Combining Uncertainties II: Application

PREVIOUS: Combining Uncertainties I: Propagation of Errors


DEFINITIONS:
If $\Delta h$ is the absolute uncertainty in a physical quantity $h$, then:
a) the quantity $h$ is properly expressed as $h \pm \Delta h$,
b) the fractional uncertainty in $h$ is given by $\frac{\Delta h}{h}$,
c) percent uncertainty in $h$ is given by $\frac{\Delta h}{h} \times 100$.

Suppose you make following measurements of lengths in a lab experiment using a meter scale with the least count of $0.1 \, cm$: $$p = (80.0 \pm 0.1) \, cm$$ $$q = (12.0 \pm 0.1) \, cm$$ $$r = (5.3 \pm 0.1) \, cm$$ Here, $0.1 \, cm$ is referred to as absolute uncertainty in $p,q,r$. According to the above definitions, fractional uncertainties in these measurements are: $$\frac{\Delta p}{p} = \frac{0.1}{80} = 0.00125$$ $$\frac{\Delta q}{q} = \frac{0.1}{12} = 0.00833$$ $$\frac{\Delta r}{r} = \frac{0.1}{5.3} = 0.01887$$

Results from the previous post

$$\begin{array}{|c|c|} \hline \\ X = aA & \Delta X = a\Delta A \,\,\,\,\,\,\, \mbox{(\(a\) is a constant)} \\ & \\ \hline \\ X = A \pm B & \Delta X = \Delta A + \Delta B \\ & \\ \hline\\ X = \displaystyle AB \mbox{ or }\frac{A}{B} & \displaystyle \frac{\Delta X}{X} = \frac{\Delta A}{A} + \frac{\Delta B}{B} \\ & \\ \hline\\ X = \displaystyle A^{n} & \displaystyle \frac{\Delta X}{X} = n\frac{\Delta A}{A} \\ & \\ \hline\\ X = \displaystyle \frac{A^{m}B^{n}}{C^{p}} & \displaystyle \frac{\Delta X}{X} = m\frac{\Delta A}{A} + n\frac{\Delta B}{B} + p\frac{\Delta C}{C} \\ & \\ \hline \end{array}$$
COMBINING UNCERTAINTIES

1. When a measured quantity is multiplied by a constant.
Suppose you want to figure out uncertainty functions $C = 5p$ and $D = 3r$. In such cases, the constant factor multiplies with the measured values and also with the absolute uncertainty in the measured values: $$C = 5(80 \pm 0.1) \, cm = (400 \pm 0.5) \, cm$$ $$D = 3(5.3 \pm 0.1) \, cm = (15.9 \pm 0.3) \, cm$$

2. When measured quantities are added or subtracted.
In case measured quantities are being added or subtracted, absolute uncertainties add: $$\begin{eqnarray} p + q &=& (92 \pm 0.2) \, cm \\ p - q &=& (68 \pm 0.2) \, cm \\ p + q - 3r &=& (92 \pm 0.2) \, cm - 3(5.3 \pm 0.1) \, cm \\ &=& (92 \pm 0.2) \, cm - (15.9 \pm 0.3) \, cm \\ &=& (76.1 \pm 0.5) \, cm \\ \end{eqnarray}$$

3. When measured quantities are multiplied (or divided).
In case measured quantities are being multiplied or divided, relative uncertainties add-up. For instance, if $f$, $g$ and $h$ are defined as: $$f = pq = 960, \, g = \frac{q}{p} = 0.15 \,\,\, \mbox{and} \,\,\, h = \frac{qr}{p} = 0.795,$$ relative errors in these quantities are obtained as: $$\begin{eqnarray} \frac{\Delta f}{f} &=& \frac{\Delta p}{p} + \frac{\Delta q}{q} = 0.00125+0.00833 = 0.00958 \\ \frac{\Delta g}{g} &=& \frac{\Delta p}{p} + \frac{\Delta q}{q} = 0.00125+0.00833 = 0.00958 \\ \frac{\Delta h}{h} &=& \frac{\Delta p}{p} + \frac{\Delta r}{r}+ \frac{\Delta q}{q} = 0.00125+0.01887+0.00833 = 0.02845 \end{eqnarray}$$ Therefore, the absolute errors in these quantities are: $$\begin{eqnarray} \Delta f &=& 0.00958 \, f = 0.00958 \times 960 = 9.1968 \, cm^{2} \\ \Delta g &=& 0.00958 \, g = 0.00958 \times 0.15 = 0.001437 \\ \Delta h &=& 0.02845 \, h = 0.02845 \times 0.795 = 0.02262 \, cm \end{eqnarray}$$ Therefore, $$\begin{eqnarray} f &=& (960 \pm 9.1968) \, cm^{2} = (960 \pm 9) \, cm^{2} \\ g &=& 0.15 \pm 0.001437 = 0.150 \pm 0.001 \\ h &=& (0.795 \pm 0.02262) \, cm = (0.79 \pm 0.02) \, cm \end{eqnarray}$$
I have rounded-off the final results in above calculations using the rules for significant figures and rounding-off discussed in one of the previous posts. For the calculations above, think of rounding-off in the following way: if the value of $f=960$ can be uncertain in the ones place by $9$ units, would numbers in tenths or hundredths place hold any significance? No. So, we don't write numbers beyond $ones$ place in the case of $f$. Following the same reasoning, we don't write numbers beyond hundredths place in case of $h$.

4. When a measured quantity has an exponent.
If a quantity $A = q^{2}$, the relative uncertainty in $A$ is multiplied by $2$: $$\begin{eqnarray} \frac{\Delta A}{A} &=& 2 \frac{\Delta q}{q} = (2 \times 0.008333) = 0.01666 \approx 16.66 \% \\ \Rightarrow \Delta A &=& 0.01666 \, A = 0.01666 \times 12^{2} \, cm^{2} = 2.4 \, cm^{2} \end{eqnarray}$$ Therefore, $$A = (144 \pm 2.4) \, cm^{2} = (144 \pm 2) \, cm^{2}$$ .

5. When the above operations are combined.
Say we want to figure out uncertainty in $$W = \frac{5qr^{2}}{p} = 21.0675.$$ It follows from the discussion above that: $$\begin{eqnarray} \frac{\Delta W}{W} &=& \frac{\Delta q}{q} + 2\frac{\Delta r}{r}+ \frac{\Delta p}{p} \\ &=& 0.00833+(2 \times 0.01887)+0.00125 \\ &=& 0.04732 \\ \Rightarrow \Delta W &=& 0.04732 \, W = 0.2366 \times 21.0675 \\ &=& 0.9969 \end{eqnarray}$$ Therefore, $$W = (21.0675 \pm 0.9979) \, cm^{2} = (21 \pm 1) \, cm^{2}$$ Note that the constant $5$ does not figure into the expression for relative uncertainty. Multiplication by a constant affects only the absolute uncertainty and not the relative uncertainty. In the second step where we try to find absolute uncertainty ($\Delta W$) from relative uncertainty by multiplying relative uncertainty with $W$, the constant $5$ figures into our calculations through $W$ (this is because $W$ has in it the constant $5$).