How to Calculate Absolute Uncertainty in HSC Physics
Learn what absolute uncertainty means, how to calculate it from instrument readings and repeated measurements, and how to report experimental results correctly. Includes worked HSC Physics examples covering rounding, measurement error, standard deviation and uncertainty intervals.
You measure a metal rod and obtain a length of 24.6 cm. Is the result simply 24.6 cm, or should you report something like \(24.6 \pm 0.1\text{ cm}\)? The second form tells us something important: the measurement is an estimate, not an exact value.
In HSC Physics, reporting a measurement with its absolute uncertainty means giving the measured value together with a positive and negative uncertainty, using the same units and appropriate decimal places.
The standard reporting format is:
\[
\boxed{x = (\text{measured value} \pm \text{absolute uncertainty})\text{ units}}
\]
For example:
\[
L = (24.6 \pm 0.1)\text{ cm}
\]
This means the estimated measurement interval is from 24.5 cm to 24.7 cm. It does not mean the actual length is equally likely to be anywhere in that interval, nor does it guarantee that the true value lies inside it.
## 1. What does absolute uncertainty actually tell us?
Imagine measuring the same table with two different rulers.
- Ruler A has markings every centimetre.
- Ruler B has markings every millimetre.
Both might give a reading of approximately 120 cm, but the second ruler allows a more precise measurement.
The difference is the resolution of the measuring instrument: the smallest change it can distinguish.
Absolute uncertainty describes the size of the uncertainty in a measurement, expressed in the same units as the measured quantity.
For example:
\[
L = (120.0 \pm 0.5)\text{ cm}
\]
Here:
- \(L\) is the measured length.
- \(120.0\text{ cm}\) is the estimated length.
- \(0.5\text{ cm}\) is the absolute uncertainty.
The corresponding measurement interval is:
\[
119.5\text{ cm} \leq L \leq 120.5\text{ cm}
\]
This interval is a useful way to communicate the range of values consistent with the stated uncertainty convention.
[IMAGE: Two side-by-side ruler measurements of the same table edge, one using a coarse centimetre scale and one using a millimetre scale. Show the reading positions, scale divisions, and different uncertainty intervals. Include a clear enlargement of the measurement endpoints.]
### Absolute uncertainty versus percentage uncertainty
These two quantities describe uncertainty in different ways.
| Quantity | What it tells you | Example |
|---|---|---|
| Absolute uncertainty | The uncertainty in the original measurement units | \(\pm 0.5\text{ cm}\) |
| Percentage uncertainty | The uncertainty relative to the measured value | \(0.42\%\) |
Absolute uncertainty is the focus of this article. Percentage uncertainty is useful when comparing the relative precision of measurements with different magnitudes.
You can read more about the distinction in [Percentage Error vs Percentage Uncertainty](https://tutorgum.com/learn/resources/hsc-physics-percentage-error-vs-percentage-uncertainty-how-to-calculate-each/).
## 2. How to calculate absolute uncertainty
The method depends on how the measurement was obtained.
There are two common situations:
- A measurement taken using a single instrument reading.
- A quantity measured repeatedly, producing a set of readings.
The conventions used to estimate uncertainty are slightly different.
### Method 1: Using the smallest scale division
For an analogue instrument, a common school-laboratory convention is:
\[
\boxed{\Delta x = \frac{\text{smallest scale division}}{2}}
\]
Here, \(\Delta x\) represents the absolute uncertainty.
The reasoning is straightforward. If an analogue scale has divisions every 1 mm, a reading between two marks cannot usually be identified more precisely than approximately half a division without additional measurement techniques.
For an instrument with 1 mm divisions:
\[
\Delta x = \frac{1}{2} = 0.5\text{ mm}
\]
Suppose a student measures a metal rod and reads 246 mm on this ruler.
The measurement is reported as:
\[
L = (246.0 \pm 0.5)\text{ mm}
\]
The corresponding interval is:
\[
245.5\text{ mm} \leq L \leq 246.5\text{ mm}
\]
The trailing zero in 246.0 aligns the measured value with the decimal place used for the uncertainty. It does not mean that the ruler can distinguish changes of 0.1 mm.
[IMAGE: Close-up of an analogue metric ruler measuring a metal rod. Show the rod's left endpoint at zero, its right endpoint at 246 mm, the 1 mm scale divisions, and a bracket illustrating the half-division uncertainty of 0.5 mm.]
**What about digital instruments?**
A digital instrument does not have an analogue scale that you estimate between markings.
A common introductory laboratory convention is to use the smallest displayed increment as the absolute uncertainty.
For example, if a digital balance displays mass in increments of 0.1 g:
\[
\Delta m = 0.1\text{ g}
\]
If the balance reads 152.4 g, report:
\[
m = (152.4 \pm 0.1)\text{ g}
\]
However, this is a stated convention, not a universal law. The manufacturer's accuracy specification may give a larger uncertainty, and that specification should take precedence when available. A display resolution alone does not establish the instrument's total accuracy.
### Worked example: Measuring a volume with a measuring cylinder
A student uses a measuring cylinder with scale divisions of 2 mL. The liquid level is read as 36 mL. Estimate and report the absolute uncertainty using the half-smallest-division convention.
**Step 1**: Identify the smallest scale division.
The cylinder has divisions of 2 mL.
**Step 2**: Calculate half the smallest division.
\[
\Delta V = \frac{2}{2} = 1\text{ mL}
\]
**Step 3**: Write the measurement with its absolute uncertainty.
\[
\boxed{V = (36 \pm 1)\text{ mL}}
\]
**Step 4**: Express the measurement interval.
\[
35\text{ mL} \leq V \leq 37\text{ mL}
\]
The estimated volume is 36 mL, with an absolute uncertainty of 1 mL under the chosen convention.
Notice that the uncertainty has the same unit as the measured volume. It would be incorrect to report an absolute uncertainty of 1% here. That would be a percentage uncertainty, not an absolute uncertainty.
[IMAGE: A measuring cylinder viewed slightly from the side, containing coloured liquid. Show the graduated scale with 2 mL divisions, the meniscus, and the 36 mL reading. Include a small scale enlargement identifying one division and half a division.]
### Method 2: Using repeated measurements
A single measurement can be affected by random variation.
Imagine timing one complete swing of a pendulum using a handheld stopwatch. Your reaction time when starting and stopping the timer might introduce an uncertainty of around 0.2 s. That's a substantial fraction of the pendulum's period.
A better approach is to time 10 complete oscillations, then divide the total time by 10.
This is a useful experimental improvement because a roughly fixed start-and-stop timing uncertainty has a smaller effect on the calculated time for one oscillation.
[IMAGE: A school physics pendulum experiment showing a clamp stand, pivot, string, spherical bob, and stopwatch. Include a metre ruler positioned vertically beside the string, with its measurement extending from the pivot to the centre of the bob. Show the bob's swing path and a reference position used for counting oscillations.]
**How does the experiment work?**
1. Suspend a pendulum bob from a fixed pivot.
2. Measure its length from the pivot to the centre of the bob.
3. Pull the bob through a small angle and release it without pushing.
4. Start the stopwatch when the bob passes a chosen reference position.
5. Count 10 complete oscillations and stop the stopwatch when the bob returns to the same position, moving in the same direction.
6. Repeat the timing several times under the same conditions.
We expect the readings to be slightly different because of reaction time and other random variations.
The average gives us a central estimate, while the spread of the readings helps us estimate the absolute uncertainty.
For repeated measurements, a common school-laboratory convention is:
\[
\boxed{\Delta x = \frac{x_{\text{max}}-x_{\text{min}}}{2}}
\]
This is called the half-range method.
Here:
- \(x_{\text{max}}\) is the largest measurement.
- \(x_{\text{min}}\) is the smallest measurement.
- \(\Delta x\) is the estimated absolute uncertainty.
The measured value is usually the arithmetic mean:
\[
\bar{x}=\frac{x_1+x_2+\cdots+x_n}{n}
\]
Here, \(n\) is the number of measurements.
### Worked example: Finding the period of a pendulum
A student times 10 complete oscillations of a pendulum five times. The recorded times are shown below.
The stopwatch has a resolution of 0.01 s, which is small compared with the spread of the readings. For this example, use the half-range method to estimate the uncertainty.
| Trial | Time for 10 oscillations (s) |
|---|---:|
| 1 | 14.2 |
| 2 | 14.4 |
| 3 | 14.3 |
| 4 | 14.5 |
| 5 | 14.1 |
Calculate the mean period of the pendulum and report it with its absolute uncertainty.
**Step 1**: Calculate the mean time.
Add all five readings and divide by five.
\[
\begin{aligned}
\bar{t}
&=\frac{14.2+14.4+14.3+14.5+14.1}{5}\\
&=\frac{71.5}{5}\\
&=14.3\text{ s}
\end{aligned}
\]
The mean time for 10 oscillations is 14.3 s.
**Step 2**: Calculate the absolute uncertainty using the half-range method.
The maximum reading is 14.5 s, and the minimum reading is 14.1 s.
\[
\begin{aligned}
\Delta t
&=\frac{14.5-14.1}{2}\\
&=\frac{0.4}{2}\\
&=0.2\text{ s}
\end{aligned}
\]
Therefore, the total time is:
\[
t=(14.3\pm0.2)\text{ s}
\]
**Step 3**: Calculate the period.
The period is the time taken for one complete oscillation.
\[
T=\frac{\bar{t}}{N}
\]
Here, \(N=10\) is the exact number of oscillations.
\[
T=\frac{14.3}{10}=1.43\text{ s}
\]
**Step 4**: Calculate the uncertainty in the period.
Because the total time is divided by an exact number, the absolute uncertainty is divided by the same number.
\[
\begin{aligned}
\Delta T
&=\frac{\Delta t}{N}\\
&=\frac{0.2}{10}\\
&=0.02\text{ s}
\end{aligned}
\]
**Step 5**: Write the final measurement.
\[
\boxed{T=(1.43\pm0.02)\text{ s}}
\]
The corresponding interval is:
\[
1.41\text{ s}\leq T\leq1.45\text{ s}
\]
The estimated period is 1.43 s, with an absolute uncertainty of 0.02 s.
Notice that we divided both the measured time and its uncertainty by 10. Dividing the measured value without dividing its uncertainty would give an incorrect result.
**Why time 10 oscillations instead of one?**
Timing multiple oscillations reduces the relative effect of reaction time on each calculated period.
However, it doesn't eliminate every source of uncertainty. If the stopwatch runs consistently slow, all five readings could be too small by a similar amount.
Repeating a measurement helps reveal random variation. It does not automatically identify a systematic error.
## 3. How to round and write your final measurement
You can calculate the correct uncertainty and still lose marks by reporting the result incorrectly.
Follow these conventions when writing your final answer.
### Rule 1: Give the uncertainty an appropriate number of significant figures
For straightforward school-laboratory reporting, absolute uncertainty is commonly rounded to one significant figure.
For example:
\[
0.047\text{ cm}\rightarrow0.05\text{ cm}
\]
Some tasks or laboratory conventions retain two significant figures, particularly when greater detail is needed. Follow the instructions given.
### Rule 2: Round the measured value to the same decimal place as the uncertainty
This is one of the easiest rules to overlook.
Consider:
\[
T=1.43\text{ s},\qquad \Delta T=0.02\text{ s}
\]
The uncertainty is expressed to two decimal places, so the measured value must also be expressed to two decimal places.
The correct report is:
\[
T=(1.43\pm0.02)\text{ s}
\]
Compare this with the following incorrect reports:
| Incorrect report | What's wrong? |
|---|---|
| \(1.4\pm0.02\text{ s}\) | The measured value has been rounded to a different decimal place. |
| \(1.43\pm0.2\text{ s}\) | The uncertainty has changed by a factor of 10. |
| \(1.43\pm0.02\%\) | This is not an absolute uncertainty. |
### Rule 3: Always include units
The measured value and its absolute uncertainty must use the same units.
Correct:
\[
L=(24.6\pm0.1)\text{ cm}
\]
Incorrect:
\[
L=24.6\text{ cm}\pm0.1\text{ mm}
\]
If you want to change units, convert both the measurement and its uncertainty.
For example:
\[
0.5\text{ mm}=0.05\text{ cm}
\]
Therefore:
\[
L=(24.60\pm0.05)\text{ cm}
\]
### Rule 4: State the convention you used
Different instruments and experimental procedures can require different uncertainty estimates.
When explaining your working, identify the method.
Examples:
- "The absolute uncertainty is half the smallest scale division."
- "The absolute uncertainty is half the range of the repeated measurements."
- "The manufacturer's stated instrument uncertainty is used."
This makes your result understandable and allows someone else to follow your reasoning.
[IMAGE: An annotated comparison of three correctly formatted measurement reports: an analogue ruler reading with half-division uncertainty, a digital balance reading with manufacturer-specified uncertainty, and a repeated pendulum measurement with half-range uncertainty. Show the decimal-place alignment and units in each.]
## 4. Absolute uncertainty is not the same as error or standard deviation
These terms are related, but they describe different things.
Confusing them can lead to incorrect conclusions about experimental results.
### Absolute uncertainty versus measurement error
Suppose you measure a rod and obtain:
\[
L=(24.6\pm0.1)\text{ cm}
\]
Now imagine that the rod has a known reference length of exactly 24.0 cm.
The signed measurement error is:
\[
\text{Error}=\text{measured value}-\text{reference value}
\]
\[
\text{Error}=24.6-24.0=+0.6\text{ cm}
\]
The positive sign means the measurement is higher than the reference value.
The absolute error is the magnitude of this difference:
\[
|\text{Error}|=0.6\text{ cm}
\]
Compare the two quantities:
| Quantity | Result |
|---|---:|
| Absolute uncertainty | 0.1 cm |
| Signed measurement error | +0.6 cm |
| Absolute error | 0.6 cm |
The measurement interval is:
\[
24.5\text{ cm}\leq L\leq24.7\text{ cm}
\]
The reference value of 24.0 cm is outside this interval.
This tells us something important: the measurement has a relatively small estimated uncertainty, but it is still inaccurate.
The likely explanation is a systematic error, such as an incorrectly calibrated instrument.
[IMAGE: A comparison of a measured interval from 24.5 cm to 24.7 cm and a reference value at 24.0 cm. Show the reference value clearly outside the uncertainty interval, illustrating a precise but inaccurate measurement.]
**The key distinction:**
- Uncertainty describes the limitations of our knowledge about a measurement.
- Error describes the difference between a measured value and the true or accepted reference value.
The true value is generally unknown, which means the actual measurement error is usually unknown too.
This distinction is consistent with formal metrology terminology. The [BIPM International Vocabulary of Metrology](https://jcgm.bipm.org/vim/en/2.26.html) defines measurement uncertainty in terms of the dispersion of values that could reasonably be attributed to a measured quantity. [NIST's measurement terminology](https://www.nist.gov/pml/nist-technical-note-1297/nist-tn-1297-appendix-d1-terminology) separately defines measurement error as the difference between the measurement result and the value of the quantity being measured.
### Absolute uncertainty versus standard deviation
Standard deviation describes how spread out a set of repeated measurements is around their mean.
A small standard deviation means the readings are clustered closely together.
A large standard deviation means they are more spread out.
For the five pendulum measurements earlier, the mean was 14.3 s.
The deviations from the mean were:
\[
-0.1,\quad +0.1,\quad 0,\quad +0.2,\quad -0.2
\]
The sample standard deviation is calculated using:
\[
s=\sqrt{\frac{\sum(t_i-\bar{t})^2}{n-1}}
\]
Here:
- \(s\) is the sample standard deviation.
- \(t_i\) is an individual reading.
- \(\bar{t}\) is the mean.
- \(n\) is the number of readings.
Substituting the deviations:
\[
\begin{aligned}
s&=\sqrt{\frac{0.01+0.01+0+0.04+0.04}{4}}\\
&=\sqrt{0.025}\\
&\approx0.16\text{ s}
\end{aligned}
\]
The results are:
| Quantity | Value |
|---|---:|
| Half-range | 0.20 s |
| Sample standard deviation | 0.16 s |
These numbers are different because the two methods measure spread differently.
The half-range uses only the largest and smallest readings. Standard deviation considers how every reading is distributed around the mean.
Standard deviation can be used when evaluating measurement uncertainty, but it is not automatically the absolute uncertainty you should report.
For example, the uncertainty of a mean can be estimated using the standard deviation divided by the square root of the number of independent readings, under suitable statistical assumptions. That is a different calculation from the half-range method.
**For HSC-style experimental questions, use the uncertainty convention specified in the question or your experimental instructions. Don't substitute standard deviation for half-range simply because both describe spread.**
## 5. A quick decision process
When you are given a measurement question, use the following guide.
| Experimental situation | Appropriate approach |
|---|---|
| One reading from an analogue instrument | Commonly half the smallest scale division |
| One reading from a digital instrument | Use the instrument specification or the stated digital-resolution convention |
| Several repeated readings | Calculate the mean and, when specified, half the range |
| A known calibration error | Apply the appropriate correction to the measured value |
| A quantity calculated from other measurements | Propagate the relevant absolute uncertainties |
One final caution: don't assume that repeating measurements automatically removes systematic errors. If an instrument consistently reads too high, taking another 20 readings may simply give you 20 consistently high results.
Once you have a reliable estimate of the uncertainty, you can use it to compare experimental results, assess the quality of a relationship, and draw error bars on a graph.
## Questions and solutions
The following questions build from straightforward instrument readings to situations where several sources of uncertainty interact.
### Question 1
A student measures the length of an aluminium strip using a ruler with smallest scale divisions of 0.2 cm.
The measured length is 18.4 cm.
Using the half-smallest-division convention:
a) Calculate the absolute uncertainty.
b) Write the measurement in the correct format.
c) State the corresponding measurement interval.
[IMAGE: An aluminium strip placed alongside a metric ruler with 0.2 cm scale divisions. Show the left endpoint aligned with zero and the right endpoint at 18.4 cm. Include clearly visible scale markings.]
### Solution 1
**The measurement should be reported as \(L=(18.4\pm0.1)\text{ cm}\), giving an interval from 18.3 cm to 18.5 cm.**
**Step 1**: Identify the smallest scale division.
The ruler has divisions of 0.2 cm.
**Step 2**: Calculate the absolute uncertainty.
\[
\Delta L=\frac{0.2}{2}=0.1\text{ cm}
\]
**Step 3**: Report the measurement.
\[
\boxed{L=(18.4\pm0.1)\text{ cm}}
\]
**Step 4**: Calculate the interval.
\[
L_{\text{min}}=18.4-0.1=18.3\text{ cm}
\]
\[
L_{\text{max}}=18.4+0.1=18.5\text{ cm}
\]
Therefore:
\[
\boxed{18.3\text{ cm}\leq L\leq18.5\text{ cm}}
\]
The uncertainty is expressed in centimetres because it is an absolute uncertainty.
### Question 2
A student measures a small metal block using a digital balance.
The five readings are:
| Trial | Mass (g) |
|---|---:|
| 1 | 83.45 |
| 2 | 83.46 |
| 3 | 83.47 |
| 4 | 83.46 |
| 5 | 83.46 |
The manufacturer's specification gives an instrument uncertainty of \(\pm0.03\text{ g}\).
For this question, the laboratory uses the following conservative convention: calculate the half-range of repeated measurements, then use the larger of that value and the manufacturer's stated uncertainty. Do not add the two values.
a) Calculate the mean mass.
b) Calculate the half-range.
c) Determine the absolute uncertainty and report the final measurement.
d) Explain why using only the half-range would be inappropriate.
[IMAGE: A small metal block sitting centrally on the weighing pan of a digital laboratory balance. Show a clear digital display reading 83.46 g and the balance's surrounding controls.]
### Solution 2
**The correct report is \(m=(83.46\pm0.03)\text{ g}\), because the manufacturer's uncertainty is greater than the half-range of the repeated readings.**
**Step 1**: Calculate the mean mass.
\[
\begin{aligned}
\bar{m}
&=\frac{83.45+83.46+83.47+83.46+83.46}{5}\\
&=\frac{417.30}{5}\\
&=83.46\text{ g}
\end{aligned}
\]
**Step 2**: Calculate the half-range.
The maximum reading is 83.47 g, and the minimum reading is 83.45 g.
\[
\begin{aligned}
\Delta m_{\text{range}}
&=\frac{83.47-83.45}{2}\\
&=0.01\text{ g}
\end{aligned}
\]
**Step 3**: Compare the uncertainty estimates.
The two values are:
- Half-range: 0.01 g.
- Manufacturer's uncertainty: 0.03 g.
According to the convention specified in the question, use the larger value.
\[
\Delta m=0.03\text{ g}
\]
**Step 4**: Report the final measurement.
\[
\boxed{m=(83.46\pm0.03)\text{ g}}
\]
**Step 5**: Explain why the half-range alone is insufficient.
The readings are closely clustered, but this only tells us about their observed spread.
The manufacturer's specification indicates that the instrument itself has a larger uncertainty. The five readings cannot establish that the balance is more accurate than its stated specification.
This is why instrument specifications should not be ignored simply because repeated measurements look consistent.
### Question 3
A student wants to estimate the average thickness of one sheet of paper.
Instead of measuring one sheet, they stack 25 sheets together and measure the total thickness using a micrometer.
The five readings are:
| Trial | Stack thickness (mm) |
|---|---:|
| 1 | 2.76 |
| 2 | 2.78 |
| 3 | 2.77 |
| 4 | 2.75 |
| 5 | 2.79 |
Use the mean and half-range method to calculate the average thickness of one sheet, including its absolute uncertainty.
Assume:
- The number of sheets is exactly 25.
- The half-range is the appropriate uncertainty estimate for the stack.
- The uncertainty in counting the sheets is negligible.
[IMAGE: A micrometer screw gauge measuring a neat stack of 25 sheets of paper between its anvil and spindle. Show the ratchet, thimble, spindle, and the paper stack, with the total thickness being measured.]
### Solution 3
**The average sheet thickness is \(0.1108\pm0.0008\text{ mm}\). The uncertainty must be divided by 25 along with the measured stack thickness.**
**Step 1**: Calculate the mean stack thickness.
\[
\begin{aligned}
\bar{d}
&=\frac{2.76+2.78+2.77+2.75+2.79}{5}\\
&=\frac{13.85}{5}\\
&=2.77\text{ mm}
\end{aligned}
\]
**Step 2**: Calculate the absolute uncertainty of the stack.
\[
\begin{aligned}
\Delta d
&=\frac{2.79-2.75}{2}\\
&=0.02\text{ mm}
\end{aligned}
\]
Therefore:
\[
d=(2.77\pm0.02)\text{ mm}
\]
**Step 3**: Calculate the average thickness of one sheet.
Divide the total thickness by 25.
\[
\begin{aligned}
d_{\text{sheet}}
&=\frac{2.77}{25}\\
&=0.1108\text{ mm}
\end{aligned}
\]
**Step 4**: Calculate the uncertainty per sheet.
Because we divide the stack thickness by an exact number, we divide its absolute uncertainty by the same number.
\[
\begin{aligned}
\Delta d_{\text{sheet}}
&=\frac{0.02}{25}\\
&=0.0008\text{ mm}
\end{aligned}
\]
**Step 5**: Report the final result.
\[
\boxed{d_{\text{sheet}}=(0.1108\pm0.0008)\text{ mm}}
\]
This is a useful example of how measuring several objects together can produce a more precise estimate of the average thickness of one object.
A tempting mistake is to report an uncertainty of \(\pm0.02\text{ mm}\) for each sheet. That would incorrectly carry the uncertainty of the entire stack into the individual measurement without dividing it by 25.
The micrometer did not directly resolve each sheet to 0.0008 mm. Rather, the average thickness was inferred by measuring the larger stack and dividing by the exact sheet count.
### Question 4
A student uses a temperature sensor to measure the temperature of water.
Five readings are recorded:
\[
24.5,\ 24.6,\ 24.6,\ 24.7,\ 24.6\ ^\circ\text{C}
\]
A calibrated reference thermometer measures the same water at exactly \(23.8^\circ\text{C}\). Assume the reference thermometer's uncertainty is negligible.
a) Calculate the mean temperature.
b) Estimate the absolute uncertainty using the half-range method.
c) Calculate the signed measurement error and absolute error.
d) Explain whether the small spread of the readings demonstrates that the temperature sensor is accurate.
e) If the calibration offset is known and corrected, what should the corrected measurement be?
[IMAGE: Two temperature probes immersed side by side in a well-stirred beaker of water. One is a digital temperature sensor displaying approximately 24.6 degrees Celsius, and the other is a calibrated reference thermometer displaying 23.8 degrees Celsius. Show both probes measuring the same water.]
### Solution 4
**The sensor reports \(24.6\pm0.1^\circ\text{C}\), but its signed error is \(+0.8^\circ\text{C}\). The readings are consistent with one another, but the sensor is systematically biased.**
**Step 1**: Calculate the mean temperature.
\[
\begin{aligned}
\bar{T}
&=\frac{24.5+24.6+24.6+24.7+24.6}{5}\\
&=\frac{123.0}{5}\\
&=24.6^\circ\text{C}
\end{aligned}
\]
**Step 2**: Calculate the half-range.
\[
\begin{aligned}
\Delta T
&=\frac{24.7-24.5}{2}\\
&=0.1^\circ\text{C}
\end{aligned}
\]
The reported measurement is:
\[
\boxed{T=(24.6\pm0.1)^\circ\text{C}}
\]
**Step 3**: Calculate the signed measurement error.
Using the sign convention:
\[
\text{Error}=\text{measured value}-\text{reference value}
\]
\[
\begin{aligned}
\text{Error}
&=24.6-23.8\\
&=+0.8^\circ\text{C}
\end{aligned}
\]
The absolute error is:
\[
\boxed{|\text{Error}|=0.8^\circ\text{C}}
\]
**Step 4**: Compare the uncertainty interval with the reference value.
The measurement interval is:
\[
24.5^\circ\text{C}\leq T\leq24.7^\circ\text{C}
\]
The reference temperature is 23.8°C, which is outside this interval.
The sensor therefore has a systematic bias, even though its repeated readings are closely grouped.
The likely issue is an instrument calibration offset.
**Step 5**: Apply the calibration correction.
The sensor reads 0.8°C too high.
Therefore, subtract 0.8°C from the measured value:
\[
T_{\text{corrected}}=24.6-0.8
\]
\[
T_{\text{corrected}}=23.8^\circ\text{C}
\]
Assuming the correction is known exactly, the random uncertainty remains 0.1°C.
The corrected result is:
\[
\boxed{T_{\text{corrected}}=(23.8\pm0.1)^\circ\text{C}}
\]
This question demonstrates why precision and accuracy must not be confused. Repeating a measurement can produce a very small spread without identifying a systematic error.
### Question 5
A student investigates the period of a simple pendulum and wants to determine whether the experimental result agrees with the theoretical model.
The experiment has two measurement issues that must be considered.
**The pendulum length**
The student measures the distance from the pivot to the top surface of the spherical bob:
\[
L_{\text{top}}=(0.800\pm0.002)\text{ m}
\]
The bob has a measured radius:
\[
r=(0.015\pm0.001)\text{ m}
\]
For a simple pendulum, the length is measured from the pivot to the centre of the bob.
Use the conservative convention of adding absolute uncertainties when adding these two lengths.
**The timing measurements**
The student times 10 complete oscillations five times. The stopwatch displays:
| Trial | Displayed time (s) |
|---|---:|
| 1 | 17.7 |
| 2 | 17.8 |
| 3 | 17.8 |
| 4 | 17.9 |
| 5 | 17.8 |
A calibration test establishes that when the true time interval is 10.00 s, the stopwatch displays 9.80 s.
Assume this calibration factor is exact for the experiment, and that the stopwatch's display resolution is negligible compared with the repeated-reading spread.
Use the half-range method to estimate the uncertainty in the timing measurements.
Take \(g=9.80\text{ m s}^{-2}\), and assume the pendulum swings through a sufficiently small angle for the simple pendulum model to apply.
Answer the following:
a) Calculate the correct pendulum length, including its absolute uncertainty.
b) Correct the stopwatch readings for the calibration error.
c) Calculate the experimental period and report it with its absolute uncertainty.
d) Calculate the theoretical period, including its uncertainty interval.
e) Determine whether the corrected experimental result is consistent with the theoretical prediction. Explain why using the uncorrected stopwatch readings would lead to a different conclusion.
[IMAGE: A simple pendulum suspended from a fixed pivot, with a spherical bob shown in side view. Clearly distinguish the measured distance from the pivot to the top surface of the bob, the bob's radius, and the pivot-to-centre distance required by the pendulum model. Include a small stopwatch beside the apparatus and show the bob's swing path. Do not include calculated results.]
### Solution 5
**The corrected experimental period is approximately \(1.82\pm0.01\text{ s}\), and it is consistent with the theoretical prediction. The key is to correct both the pendulum's geometric length and the stopwatch's systematic timing error before comparing the results.**
This question combines several ideas. Work through them in order.
**Step 1**: Calculate the correct pendulum length.
The simple pendulum length is measured from the pivot to the centre of the bob.
The student measured only to the top surface, so the bob's radius must be added.
\[
\begin{aligned}
L&=L_{\text{top}}+r\\
&=0.800+0.015\\
&=0.815\text{ m}
\end{aligned}
\]
The absolute uncertainties are added using the conservative convention specified in the question.
\[
\begin{aligned}
\Delta L&=0.002+0.001\\
&=0.003\text{ m}
\end{aligned}
\]
Therefore:
\[
\boxed{L=(0.815\pm0.003)\text{ m}}
\]
Using 0.800 m directly would underestimate the pendulum length because the bob's centre is below its top surface.
**Step 2**: Calculate the mean displayed time.
\[
\begin{aligned}
\bar{t}
&=\frac{17.7+17.8+17.8+17.9+17.8}{5}\\
&=\frac{89.0}{5}\\
&=17.8\text{ s}
\end{aligned}
\]
This is the mean displayed time for 10 oscillations.
**Step 3**: Calculate the half-range.
The largest reading is 17.9 s, and the smallest is 17.7 s.
\[
\begin{aligned}
\Delta t
&=\frac{17.9-17.7}{2}\\
&=0.1\text{ s}
\end{aligned}
\]
The displayed time is therefore:
\[
t=(17.8\pm0.1)\text{ s}
\]
However, this is not yet the corrected time.
**Step 4**: Correct the stopwatch calibration error.
The stopwatch displays 9.80 s for a true interval of 10.00 s.
This means:
\[
t_{\text{display}}=0.98t_{\text{true}}
\]
To find the true time, divide the displayed time by 0.98.
\[
t_{\text{true}}=\frac{t_{\text{display}}}{0.98}
\]
Therefore, the corrected mean time is:
\[
\begin{aligned}
t_{\text{true}}
&=\frac{17.8}{0.98}\\
&=18.163\text{ s}
\end{aligned}
\]
The uncertainty must also be scaled by the same correction factor:
\[
\begin{aligned}
\Delta t_{\text{true}}
&=\frac{0.1}{0.98}\\
&\approx0.102\text{ s}
\end{aligned}
\]
Thus:
\[
t_{\text{true}}=(18.163\pm0.102)\text{ s}
\]
The important point is that calibration changes the measured value. It does not make the spread of the repeated readings disappear.
**Step 5**: Calculate the corrected experimental period.
The period is the time for one complete oscillation.
There are 10 oscillations, so:
\[
T=\frac{18.163}{10}
\]
\[
T\approx1.816\text{ s}
\]
The absolute uncertainty is:
\[
\begin{aligned}
\Delta T
&=\frac{0.102}{10}\\
&\approx0.0102\text{ s}
\end{aligned}
\]
Rounding the uncertainty to one significant figure and matching the decimal place of the measured value:
\[
\boxed{T_{\text{experimental}}=(1.82\pm0.01)\text{ s}}
\]
The corresponding rounded interval is approximately:
\[
1.81\text{ s}\leq T\leq1.83\text{ s}
\]
**Step 6**: Calculate the theoretical period.
The theoretical period of a simple pendulum undergoing small oscillations is:
\[
T=2\pi\sqrt{\frac{L}{g}}
\]
Here:
- \(T\) is the theoretical period in seconds.
- \(L\) is the pendulum length in metres.
- \(g\) is gravitational acceleration in metres per second squared.
Substitute the corrected length and gravitational acceleration:
\[
\begin{aligned}
T
&=2\pi\sqrt{\frac{0.815}{9.80}}\\
&\approx1.812\text{ s}
\end{aligned}
\]
We also need to account for the uncertainty in the length.
The length interval is:
\[
0.812\text{ m}\leq L\leq0.818\text{ m}
\]
Because the theoretical period increases with pendulum length, we can calculate the corresponding extreme periods.
At the minimum length:
\[
T_{\text{min}}=2\pi\sqrt{\frac{0.812}{9.80}}
\]
\[
T_{\text{min}}\approx1.809\text{ s}
\]
At the maximum length:
\[
T_{\text{max}}=2\pi\sqrt{\frac{0.818}{9.80}}
\]
\[
T_{\text{max}}\approx1.815\text{ s}
\]
Therefore, the theoretical period lies approximately within:
\[
\boxed{1.809\text{ s}\leq T_{\text{theoretical}}\leq1.815\text{ s}}
\]
**Step 7**: Compare the experimental and theoretical intervals.
The corrected experimental interval is approximately:
\[
[1.81,\ 1.83]\text{ s}
\]
The theoretical interval is approximately:
\[
[1.809,\ 1.815]\text{ s}
\]
These intervals overlap.
Therefore, the experimental result is consistent with the theoretical prediction under this simple interval comparison.
However, overlapping intervals do not prove that the model is correct. They simply indicate that the results are not inconsistent within the stated uncertainty estimates.
**Step 8**: Explain why the uncorrected result gives a different conclusion.
Suppose the student ignores the stopwatch calibration.
The uncorrected period would be:
\[
T_{\text{uncorrected}}=\frac{17.8}{10}
\]
\[
T_{\text{uncorrected}}=1.78\text{ s}
\]
The corresponding uncertainty would be:
\[
\Delta T_{\text{uncorrected}}=\frac{0.1}{10}=0.01\text{ s}
\]
The uncorrected result would be:
\[
T_{\text{uncorrected}}=(1.78\pm0.01)\text{ s}
\]
Its interval is:
\[
1.77\text{ s}\leq T\leq1.79\text{ s}
\]
This does not overlap with the theoretical interval.
A student might conclude that the pendulum model disagrees with the experiment.
That conclusion would be premature. The stopwatch has a known calibration error, and the repeated readings cannot reveal that error because it affects every reading in the same direction.
The correction changes the central measurement and restores consistency with the theoretical prediction.
The deeper lesson is that a small half-range does not guarantee a reliable measurement. You must consider the instrument, the experimental arrangement, and the physical quantity being measured.
---
Reporting a measurement with absolute uncertainty is the starting point for analysing experimental results, not the end of the process.
Once you can write a result correctly as \(x\pm\Delta x\), the next step is to compare measurements and decide what their uncertainty intervals actually tell you.
For example, [Do Overlapping Error Bars Mean Two Results Agree?](https://tutorgum.com/learn/resources/hsc-physics-do-overlapping-error-bars-prove-two-results-are-the-same/) explores how to interpret uncertainty when comparing experimental data. This becomes particularly useful when you start drawing error bars and calculating maximum and minimum gradients on HSC Physics graphs.