# Homework Help: Estimating time intervals

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1. Oct 29, 2016

### alisdfd

1. The problem statement, all variables and given/known data
Question:
A student used a digital stopwatch three time 10 oscillations of a pendulum. The timings were 13.52s, 13.64s and 13.58s
a) Calculate i the average time for 10 oscillations, ii the time period of the oscillations.
b) Estimate the accuracy of the timings, giving a reason for your estimate.

The problem i have is with part b. In maths to estimate is round up ideally to 2 sf.(my answer is 1.4 to s.f)
if i give my estimate my answer how do i give a reason. I feel to give my estimate to 2 sf because it seems to be the standard in physics (i could be wrong)
but also knowing that gate timers and other forms of measuring time make be more effective than a stopwatch am i suppose to estimate the time on those and give it as my answer since it will be more accurate?

I just a bit confused and need a second opinion. This very easy but need someone to please clarify this for me as I do not full understand and I have started to learn physics

Thank you

2. Relevant equations

3. The attempt at a solution

2. Oct 29, 2016

### BvU

Hello Alis,

Please be a bit more complete. The average of 13.52s, 13.64s and 13.58s is easily calculated to be 13.58 s. Your
is a total misconception. Such a standard does not exist at all. What you do is quote some result and a best estimate for the accuracy. The number of digits is determined by the accuracy (and the number of digits for the accuracy is most often one, and sometimes 2 but then it's based on an extensive error analysis and considerable statistics). Example: 314.15 s $\pm$ 0.08 s.

To go from the average for 10 oscillations to the value for a single oscillation you divide by 10, so you get 1.358 s.

Now for the error. Statistics is poor with only three measurements, so you need some judgment.
First of all you want to ask: what do I expect? You expect that every single full oscillation takes the same amount of time. Not trivial, but acceptable if your measurements aren't hours apart without re-starting the pendulum.

There are two contributions: the error in the 13.58 and the error in the 10. Difficult to distinguish, but a coarse calculation will probably show the error in the 10 can be ignored. But you still have to mention that in your report.

The deviations from average are 0.04 s on average. Three out of three measurements fall within 0.06 s of the average.
As I said, statistics isn't really in order here; but perhaps you already know that the error in an average from n measurements is approximately $1/\sqrt n$ or $1/\sqrt{n-1}$ times smaller than the error in one measurement, so 0.1 s is a good guess for your stopwatching accuracy on a single measurement (pretty good work !) and that would mean 0.01 s for the period of 1 oscillation. Note the word guess: not a wild guess, but also not a strict and secure determination. So just one significant figure is the best that can be offered here. Saying the error is 0.0094 s is nonsense (you don't know it that well) and saying 0.02 s to be on the safe side is short-selling the accurate work you did.

A long story, but I hope it helps. Don't worry if some of it is vague. And I hope you have lots of fun in learning physics !

3. Oct 30, 2016

### alisdfd

thank you very much

4. Oct 30, 2016

### haruspex

As I read the question, there is no mention of the time for one oscillation. We are only interested in the estimated time for ten.