Consideration of number of significant figures in experimental quantities

[f3sicA_A]

Homework Statement

I measure time accurate to the sixth decimal place using a Photogate (seventh digit is the uncertain digit). How many decimal places should I round of this time value (and why?) given that I also measure length with an uncertainty of ##\pm0.1## (cm), and I want to use the two experimental quantities to calculate velocity of an object? To calculate uncertainty in the final velocity that I calculate, I use summation in quadrature.

Relevant Equations

$$d=s\times t$$
$$\frac{\delta a}{a}=\sqrt{\left(\frac{\delta b}{b}\right)^2+\left(\frac{\delta c}{c}\right)^2}$$

I am not sure how to approach this problem. I know that there really is no use taking time values accurate up to the sixth decimal place if my length values are accurate only to the first decimal place, after all errors should be comparable. So I wanted to know how I should quote my time values up to the appropriate decimal places, and is there a general rule one can follow in such situations?

 

[kuruman]

What's wrong with using $$\frac{\delta d}{d}=\sqrt{\left(\frac{\delta s}{s}\right)^2+\left(\frac{\delta t}{t}\right)^2}~?$$ If ##~\dfrac{\delta s}{s}>>\dfrac{\delta t}{t}##, $$\frac{\delta d}{d}\approx \left(\frac{\delta s}{s}\right)\implies \delta d=(st)\frac{\delta s}{s}=t\delta s.$$

 

[f3sicA_A]

What's wrong with using $$\frac{\delta d}{d}=\sqrt{\left(\frac{\delta s}{s}\right)^2+\left(\frac{\delta t}{t}\right)^2}~?$$ If ##~\dfrac{\delta s}{s}>>\dfrac{\delta t}{t}##, $$\frac{\delta d}{d}\approx \left(\frac{\delta s}{s}\right)\implies \delta d=(st)\frac{\delta s}{s}=t\delta s.$$

I don't see anything wrong with it, but my professor mentions that I need to round of the uncertainty in time to keep consistent with significant figures and decimal places. For instance, let us say I have a length measure of 22.1 cm and a time measure of 0.569824 s (just arbitrary values), then speed calculation as follows:

$$s=\frac{22.1}{0.569824}$$

This seems to me like an absurd way of dealing with significant figures/decimal places?

 

[Frabjous]

As an exercise, calculate the total error if the time is rounded to .570, .5698, .56982 and .569824
Note: since you are rounding, the error for each of these numbers is different.

 

[Steve4Physics]

$$s=\frac{22.1}{0.569824}$$
This seems to me like an absurd way of dealing with significant figures/decimal places?

It may look a little strange but IMO it's OK. If ##t=0.569824 s##, then I’d say that’s what you should record and use. Presumably the associated uncertainty is 0.000001s.

The symbol ‘s’ is sometimes used for displacement so let’s avoid any confusion and use ##x## for distance and ##v## for speed.

In this case ##v = \frac {22.1cm}{0.569824 s} = 38.7839cm/s##, which now needs rounding.

As already noted, the fractional uncertainty in ##t## is negligible compared to the fractional uncertainty in ##x##. So here we can use:

##\delta v = v \frac {\delta x}{x}= 38.7839 cm/s \times \frac {0.1cm}{22.1cm}##

## = 0.2cm/s## to 1 sig. fig. (or ##0.18cm/s## to 2 sig. figs.)

The value of ##v## should then be rounded to the same number of decimal places as ##\delta v## so we end up with ##v=(38.8 \pm 0.2) cm/s##.

Or if we prefer to use 2 sig. figs. for ##\delta v## we get ##v = (38.78 \pm 0.18) cm/s##.

Note: if we want to use ##v## in subsequent calculations, we should use its unrounded value to avoid introducing unnecessary rounding errors.

 

[kuruman]

I don't see anything wrong with it, but my professor mentions that I need to round of the uncertainty in time to keep consistent with significant figures and decimal places. For instance, let us say I have a length measure of 22.1 cm and a time measure of 0.569824 s (just arbitrary values), then speed calculation as follows:

$$s=\frac{22.1}{0.569824}$$

This seems to me like an absurd way of dealing with significant figures/decimal places?

First you do the exact calculation $$s=\frac{22.1~\text{cm}}{0.569824~\text{s}}=38.783905205817937~\text{cm/s}.$$I deliberately kept a ridiculous number of significant figures to make a point.
Now for the calculation of the uncertainties. Use exponential notation, one decimal place is sufficient.
##\dfrac{\delta d}{d}=\dfrac{0.1}{22.1}=4.5\times 10^{-3} \implies \left(\dfrac{\delta d}{d} \right)^2=2.0\times 10^{-5}##
##\dfrac{\delta t}{t}=\dfrac{0.1}{22.1}=8.8\times 10^{-6}\implies \left(\dfrac{\delta t}{t} \right)^2=7.7\times 10^{-11}##
##\left(\dfrac{\delta d}{d} \right)^2+\left(\dfrac{\delta t}{t} \right)^2=2.0\times 10^{-5}.## The fractional uncertainty in time doesn't contribute anything and can be ignored.
Then $$\delta s=s*\sqrt{\left(\dfrac{\delta d}{d} \right)^2}=38.783905205817937~(\text{cm/s})\times 4.5\times 10^{-3}=2.8\times 10^{-2}~\text{cm/s}.$$ How to report the value of ##s##? Note that the uncertainties in exponential form were carried to two decimal places to avoid too much roundoff. At this point you round the uncertainty to one sig-fig: ##\delta s = 0.03~\text{cm/s}.## That's two decimal places. Your reported value for ##s## is truncated accordingly to two decimal places, $$s=38.78\pm 0.03 ~\text{cm/s}.$$

 

[haruspex]

my professor mentions that I need to round of the uncertainty in time to keep consistent with significant figures and decimal places. For instance, let us say I have a length measure of 22.1 cm and a time measure of 0.569824 s (just arbitrary values), then speed calculation as follows:

$$s=\frac{22.1}{0.569824}$$

I do not understand. The equation you quote does not seem to be an example of rounding the time to be consistent with the distance accuracy. That would look like $$s=\frac{22.1}{0.570}$$