when I first calculated the error as $\sqrt{(\sigma_A /20)^2+\sigma^2_B}$, the error was of course huge ...

 

[kuruman]

Question: Why did you time 20 oscillations and not just 1?
Answer: So that you can divide the time you measured by 20 (which is exact) and get a value for the period $T$. This minimizes the effects of your reaction time which is a smaller fraction of 20 T than T. If you measured T by timing one period So find the standard deviation for the 10 measurements of $T$.

Please fill in the blanks.

Quantity $\sigma_A$ is the uncertainty in the period because of the uncertainty in __________ .

Quantity $\sigma_B$ is the uncertainty in the period because of the uncertainty in __________ .

 

[haruspex]

There are two significant sources of error in this experiment: your reaction time (which varies) and the granularity of the measurements.
We can take your reaction times as normally distributed about some nonzero mean. The systematic error is that mean*.
Is there any reason your reaction time distribution for start would differ from that for stop?
If not, what does that imply about the effect of the systematic error on your result?

*Edit:
There will also be a systematic error from the granularity, but it is complicated.
First, is the value read for 20T rounded down or to nearest? If rounded down then there is systematic error of half a measurement grain (plus a random error distributed evenly around that).
Secondly, consider what would happen if there were no other error and the actual value for 20 swings is constant at 10.966, say, and the measured value is to two decimal places, rounded to nearest. The value read would be consistently 10.97, a systematic error of +.004.
However, if we now introduce an error from some other cause of $\pm 0.1$, smoothly distributed (uniformly, say), when you average over a number of runs the systematic error of +.004 disappears! (I have exploited this.)
So assume until proven otherwise that any systematic error from a rounding to nearest is too small to matter.

 

[kuruman]

We can take your reaction times as normally distributed about some nonzero mean. The systematic error is that mean.

I think that "reaction time" is a misnomer. "Synchronization time" would be more appropriate. If this experiment is done by operating a timer manually, the experimenter is not reacting to anything. All one needs to do is start and stop the clock when the pendulum crosses and recrosses the vertical after $N$ oscillations. One can anticipate when that is going to happen with good accuracy especially if one has extensive experience playing video games.

If the timer is started at $t=t_0$ as displayed by a wall clock, we expect that the actual starting time of the timer would be at clock time $t_{\text{start}}=t_0\pm\Delta t$ where $\Delta t$ is what I call synchronization time. We can assume that it is a normally-distributed random error with a mean value of zero (not to be confused with $t=0.$) The same considerations apply to the stopping time and there is no reason to believe that the stopping time $\Delta t$ has a different distribution.

However, there could be a systematic error component to $\Delta t$ due to parallax if the observer lines up the string with the vertical support rod when the line of sight is angled with respect to the perpendicular to the plane of the motion.

I can only speculate without the knowledge of exactly how the 20 pendulum oscillations were measured by the OP.

Edited for typos and clarifications.

 

[haruspex]

I think that "reaction time" is a misnomer. "Synchronization time" would be more appropriate.

Fair enough, but I don’t think that affects my argument because I disagree with:

We can assume that it is a normally-distributed random error with a mean value of zero

It is entirely possible that the experimenter has a consistent tendency to over- or under-anticipate. My point was that such a systematic error does not matter if it is the same for starting and stopping.
OTOH, as discussed in my edit to post #5, systematic error from the measurement granularity is trickier.