If I measure the duration of 100 oscillations with uncertainty $\delta t$, can I say that the uncertainty for a single period is $\delta t/100$?

Question

I am really very annoyed by how there are multiple perspectives about how we treat uncertainty in measurements and I want to get rid of the misconceptions once and for all.

The Problem

I was finding the value of 'g'(acceleration due to gravity) using Kater's pendulum.

I measured the time taken for 100 oscillations of the pendulum with a stopwatch of least count 0.01 seconds, let's call it $t_1$ I obtained the time period of the pendulum by the relation, $T_1=t_1/100$

While finding the relative error in g I have to know the mean absolute error in $T_1$ Professor says that $\delta T_1=\delta t_1/100=0.0001 \: \rm s$

My argument, $\delta T_1=0.01 \: \rm s$ since the measurement is done only for 100 oscillations and you can't really consider the error in 100 oscillations as 100 times the error in 1 because if I were to take one oscillation I would never get such small error because of the low precision of stopwatch.

Professor says that taking 100 oscillations enables us to have a smaller error in the result.

But I feel this is wrong somehow

Question

Even if we consider an ideal situation where the only error in measurement is due to the least count of the stopwatch, should you be able to measure a value of much higher accuracy than that enabled by the precision of the instrument?

Answer 1

Let's say for example the period is around 1 second and you measure for 100 periods, with a timing precision of $0.1$ seconds. So your $\delta t_1 = 0.1$ s. I claim you then know the period to a precision around $\delta t_1/100 = 0.001$ s. That's 1 millisecond. And this seems surprising to you.

But suppose we take your proposal that really we only know the period to about $0.1$ s accuracy. On this argument there is a good chance that the period is somewhere between $0.9$ s and $1.1$s. If this were so then 100 oscillations would take between $90$ and $110$ seconds. But your measurement is much more accurate than that! You already know that 100 oscillations takes not 90 nor 110 seconds but somewhere between $99.9$ and $100.1$ seconds. So it must be that you can now deduce the period to a precision around 1 millisecond.

Here is another example. I walk along a wall, counting bricks. And let's assume the bricks are all the same size as each other. I count 100 bricks. Say the whole wall has a length 20 metres plus or minus 10 cm. So how accurately do I now know the length of 1 brick? The answer is that I now know the length of one brick is 20 cm plus or minus 1 millimetre. For if the bricks each had a length of, say 21 cm, then after 100 bricks the total length would have been 21 metres. But my measurement has ruled that out.

All this is just repeated examples to help you see that it really does make sense that if $T = t/100$ then $\delta T = \delta t / 100$. This concept is used throughout experimental physics, because to measure some quantity $A$ we almost always find ourselves measuring some other quantity $B$ which is proportional to $A$. So if the relationship is $A = kB$ where $k$ is the proportionality constant then the precision $\delta B$ in $B$ gives a precision in $A$ of $\delta A = k \; \delta B$. This is so common a situation that there is hardly any published experimental precision estimate that does not make use of it at some stage.

Answer 2

Your professor's position is correct. If you measure $T_1$ over $N$ periods to $t_1=NT$, then the uncertainty $\delta t$ is only for the combined duration, and it gets divided by $N$ to give the uncertainty of your measurement of each individual period.

It's hard to explain it in more detail, and your argument against this isn't terribly clear. To the extent that you say

if I were to take one oscillation I would never get such small error because the low precision of stop watch

then that is correct (if you were to measure each oscillation individually you would never be able to achieve this precision), but the point is that you're not measuring each oscillation individually. You're clicking the stop button exactly once (per $N=100$-period run), and this incurs the least-count uncertainty exactly once.

Answer 3

Should you be able to measure a value of much higher accuracy than that enabled by the precision of the instrument?

Do not confuse accuracy and precision. They are different concepts. Precision is a measure of random errors; accuracy is a measure of systematic errors. You can the increase the precision of your timing by taking the mean of multiple measurements, but this will not improve its accuracy.

Answer 4

But you are not making a measurement with much better precision than the instrument is capable of. You are making one measurement of a time interval, with a precision limited by the instrument. You are then dividing that by the number 100.

Your Prof is correct.

Answer 5

If we are only interested in the order of magnitude of the uncertainty, your professor is correct. But if you are interested in a more detail description ...

Let's assume we have a perfect stop watch, but its display provides only two decimal places. In this case we say that the resolution of our stop watch is $\delta t_{res} = 0.01s$. The resolution is associated with the quantity we measure (this is the key point your professor made). Hence,

• if we measure the period of a single oscillation $t_{1}$, we get $\delta_{t_{1}} = \delta t_{res}$.
• if we measure the period of one hundred oscillation $t_{100} = 100 t_1$, we still get $\delta_{t_{100}} = \delta t_{res}$.
• In general: If we measure the period of $N$ oscillations $t_{N} = N t_1$, we get $\delta_{t_{N}} = \delta t_{res}$.

Thus, the resolution of the measurement is the same, no matter which duration we are measuring. This makes sense -- just recall that we assumed a perfect stop watch.

What your professor did not explain is that the resolution is not equal to the uncertainty (standard deviation) of the measurement. If we read a number, say $t_N = 31.45s$, we know that the true value of $t_N$ lies in the interval $[31.45, 31.46)s$. By assuming that any value within this interval has the same probability to be the true value, we describe our measured value as a random variable drawn from a uniform probability distribution with width $w=0.01s$. The standard deviation of such a probability distribution is given by $\sigma_{t_N} = w/\sqrt{12}$, see link. The factor $1/\sqrt{12}$ is what your prof omitted.

In conclusion: What we are interested in is the uncertainty of $$t_1 = \frac{t_N}{N} $$ where $N$ is a constant. Applying the standard error formula yields $$ \sigma_{t_1} = \frac{\sigma_{t_N}}{N} = \frac{0.01s}{N\sqrt{12}} $$