How to Distinguish Instrument Resolution from Measurement Uncertainty

A ruler has markings every 1 mm. You measure a metal rod and record its length as 84 mm. What is the uncertainty: ±1 mm, ±0.5 mm, or zero?

Under the common HSC convention of reading to the nearest scale marking, the ruler has a resolution of 1 mm and an estimated reading uncertainty of ±0.5 mm. Your measurement interval is therefore 83.5 mm to 84.5 mm.

But here’s the important distinction: instrument resolution and measurement uncertainty are not the same thing. The resolution describes the smallest change an instrument can distinguish. Measurement uncertainty describes how widely the actual value could reasonably differ from your reported result, given the instrument and the experimental method.

That second quantity can be larger than the uncertainty suggested by the instrument’s scale.

01Resolution and uncertainty answer different questions

Start by separating the two ideas.

Instrument resolution

Resolution is the smallest change in a measured quantity that produces a detectable change in the instrument’s indication.

For a typical HSC experiment, you can usually identify resolution from the smallest readable scale division or the smallest increment on a digital display.

Examples:

Resolution is a property of the measuring instrument under the conditions in which it is used.

There is a small complication. In formal metrology, resolution can also depend on noise, friction, and other properties of the measuring system. A display that changes in 0.01-unit increments does not necessarily mean the complete instrument can reliably distinguish every change of that size.

The formal definition is given in the International Vocabulary of Metrology (BIPM).

Measurement uncertainty

Measurement uncertainty describes the range or dispersion of values that can reasonably be attributed to a measurement, based on the available information.

Unlike resolution, uncertainty depends on more than the instrument.

Consider measuring the length of a metal rod. Your uncertainty might be affected by:

• The ruler’s resolution.
• Whether the ruler is aligned correctly.
• Whether the rod’s endpoints are clearly defined.
• Whether you read the scale at eye level.
• Whether the ruler’s zero mark is accurate.
• How consistently you can repeat the measurement.

Imagine two students using the same ruler.

Both record 84 mm. However, one measures a rod with sharp, clearly defined ends, while the other measures an object with a rough, rounded edge.

They have used an instrument with the same resolution, but their measurements need not have the same uncertainty.

This is why you cannot automatically equate resolution with measurement uncertainty.

How resolution leads to an uncertainty estimate

For an analogue instrument, a common HSC classroom convention is to estimate reading uncertainty as half the smallest scale division.

If the smallest division is \(r\), then:

\[ U=\frac{r}{2} \]

Here:

• \(U\) is the estimated absolute uncertainty.
• \(r\) is the instrument’s resolution.

For a ruler with 1 mm divisions:

\[ U=\frac{1}{2}=0.5\ \mathrm{mm} \]

This gives an instrument-reading uncertainty of ±0.5 mm.

It is a useful estimate, not a universal rule for every measurement. Other sources of uncertainty may increase the uncertainty of the complete experiment.

Common uncertainty conventions

Different experimental situations require different conventions.

Be particularly careful with digital instruments.

A display showing 4.82 s has a display resolution of 0.01 s. If the value is rounded to the nearest increment, the rounding-only interval is:

\[ 4.815\leq t<4.825\ \mathrm{s} \]

That does not automatically mean the complete measurement uncertainty is ±0.005 s.

The instrument might have additional calibration limitations or other sources of uncertainty.

If an examination question specifies an uncertainty convention, use it. If it does not, state your assumption clearly.

02Measurement intervals: what does the ± actually mean?

Suppose a measurement is reported as:

\[ x=(12.4\pm0.3)\ \mathrm{cm} \]

The central value is 12.4 cm, and the stated uncertainty is 0.3 cm.

If the uncertainty represents a maximum estimated bound, the measurement interval is:

\[ x-U\leq X\leq x+U \]

Substituting:

\[ 12.4-0.3\leq X\leq12.4+0.3 \]

Therefore:

\[ 12.1\leq X\leq12.7\ \mathrm{cm} \]

The interval extends 0.3 cm below and above the measured value.

However, there is another important distinction.

Not every uncertainty expressed using ± represents a maximum boundary.

Two conventions you should recognise are:

1. Maximum uncertainty bound

The stated uncertainty represents an estimated maximum deviation under the assumptions of the experiment.

For example:

\[ x=(12.4\pm0.3)\ \mathrm{cm} \]

The corresponding interval is 12.1-12.7 cm.

2. Standard uncertainty

A standard uncertainty is expressed as a standard deviation.

If a measurement is written as \(x\pm u\), where \(u\) is the standard uncertainty, the ± value is not a guaranteed boundary.

For an approximately normal distribution, one standard uncertainty corresponds to roughly 68% coverage.

These conventions are not interchangeable. The uncertainty convention determines how you interpret the interval.

The National Institute of Standards and Technology (NIST) explains how measurement uncertainty can be evaluated using both statistical and non-statistical information.

04Digital instruments: why more decimal places do not guarantee greater certainty

Digital instruments create a particularly tempting misconception.

Suppose two voltmeters measure the same potential difference.

• Voltmeter A displays 4.8 V.
• Voltmeter B displays 4.82 V.

Which instrument gives the more certain result?

You might immediately choose B because it displays an additional decimal place.

That conclusion is not necessarily correct.

Voltmeter B has a finer display resolution. It can display changes in smaller increments.

However, its measurement uncertainty might still be larger because of calibration limitations, electrical noise, or other sources of error.

Consider an illustrative situation:

Voltmeter B has finer resolution, but Voltmeter A has the smaller stated uncertainty.

The extra decimal place does not make B the more certain instrument.

What should you do with a digital instrument?

Follow this decision rule:

• If the question supplies a measurement uncertainty, use it.
• If the manufacturer or calibration information supplies an uncertainty specification, use the relevant specification.
• If the question explicitly asks for rounding uncertainty, use half the display increment when rounding to the nearest increment.
• If your teacher or assessment specifies ±1 last displayed digit, follow that convention instead.

Do not automatically assign ±0.01 s to every stopwatch displaying hundredths of a second.

The displayed increment tells you the resolution. It does not, by itself, establish the total uncertainty.

06Measurement error, uncertainty, and standard deviation

These three terms are closely related, but they describe different things.

Confusing them can lead to incorrect conclusions about experimental results.

Measurement error

Measurement error is the difference between a measured value and an accepted reference value.

We will use the sign convention:

\[ e=x_{\mathrm{measured}}-x_{\mathrm{reference}} \]

A positive error means the measurement is higher than the reference.

A negative error means the measurement is lower than the reference.

For example, suppose a student measures a resistance as 48.5 Ω, while the accepted reference value is 50.0 Ω.

The measurement error is:

\[ e=48.5-50.0 \]

\[ \boxed{e=-1.5\ \Omega} \]

The student has underestimated the resistance by 1.5 Ω.

Notice that this is not an uncertainty calculation.

The error is a difference from a reference value. Uncertainty is an estimate of the dispersion of values consistent with the measurement information.

In a real experiment, the exact value is usually unknown, so the actual measurement error is often unknown too.

Standard deviation

Standard deviation describes how much a set of repeated measurements varies around its mean.

A small standard deviation indicates that the readings are closely grouped.

A large standard deviation indicates that they are more widely spread.

However, a small standard deviation does not prove that the measurement is accurate.

Standard deviation can be used to represent a component of measurement uncertainty when an appropriate statistical model is used.

But you must distinguish between:

• The standard deviation of individual measurements, \(s\).
• The standard error of the mean, \(s/\sqrt n\).
• The total uncertainty, which may include additional contributions.

These are not interchangeable.

A contrasting case: Can a consistent instrument still be inaccurate?

Imagine two digital temperature probes measuring the same water bath.

Both display readings in increments of 0.1 °C.

An independent reference thermometer gives:

\[ T_{\mathrm{ref}}=20.50^\circ\mathrm C \]

The probes produce these illustrative measurements:

Both probes have the same resolution.

Their sample standard deviations are also identical:

\[ s_A=s_B=0.05^\circ\mathrm C \]

But their mean readings are different.

For Probe A:

\[ \bar T_A=\frac{20.0+20.0+20.1+20.0}{4} \]

\[ \bar T_A=20.025^\circ\mathrm C \]

Its error is:

\[ e_A=20.025-20.50 \]

\[ e_A=-0.475^\circ\mathrm C \]

For Probe B:

\[ \bar T_B=\frac{20.4+20.4+20.5+20.4}{4} \]

\[ \bar T_B=20.425^\circ\mathrm C \]

Its error is:

\[ e_B=20.425-20.50 \]

\[ e_B=-0.075^\circ\mathrm C \]

Probe B is much closer to the reference value.

Yet both probes have identical resolution and repeatability in this dataset.

What does this tell us?

Probe A produces consistent readings, but those readings are systematically lower than the reference.

Its small standard deviation does not reveal that bias.

Probe B is closer to the reference, even though its standard deviation is the same.

This is why precision, accuracy, resolution, and uncertainty must be considered separately.

07A reliable decision process for HSC Physics

When an examination question asks you to identify resolution or calculate measurement uncertainty, work through the following sequence.

Step 1: Identify the instrument resolution.

Find the smallest scale division or displayed increment.

Do not confuse this with the uncertainty.

Step 2: Identify the uncertainty convention.

Look for instructions such as:

• Half the smallest division.
• One least significant digit.
• Manufacturer’s stated uncertainty.
• Half the range of repeated measurements.
• Standard deviation.

If no convention is supplied, state and justify your assumption.

Step 3: Calculate the uncertainty.

Use the appropriate method.

For an analogue instrument read to the nearest graduation:

\[ U=\frac{r}{2} \]

For repeated measurements using the half-range convention:

\[ U=\frac{x_{\max}-x_{\min}}{2} \]

For the standard error of the mean:

\[ u_{\bar x}=\frac{s}{\sqrt n} \]

Step 4: Write the measurement interval if required.

For a maximum uncertainty bound:

\[ x_{\min}=x-U \]

\[ x_{\max}=x+U \]

Make sure the units are consistent.

Step 5: Check for other sources of uncertainty.

Consider calibration, zero offsets, alignment, reaction time, and repeatability.

Do not assume the instrument’s resolution accounts for all of them.

A useful final check is to ask yourself:

• Am I describing the instrument’s resolution?
• Am I estimating the uncertainty of a measurement?
• Am I calculating the difference from a reference value?
• Am I describing the spread of repeated measurements?

If you can answer those four questions, you’re much less likely to confuse the quantities.

08Questions and solutions

Question 1

A plastic strip is measured using a ruler with graduations every 1 mm.

Its left edge is aligned with the zero mark, and its right edge is read as 63 mm to the nearest graduation.

Assume the edges are clearly defined and there are no other significant sources of uncertainty.

Calculate:

a) The ruler’s resolution.

b) The estimated absolute uncertainty using the half-smallest-division convention.

c) The measurement interval.

Solution 1

The resolution is 1 mm, the estimated absolute uncertainty is ±0.5 mm, and the measurement interval is 62.5 mm to 63.5 mm.

Step 1: Identify the resolution.

The smallest graduation is 1 mm.

Therefore:

\[ r=1\ \mathrm{mm} \]

Step 2: Calculate the uncertainty.

Using the half-division convention:

\[ U=\frac{1}{2} \]

\[ \boxed{U=0.5\ \mathrm{mm}} \]

Step 3: Write the measurement.

\[ L=(63.0\pm0.5)\ \mathrm{mm} \]

Step 4: Calculate the interval.

\[ L_{\min}=63.0-0.5=62.5\ \mathrm{mm} \]

\[ L_{\max}=63.0+0.5=63.5\ \mathrm{mm} \]

Therefore:

\[ \boxed{62.5\leq L<63.5\ \mathrm{mm}} \]

The resolution is the full 1 mm division, while the uncertainty estimate is half that division.

Do not report ±1 mm unless the question specifies that convention.

Question 2

Two instruments are used to measure the same steady force.

The digital force sensor displays:

\[ F_D=3.46\ \mathrm N \]

Its display resolution is 0.01 N, and its calibration information gives a maximum measurement uncertainty of ±0.06 N.

An analogue spring balance displays:

\[ F_A=3.5\ \mathrm N \]

Its smallest scale division is 0.1 N. Use the half-smallest-division convention.

Answer the following:

a) Determine the resolution and uncertainty of each instrument.

b) Calculate the measurement interval for each result.

c) Which instrument has finer resolution?

d) Which instrument has the smaller stated uncertainty?

e) Evaluate the claim that the digital instrument must give a more certain result because it displays more decimal places.

Solution 2

The digital sensor has finer resolution, but the analogue spring balance has the smaller stated uncertainty. More decimal places do not guarantee a more certain measurement.

Step 1: Calculate the digital instrument’s resolution and uncertainty.

The display increments are 0.01 N.

Therefore:

\[ r_D=0.01\ \mathrm N \]

The question supplies the uncertainty:

\[ U_D=0.06\ \mathrm N \]

Step 2: Calculate the digital measurement interval.

\[ F_{D,\min}=3.46-0.06 \]

\[ F_{D,\min}=3.40\ \mathrm N \]

\[ F_{D,\max}=3.46+0.06 \]

\[ F_{D,\max}=3.52\ \mathrm N \]

Therefore:

\[ \boxed{3.40\leq F_D\leq3.52\ \mathrm N} \]

Step 3: Calculate the analogue instrument’s resolution and uncertainty.

The smallest scale division is 0.1 N.

Therefore:

\[ r_A=0.1\ \mathrm N \]

Using the half-division convention:

\[ U_A=\frac{0.1}{2} \]

\[ \boxed{U_A=0.05\ \mathrm N} \]

Step 4: Calculate the analogue measurement interval.

\[ F_{A,\min}=3.5-0.05=3.45\ \mathrm N \]

\[ F_{A,\max}=3.5+0.05=3.55\ \mathrm N \]

Therefore:

\[ \boxed{3.45\leq F_A\leq3.55\ \mathrm N} \]

Step 5: Compare the instruments.

The digital sensor has finer resolution because it displays smaller increments.

However, its stated uncertainty is larger.

The analogue balance therefore has the smaller uncertainty interval in this example.

The important lesson is that resolution and uncertainty describe different properties. A finer display does not automatically produce a more reliable measurement.

Question 3

A student uses two photogates connected to an electronic timer to measure the time taken for a trolley to travel between the gates.

The timer has a resolution of 0.01 s.

Five trials produce the following results:

Assume the same distance and experimental conditions are maintained for every trial.

Calculate:

a) The mean time.

b) The half-range uncertainty estimate.

c) The sample standard deviation.

d) The standard error of the mean.

e) Explain why these quantities should not be treated as interchangeable.

Solution 3

The mean time is 1.43 s, the half-range is 0.02 s, the sample standard deviation is approximately 0.0158 s, and the standard error is approximately 0.0071 s.

Step 1: Calculate the mean.

\[ \bar t=\frac{1.42+1.45+1.43+1.44+1.41}{5} \]

\[ \boxed{\bar t=1.43\ \mathrm s} \]

Step 2: Calculate the half-range uncertainty.

The maximum reading is 1.45 s.

The minimum reading is 1.41 s.

Using the half-range convention:

\[ U=\frac{t_{\max}-t_{\min}}{2} \]

\[ U=\frac{1.45-1.41}{2} \]

\[ \boxed{U=0.02\ \mathrm s} \]

Under this classroom convention, the result is:

\[ t=(1.43\pm0.02)\ \mathrm s \]

Step 3: Calculate the sample standard deviation.

Use:

\[ s=\sqrt{\frac{\sum(t_i-\bar t)^2}{n-1}} \]

Substituting:

\[ s= \sqrt{ \frac{ (-0.01)^2+(0.02)^2+0^2+ (0.01)^2+(-0.02)^2 }{4} } \]

\[ s=\sqrt{\frac{0.0010}{4}} \]

\[ \boxed{s=0.0158\ \mathrm s} \]

This describes the spread of individual measurements around the mean.

Step 4: Calculate the standard error.

The standard error estimates the random uncertainty in the mean:

\[ u_{\bar t}=\frac{s}{\sqrt n} \]

\[ u_{\bar t}=\frac{0.0158}{\sqrt5} \]

\[ \boxed{u_{\bar t}=0.0071\ \mathrm s} \]

Step 5: Explain the differences.

The four relevant quantities are:

• Resolution: 0.01 s.
• Half-range: 0.02 s.
• Sample standard deviation: 0.0158 s.
• Standard error: 0.0071 s.

The resolution describes the timer’s displayed increments.

The half-range provides a simple estimate of the spread of the readings.

The standard deviation describes the statistical spread of individual measurements.

The standard error estimates the random uncertainty in the mean, assuming independent measurements.

If an examination asks for half-range uncertainty, you should not substitute the standard error simply because it is smaller.

Always identify the convention before reporting an uncertainty.

Question 4

An electronic balance has a display resolution of 0.01 g.

Before a sample is placed on the balance, its empty reading is:

\[ m_0=+0.20\ \mathrm g \]

A student then measures a sample three times without resetting the balance.

The recorded values are:

Assume the zero offset is stable and known exactly.

a) Calculate the mean displayed mass.

b) Determine the corrected mean mass of the sample.

c) Calculate the sample standard deviation and standard error of the mean.

d) Explain whether repeating the measurements removes the zero offset.

e) Has the balance’s resolution changed after correcting the zero offset?

Solution 4

The corrected mean mass is 12.21 g, the sample standard deviation is 0.01 g, and the standard error is approximately 0.0058 g. Correcting the zero offset changes the measured value but does not change the balance’s resolution or the spread of the readings.

Step 1: Calculate the mean displayed mass.

\[ \bar m=\frac{12.40+12.42+12.41}{3} \]

\[ \boxed{\bar m=12.41\ \mathrm g} \]

This is the average displayed mass before correcting the zero offset.

Step 2: Correct the zero offset.

The empty balance reads +0.20 g.

This means the balance overreads by 0.20 g.

Therefore, subtract the offset:

\[ m_{\mathrm{corrected}}=m_{\mathrm{displayed}}-m_0 \]

\[ m_{\mathrm{corrected}}=12.41-0.20 \]

\[ \boxed{m_{\mathrm{corrected}}=12.21\ \mathrm g} \]

Step 3: Calculate the sample standard deviation.

The deviations from the original mean are:

\[ -0.01,\quad +0.01,\quad 0 \]

Therefore:

\[ s=\sqrt{\frac{(-0.01)^2+(0.01)^2+0^2}{3-1}} \]

\[ s=\sqrt{\frac{0.0002}{2}} \]

\[ \boxed{s=0.01\ \mathrm g} \]

Subtracting the same constant offset from every reading does not change the deviations from the mean.

Therefore, the standard deviation remains unchanged after correction.

Step 4: Calculate the standard error.

\[ u_{\bar m}=\frac{s}{\sqrt n} \]

\[ u_{\bar m}=\frac{0.01}{\sqrt3} \]

\[ \boxed{u_{\bar m}=0.0058\ \mathrm g} \]

This is the estimated random uncertainty in the mean due to the observed repeatability.

It is not necessarily the complete uncertainty of the corrected mass.

Step 5: Explain the zero offset and resolution.

Repeating the measurement does not remove the zero offset.

All three readings contain the same positive offset of 0.20 g. Averaging them preserves that offset.

The correction must be applied separately.

The balance’s resolution also remains 0.01 g. Subtracting a constant changes the central value, not the smallest displayed increment.

If the zero correction were uncertain rather than known exactly, that uncertainty would also need to be considered.

Question 5

A student wants to measure the period of a simple pendulum.

They use two different timing methods, each measuring ten complete oscillations. The pendulum swings through a small angle, so assume its period remains constant.

Method A: Manual stopwatch

The stopwatch has a resolution of 0.01 s.

The student starts and stops the stopwatch when the bob passes through the equilibrium position in the same direction.

Five recorded elapsed times are:

An independent video analysis establishes that:

• The student presses start 0.18 s after the initial crossing.
• The student presses stop 0.12 s after the final crossing.

Assume these delays are stable and treat their stated values as exact for the numerical correction.

Method B: Electronic photogate

The photogate has a resolution of 0.001 s.

It triggers 0.020 s after every equilibrium crossing. This delay is identical at the start and stop events and does not drift.

The photogate measures the elapsed time between crossings in the same direction, ten complete oscillations apart.

Answer the following:

a) Calculate the mean recorded time and uncorrected period using Method A.

b) Determine the corrected mean period using the reaction-time information.

c) Calculate the standard error of the mean period from the five manual measurements.

d) Calculate the rounding-only uncertainty in the period arising from the stopwatch’s resolution.

e) Does the photogate’s 0.020 s trigger delay bias its measured period? Explain.

f) A student claims that Method B must have exactly one-tenth the total uncertainty of Method A because its resolution is ten times finer. Another student claims that repeating Method A five times eliminates the reaction-time error. Evaluate both claims.

Solution 5

The corrected manual period is 1.4268 s. The manual stopwatch has a systematic timing bias that does not disappear through repetition. The photogate’s identical fixed trigger delays cancel when its start and stop events are subtracted. However, its finer resolution alone does not establish its total measurement uncertainty.

Step 1: Calculate the mean recorded time for Method A.

\[ \bar t= \frac{14.21+14.20+14.22+14.21+14.20}{5} \]

\[ \boxed{\bar t=14.208\ \mathrm s} \]

The student timed ten complete oscillations.

Therefore, the uncorrected period is:

\[ T_{\mathrm{uncorrected}}=\frac{14.208}{10} \]

\[ \boxed{T_{\mathrm{uncorrected}}=1.4208\ \mathrm s} \]

Step 2: Correct for the difference in reaction delays.

This is the most important part of the question.

The student presses start 0.18 s after the initial crossing.

This makes the recorded interval shorter than the actual interval.

The student presses stop 0.12 s after the final crossing.

This makes the recorded interval longer than it would have been if the stopwatch had stopped exactly at the crossing.

The net effect is:

\[ t_{\mathrm{recorded}} = t_{\mathrm{actual}}+0.12-0.18 \]

Therefore:

\[ t_{\mathrm{recorded}} = t_{\mathrm{actual}}-0.06 \]

The stopwatch underestimates the actual elapsed time by 0.06 s.

The correction is:

\[ t_{\mathrm{actual}}=t_{\mathrm{recorded}}+0.06 \]

Applying this to the mean:

\[ \bar t_{\mathrm{corrected}}=14.208+0.06 \]

\[ \bar t_{\mathrm{corrected}}=14.268\ \mathrm s \]

Divide by ten oscillations:

\[ T_{\mathrm{corrected}}=\frac{14.268}{10} \]

\[ \boxed{T_{\mathrm{corrected}}=1.4268\ \mathrm s} \]

The corrected period is higher than the uncorrected period because the stopwatch was started later than it was stopped, relative to the corresponding physical events.

Step 3: Calculate the standard error of the mean period.

First, calculate the sample standard deviation of the five recorded times.

The deviations from the mean of 14.208 s are:

\[ 0.002,\ -0.008,\ 0.012,\ 0.002,\ -0.008 \]

Therefore:

\[ s_t= \sqrt{ \frac{ 0.002^2+(-0.008)^2+0.012^2+ 0.002^2+(-0.008)^2 }{4} } \]

\[ s_t=\sqrt{0.000070} \]

\[ s_t=0.00837\ \mathrm s \]

Now calculate the standard error of the mean elapsed time:

\[ u_{\bar t}=\frac{0.00837}{\sqrt5} \]

\[ u_{\bar t}=0.00374\ \mathrm s \]

Since the period is the elapsed time divided by ten, divide this uncertainty by ten as well:

\[ u_T=\frac{0.00374}{10} \]

\[ \boxed{u_T=0.000374\ \mathrm s} \]

This is the repeatability-based standard error of the corrected mean period.

It describes the random variation in the repeated timing results, under the assumptions given.

Step 4: Calculate the rounding-only uncertainty from stopwatch resolution.

The stopwatch resolution is:

\[ r=0.01\ \mathrm s \]

Assuming rounding to the nearest displayed increment:

\[ U_{\mathrm{round}}=\frac{0.01}{2} \]

\[ U_{\mathrm{round}}=0.005\ \mathrm s \]

This is the rounding uncertainty in the total elapsed time.

We measured ten oscillations, so the corresponding rounding-only uncertainty in the period is:

\[ U_T=\frac{0.005}{10} \]

\[ \boxed{U_T=0.0005\ \mathrm s} \]

The distinction is important.

The stopwatch still has a resolution of 0.01 s for the total elapsed time. Dividing that elapsed time by ten produces period estimates in increments of 0.001 s.

The derived period’s rounding-only interval is therefore approximately ±0.0005 s.

This does not mean every other source of uncertainty has disappeared.

The standard error and rounding-only uncertainty describe different contributions. They should not automatically be combined, particularly because the observed scatter may already include some effects of display quantisation.

Step 5: Examine the photogate’s fixed trigger delay.

The photogate triggers 0.020 s after each crossing.

At first, this might appear to introduce a systematic error.

However, consider what happens when both events have the same delay.

The recorded start time is:

\[ t_{\mathrm{start,recorded}} = t_{\mathrm{start,actual}}+0.020 \]

The recorded stop time is:

\[ t_{\mathrm{stop,recorded}} = t_{\mathrm{stop,actual}}+0.020 \]

The measured interval is the difference:

\[ \begin{aligned} \Delta t_{\mathrm{recorded}} &=t_{\mathrm{stop,recorded}} -t_{\mathrm{start,recorded}}\\ &=(t_{\mathrm{stop,actual}}+0.020)\\ &\quad -(t_{\mathrm{start,actual}}+0.020) \end{aligned} \]

The two delays cancel:

\[ \boxed{\Delta t_{\mathrm{recorded}}= \Delta t_{\mathrm{actual}}} \]

Therefore, the photogate’s identical fixed latency does not bias the measured elapsed time.

This is a subtle but important point: a delay in recording an event does not necessarily produce an error in a measured time interval. What matters is the difference between the delays at the two endpoints.

The photogate’s 0.001 s display resolution gives a rounding-only half-increment of:

\[ U_{\mathrm{round}}=\frac{0.001}{2} =0.0005\ \mathrm s \]

For ten oscillations, the corresponding period contribution is:

\[ U_T=\frac{0.0005}{10} \]

\[ \boxed{U_T=0.00005\ \mathrm s} \]

This is ten times smaller than the manual stopwatch’s rounding-only contribution.

Step 6: Evaluate the two students’ claims.

The first student claims that the photogate must have exactly one-tenth the total uncertainty because its resolution is ten times finer.

This is incorrect.

The photogate has ten times finer display resolution, and its fixed trigger latency cancels. However, we have not been given information about its calibration, timing jitter, or other sources of uncertainty.

We can compare the rounding-only contributions, but we cannot calculate the complete uncertainty of the photogate measurement from resolution alone.

The second student claims that repeating the manual measurement five times eliminates reaction-time error.

This is also incorrect.

The difference between the start and stop reaction delays is the same in every trial.

Each recorded time therefore contains the same systematic offset of -0.06 s.

Averaging the five measurements reduces the effect of random variation on the mean, but it does not remove a common systematic bias.

There is one final detail worth noticing.

If the student timed only one oscillation using the same start and stop delays, the period would be underestimated by 0.06 s.

By timing ten oscillations, the same 0.06 s offset is spread across ten periods:

\[ \text{Bias per period}=\frac{-0.06}{10} \]

\[ \boxed{-0.006\ \mathrm s} \]

Timing more oscillations reduces the reaction-time contribution to the uncertainty of an individual period estimate, but it does not make the timing bias disappear.

The correct conclusion is that instrument resolution, repeatability, and systematic effects must be considered separately. A more finely graduated display is useful, but it is not a complete measure of how well an experiment determines a physical quantity.

Once you can distinguish resolution from uncertainty, the next step is understanding how uncertainty affects calculated results and experimental graphs.

For the related calculations, see Percentage Error vs Percentage Uncertainty: HSC Physics Guide.

When you’re ready to represent uncertainty graphically, move on to Error Bars and Maximum-Minimum Gradients: HSC Physics Guide.