Error Bars and Maximum-Minimum Gradients: HSC Physics Guide

A spring stretches as you add masses. You measure its extension, calculate the applied force, and plot a graph. The points sit close to a straight line, and the best-fit gradient gives a spring constant of \(50\ \mathrm{N\,m^{-1}}\).

But how certain is that value? Could the gradient reasonably be \(45\ \mathrm{N\,m^{-1}}\), or perhaps \(55\ \mathrm{N\,m^{-1}}\)? A best-fit line gives you one estimate, but it doesn’t show how much the result could change within the uncertainty of your measurements.

This is where error bars and maximum-minimum gradient lines become useful. Together, they let you estimate not just a physical quantity, but also how reliable your estimate is.

01The short answer: what do error bars and maximum-minimum gradient lines show?

There are two connected ideas:

• Error bars show the stated uncertainty interval around a measured data point. They indicate the range of values allowed by the uncertainty convention being used.
• Maximum and minimum gradient lines are the steepest and shallowest straight lines that remain consistent with those uncertainty intervals. Their gradients estimate the range of plausible gradients.

The gradient of the best-fit line is your central estimate. The maximum and minimum gradients help you estimate the uncertainty in that result.

The usual calculation is:

\[ \Delta m = \frac{m_{\max}-m_{\min}}{2} \]

where:

• \(m_{\max}\) is the maximum numerical gradient.
• \(m_{\min}\) is the minimum numerical gradient.
• \(\Delta m\) is the estimated absolute uncertainty in the best-fit gradient.

This is the worst-fit-line method described in the University of New South Wales’ Data Processing and Analysis Guide for Stage 6 Physics.

The important distinction is that the best-fit line estimates the relationship, while the extreme lines estimate how much uncertainty there is in its gradient.

02Start with the experiment: why do we need these lines?

Imagine investigating whether a spring obeys Hooke’s law.

You attach a spring to a clamp stand. A ruler is fixed vertically alongside it, and a mass hanger hangs from the bottom of the spring. As you add masses, the spring stretches.

The experimental question is straightforward:

Does the spring’s extension increase in proportion to the applied force?

The prediction comes from Hooke’s law:

\[ F=kx \]

where:

• \(F\) is the applied force in newtons (N).
• \(k\) is the spring constant in newtons per metre (\(\mathrm{N\,m^{-1}}\)).
• \(x\) is the extension in metres (m).

For a hanging mass, the applied force is its weight:

\[ F=mg \]

Here, \(m\) is the total hanging mass in kilograms and \(g\) is gravitational field strength, approximately \(9.8\ \mathrm{m\,s^{-2}}\).

If Hooke’s law holds, doubling the applied force should double the extension, provided the spring remains within its proportional limit.

Plot force on the vertical axis and extension on the horizontal axis. The prediction is a straight line through the origin, with gradient equal to the spring constant.

However, real measurements aren’t perfectly exact.

A ruler reading might be slightly high or low. The mass measurement might have limited resolution. The spring might still be oscillating when its position is recorded.

The first experimental design might involve holding a ruler by hand next to the spring. That introduces two problems: the ruler can move, and viewing it from an angle creates parallax error.

A better design clamps the ruler to the stand, attaches a pointer to the spring, and takes readings at eye level. The spring is allowed to settle before each measurement.

The improved arrangement reduces avoidable uncertainty. It doesn’t eliminate every source of uncertainty, which is exactly why we need to represent the remaining uncertainty on the graph.

03What does an error bar actually mean?

Suppose you measure an extension and obtain:

\[ x=2.40\pm0.10\ \mathrm{cm} \]

This means your estimated extension is \(2.40\ \mathrm{cm}\), with an uncertainty of \(0.10\ \mathrm{cm}\).

The corresponding measurement interval is:

\[ 2.30\ \mathrm{cm}\leq x\leq2.50\ \mathrm{cm} \]

An error bar represents this interval visually.

Notice that the uncertainty is not the same thing as saying the measurement is wrong by \(0.10\ \mathrm{cm}\).

It tells you how precisely the value is known according to the stated measurement convention.

Choosing an uncertainty convention

This is where students sometimes lose marks. The symbol \(\pm\) does not automatically tell you how the uncertainty was calculated.

You need to know where the uncertainty came from.

The analogue instrument convention is particularly useful for HSC practical questions.

For example, if a ruler has markings every \(1\ \mathrm{mm}\), and you can confidently read to half a division, the reading uncertainty is:

\[ \Delta L=\pm0.5\ \mathrm{mm} \]

For a digital instrument, don’t automatically assume that the last displayed digit gives the complete measurement uncertainty. The instrument’s accuracy specification or the convention stated in your task takes priority.

The UNSW Stage 6 Physics guide also distinguishes between half-range estimates for small sets of repeated trials and standard deviation methods for larger sets. These are useful conventions, not a universal rule that every practical task must use identically.

Uncertainty, error, and standard deviation are different

These three terms are related, but they don’t mean the same thing.

Strictly, measurement error is the measured result minus the value of the quantity being measured. It is usually unknown because the exact reference value is not known.

Measurement uncertainty can still be estimated even when the actual error is unknown.

This distinction is also made in NIST’s measurement guidance.

One subtle point: standard deviation can itself be used as a way of expressing measurement uncertainty. However, it is not automatically interchangeable with instrument uncertainty, half-range uncertainty, or uncertainty in the mean.

Consider a stopwatch experiment.

Five measurements might be tightly grouped around \(1.90\ \mathrm{s}\), but the stopwatch could have a systematic calibration error that makes every reading \(0.10\ \mathrm{s}\) too high.

The standard deviation could be very small, even though every measurement is biased.

A small standard deviation suggests consistent readings. It does not prove that those readings are accurate.

04How to draw maximum and minimum gradient lines

Now we can use the uncertainty intervals to estimate the uncertainty in a graph’s gradient.

Suppose you plot force against extension.

Each point has an uncertainty interval. If both the horizontal and vertical measurements have uncertainty, each point represents a rectangular region of possible values.

A candidate straight line is consistent with a point if it passes through its uncertainty region.

This gives you two extreme possibilities.

The maximum gradient line

For a positive relationship, the maximum gradient line is the steepest acceptable line.

It generally connects:

• The lower-force side of the first uncertainty region.
• The higher-force side of the last uncertainty region.

If horizontal uncertainty is also included, the line should use the corners that minimise the horizontal separation.

The minimum gradient line

The minimum gradient line is the shallowest acceptable line.

It generally connects:

• The higher-force side of the first uncertainty region.
• The lower-force side of the last uncertainty region.

Again, account for horizontal uncertainty if it is significant.

There is one important condition: the extreme lines must also be checked against the intermediate error bars.

You cannot simply join two convenient endpoints if the resulting line misses most of the other uncertainty regions.

The aim is to find the steepest and shallowest straight lines that remain consistent with as many uncertainty bars as possible.

Calculating the uncertainty in the gradient

Once you have drawn the lines, calculate their gradients using:

\[ m=\frac{\Delta y}{\Delta x} \]

where \(m\) is the graph’s gradient.

Then calculate:

\[ \boxed{\Delta m=\frac{m_{\max}-m_{\min}}{2}} \]

This gives an estimate of the absolute uncertainty in the gradient.

For a positive gradient:

\[ m_{\min}<m_{\mathrm{best}}<m_{\max} \]

The best-fit gradient is the central estimate, while the extreme gradients provide an approximate range.

If the best-fit gradient is not close to the midpoint of the extreme gradients, report the full interval or asymmetric uncertainties rather than assuming that a symmetric uncertainty around the best-fit value captures both limits.

Be careful with negative gradients

There is a small terminology trap here.

Maximum and minimum gradient refer to numerical values, not simply how steep a line looks.

For example:

\[ -1.2>-1.8 \]

Therefore, \(-1.2\) is the maximum numerical gradient, even though \(-1.8\) is steeper in magnitude.

For a negative relationship, make sure you preserve the signs when calculating the uncertainty.

06A practical decision process for HSC questions

When a question asks you to draw error bars and calculate maximum and minimum gradients, follow this sequence.

Two mistakes are particularly common.

First, students calculate the maximum gradient using two original data points rather than points on their maximum-gradient line. You must calculate the gradient of the line you have drawn.

Second, students forget that the gradient has units. A force-extension graph measured in newtons and centimetres gives a gradient in \(\mathrm{N\,cm^{-1}}\), not \(\mathrm{N\,m^{-1}}\).

Always check the axis units before reporting a physical constant.

07Questions and solutions

Question 1

A student measures the time taken for a trolley to travel along a track.

The measured time is:

\[ t=2.46\pm0.01\ \mathrm{s} \]

Write the measurement interval and explain what the uncertainty means.

Solution 1

The measurement interval is \(2.45\) to \(2.47\ \mathrm{s}\).

The central estimate is \(2.46\ \mathrm{s}\), and the stated uncertainty is \(0.01\ \mathrm{s}\).

Calculate the limits:

\[ t_{\min}=2.46-0.01=2.45\ \mathrm{s} \]

\[ t_{\max}=2.46+0.01=2.47\ \mathrm{s} \]

Therefore:

\[ 2.45\leq t\leq2.47\ \mathrm{s} \]

The uncertainty describes the estimated measurement interval under the stated convention.

It does not mean that the actual measurement error is definitely \(0.01\ \mathrm{s}\), nor does it automatically represent a 95% confidence interval.

Question 2

A student investigates the resistance of a resistor using a variable DC power supply.

An ammeter is connected in series with the resistor, while a voltmeter is connected in parallel across it.

The student plots current \(I\) against potential difference \(V\).

The following gradients are obtained:

• Best-fit gradient: \(0.080\ \mathrm{A\,V^{-1}}\)
• Maximum gradient: \(0.086\ \mathrm{A\,V^{-1}}\)
• Minimum gradient: \(0.074\ \mathrm{A\,V^{-1}}\)

Calculate:

a. The uncertainty in the gradient.

b. The resistance calculated from the best-fit gradient.

c. The minimum and maximum possible resistance based on the extreme gradients.

d. Whether a resistance of \(13.0\ \Omega\) is consistent with the experimental results.

Solution 2

The resistance is approximately \(12.5\ \Omega\), with an estimated interval of \(11.6\) to \(13.5\ \Omega\). Therefore, \(13.0\ \Omega\) is consistent with the experimental results.

Step 1: Calculate the gradient uncertainty.

\[ \Delta m=\frac{m_{\max}-m_{\min}}{2} \]

\[ \Delta m=\frac{0.086-0.074}{2} \]

\[ \boxed{\Delta m=0.006\ \mathrm{A\,V^{-1}}} \]

Step 2: Identify what the gradient represents.

The graph has current on the vertical axis and voltage on the horizontal axis.

Ohm’s law is:

\[ V=IR \]

Rearrange:

\[ I=\frac{V}{R} \]

Therefore:

\[ \frac{I}{V}=\frac{1}{R} \]

The gradient represents reciprocal resistance, also called conductance.

This is an important distinction. If the axes were reversed, the gradient would represent resistance directly.

Step 3: Calculate the best-estimate resistance.

\[ R=\frac{1}{m_{\mathrm{best}}} \]

\[ R=\frac{1}{0.080} \]

\[ \boxed{R=12.5\ \Omega} \]

Step 4: Calculate the resistance limits.

The largest gradient gives the smallest resistance:

\[ R_{\min}=\frac{1}{0.086} \]

\[ R_{\min}=11.63\ \Omega \]

The smallest gradient gives the largest resistance:

\[ R_{\max}=\frac{1}{0.074} \]

\[ R_{\max}=13.51\ \Omega \]

Therefore:

\[ \boxed{11.6\leq R\leq13.5\ \Omega} \]

Notice that we inverted the extreme gradients rather than simply transferring the gradient uncertainty to resistance.

The relationship is reciprocal, so the resistance interval is not perfectly symmetrical around \(12.5\ \Omega\).

Step 5: Check the proposed resistance.

The proposed value is:

\[ R=13.0\ \Omega \]

This lies between \(11.6\ \Omega\) and \(13.5\ \Omega\).

Therefore, the value is consistent with the experimental uncertainty.

It isn’t proof that the resistor has exactly this resistance. It means the experimental result does not rule it out.

Question 3

A student repeats a timing experiment five times and records:

\[ 1.82,\ 1.91,\ 1.88,\ 1.95,\ 1.84\ \mathrm{s} \]

The teacher specifies that the graph should use error bars of \(\pm1\) sample standard deviation.

The stopwatch has a stated resolution uncertainty of \(\pm0.01\ \mathrm{s}\).

Calculate:

a. The mean time.

b. The sample standard deviation.

c. The error-bar interval required by the teacher.

d. Explain why the standard deviation and stopwatch resolution uncertainty are not interchangeable.

Solution 3

The mean time is \(1.88\ \mathrm{s}\), and the sample standard deviation is approximately \(0.052\ \mathrm{s}\).

The required error bars are therefore \(\pm0.052\ \mathrm{s}\), because the teacher has specifically requested one standard deviation.

Step 1: Calculate the mean.

\[ \bar{t}=\frac{\sum t}{n} \]

Substitute the five measurements:

\[ \bar{t}= \frac{1.82+1.91+1.88+1.95+1.84}{5} \]

\[ \bar{t}=\frac{9.40}{5} \]

\[ \boxed{\bar{t}=1.88\ \mathrm{s}} \]

Step 2: Calculate the sample standard deviation.

Use:

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

The squared deviations from the mean are:

\[ \begin{aligned} (1.82-1.88)^2&=0.0036\\ (1.91-1.88)^2&=0.0009\\ (1.88-1.88)^2&=0\\ (1.95-1.88)^2&=0.0049\\ (1.84-1.88)^2&=0.0016 \end{aligned} \]

Adding them:

\[ \sum(t_i-\bar{t})^2=0.0110\ \mathrm{s^2} \]

There are five measurements, so:

\[ s=\sqrt{\frac{0.0110}{5-1}} \]

\[ s=\sqrt{0.00275} \]

\[ \boxed{s=0.0524\ \mathrm{s}} \]

Step 3: Calculate the required error-bar interval.

The teacher has specified one sample standard deviation.

Therefore:

\[ t=1.88\pm0.052\ \mathrm{s} \]

The corresponding interval is approximately:

\[ 1.828\leq t\leq1.932\ \mathrm{s} \]

Step 4: Explain the difference between the two uncertainties.

The standard deviation describes the spread of the repeated measurements around their mean.

The stopwatch resolution uncertainty describes the limitation associated with reading the instrument.

They measure different things.

The teacher has requested standard-deviation error bars, so using \(\pm0.01\ \mathrm{s}\) instead would be incorrect.

Also, a one-standard-deviation interval is not automatically a 95% confidence interval, and it does not need to contain every individual reading.

A systematic timing bias could still affect all five measurements without producing a large standard deviation.

Question 4

A student investigates the behaviour of a spring.

The following illustrative data are obtained:

Every extension has an uncertainty of \(\pm0.10\ \mathrm{cm}\), and every force has an uncertainty of \(\pm0.05\ \mathrm{N}\).

The first four points follow an approximately linear relationship.

The fifth point is repeated, producing the same result. When the spring is unloaded, it returns to its original length.

Should the student draw one maximum and minimum gradient line through all five points?

Explain what the measurements suggest about the spring’s behaviour.

Solution 4

No. A single straight-line model cannot adequately represent all five measurements within their uncertainty intervals.

The first four measurements are consistent with Hooke’s law, but the fifth indicates that the spring’s force-extension relationship has become nonlinear.

Step 1: Examine the first four points.

The first four measurements follow:

\[ F\approx0.50x \]

where force is in newtons and extension is in centimetres.

The approximate gradient is:

\[ m_{\mathrm{best}}=0.50\ \mathrm{N\,cm^{-1}} \]

Step 2: Check the steepest acceptable line through the first four points.

Using the first and fourth uncertainty regions, the steepest line can pass through:

\[ (1.10,\ 0.45) \]

and

\[ (3.90,\ 2.05) \]

Its gradient is:

\[ m_{\max}=\frac{2.05-0.45}{3.90-1.10} \]

\[ m_{\max}=\frac{1.60}{2.80} \]

\[ m_{\max}=0.571\ \mathrm{N\,cm^{-1}} \]

Step 3: Extrapolate this line to the fifth measurement.

The largest possible extension for the fifth measurement is:

\[ x=5.10\ \mathrm{cm} \]

Using the steepest acceptable line, the predicted force is approximately:

\[ F_{\max}=0.45+0.571(5.10-1.10) \]

\[ F_{\max}=2.74\ \mathrm{N} \]

However, the fifth measurement has a minimum possible force of:

\[ F_{\min}=2.90-0.05 \]

\[ F_{\min}=2.85\ \mathrm{N} \]

Even the steepest acceptable line through the first four measurements cannot reach the fifth uncertainty region.

Step 4: Interpret the physical behaviour.

The fifth point is not consistent with the original linear model.

Because the spring returns to its original length after unloading, it is still behaving elastically.

However, it has moved beyond its limit of proportionality: force is no longer directly proportional to extension.

This illustrates why you shouldn’t force maximum and minimum gradient lines through every point just to obtain a numerical uncertainty.

First establish whether a straight-line relationship is physically justified. Then calculate gradient uncertainty over the appropriate linear region.

Question 5

A student investigates a spring using a ruler.

Instead of measuring the unloaded spring length separately before every trial, the student measures it once and reuses that reference reading for all three loads.

The measurements are:

Unloaded length:

\[ L_0=10.00\pm0.05\ \mathrm{cm} \]

Loaded measurements:

Assume the three loaded-length readings and the force measurements have independent uncertainties. The same unloaded-length measurement is used for every extension.

A student calculates each extension using \(x=L-L_0\), assigns each extension an uncertainty of \(\pm0.10\ \mathrm{cm}\), and then treats the three horizontal error bars as completely independent.

They calculate the maximum and minimum gradients by moving the first and last horizontal error bars in opposite directions.

Is this method valid?

Calculate the gradient uncertainty using:

a. The student’s independent-error-bar assumption.

b. The correct treatment of the shared unloaded-length measurement.

Explain why the two answers differ.

Solution 5

The student’s method overestimates the uncertainty in the gradient because the unloaded-length uncertainty is shared by all three extension measurements.

The shared reference shifts every extension by the same amount. It cannot independently move the first and last extension measurements in opposite directions.

This is a subtle distinction between individual measurement intervals and the relationship between measurements.

Step 1: Calculate the nominal extensions.

For each measurement:

\[ x=L-L_0 \]

Therefore:

\[ x_1=12.00-10.00=2.00\ \mathrm{cm} \]

\[ x_2=14.00-10.00=4.00\ \mathrm{cm} \]

\[ x_3=16.00-10.00=6.00\ \mathrm{cm} \]

The nominal gradient, using the first and last measurements, is:

\[ m_{\mathrm{best}}= \frac{3.00-1.00}{6.00-2.00} \]

\[ m_{\mathrm{best}}=0.50\ \mathrm{N\,cm^{-1}} \]

Step 2: Calculate the student’s result.

Each extension is assigned an uncertainty of:

\[ \Delta x=0.05+0.05=0.10\ \mathrm{cm} \]

If the student treats these intervals as independent, the smallest possible horizontal separation is:

\[ \Delta x_{\min}=5.90-2.10 \]

\[ \Delta x_{\min}=3.80\ \mathrm{cm} \]

The largest possible force difference is:

\[ \Delta F_{\max}=3.02-0.98 \]

\[ \Delta F_{\max}=2.04\ \mathrm{N} \]

Therefore:

\[ m_{\max,\mathrm{naive}}= \frac{2.04}{3.80} \]

\[ m_{\max,\mathrm{naive}}=0.537\ \mathrm{N\,cm^{-1}} \]

For the minimum gradient:

\[ \Delta x_{\max}=6.10-1.90 \]

\[ \Delta x_{\max}=4.20\ \mathrm{cm} \]

The smallest force difference is:

\[ \Delta F_{\min}=2.98-1.02 \]

\[ \Delta F_{\min}=1.96\ \mathrm{N} \]

Therefore:

\[ m_{\min,\mathrm{naive}}= \frac{1.96}{4.20} \]

\[ m_{\min,\mathrm{naive}}=0.467\ \mathrm{N\,cm^{-1}} \]

The student’s estimated gradient uncertainty is:

\[ \Delta m_{\mathrm{naive}}= \frac{0.537-0.467}{2} \]

\[ \boxed{\Delta m_{\mathrm{naive}}=0.035\ \mathrm{N\,cm^{-1}}} \]

Step 3: Identify the shared measurement.

The key is that the same \(L_0\) appears in every extension calculation.

Consider the difference between the first and third extensions:

\[ x_3-x_1=(L_3-L_0)-(L_1-L_0) \]

Expand:

\[ x_3-x_1=L_3-L_0-L_1+L_0 \]

The two reference-length terms cancel:

\[ \boxed{x_3-x_1=L_3-L_1} \]

The uncertainty in the shared unloaded length does not contribute to the separation between these two extension values.

It shifts both horizontal positions together, rather than moving them independently.

Step 4: Calculate the correct horizontal interval.

The first and third loaded-length readings each have independent uncertainties of \(\pm0.05\ \mathrm{cm}\).

Therefore, the difference between their lengths has a worst-case uncertainty of:

\[ \Delta x_{\mathrm{difference}}=0.05+0.05 \]

\[ \Delta x_{\mathrm{difference}}=0.10\ \mathrm{cm} \]

The nominal separation is \(4.00\ \mathrm{cm}\).

Thus:

\[ 3.90\leq\Delta x\leq4.10\ \mathrm{cm} \]

The common uncertainty in \(L_0\) has cancelled.

Step 5: Recalculate the maximum and minimum gradients.

The force difference still has the same uncertainty:

\[ 1.96\leq\Delta F\leq2.04\ \mathrm{N} \]

For the maximum gradient:

\[ m_{\max}=\frac{2.04}{3.90} \]

\[ \boxed{m_{\max}=0.523\ \mathrm{N\,cm^{-1}}} \]

For the minimum gradient:

\[ m_{\min}=\frac{1.96}{4.10} \]

\[ \boxed{m_{\min}=0.478\ \mathrm{N\,cm^{-1}}} \]

Step 6: Calculate the corrected gradient uncertainty.

\[ \Delta m=\frac{0.523-0.478}{2} \]

\[ \boxed{\Delta m=0.0225\ \mathrm{N\,cm^{-1}}} \]

The corrected gradient estimate is approximately:

\[ m_{\mathrm{best}}=0.500\ \mathrm{N\,cm^{-1}} \]

with an uncertainty of approximately:

\[ 0.023\ \mathrm{N\,cm^{-1}} \]

The corresponding spring constant is approximately:

\[ k=(50.0\pm2.3)\ \mathrm{N\,m^{-1}} \]

The student’s original calculation gave an uncertainty of \(0.035\ \mathrm{N\,cm^{-1}}\), compared with the corrected estimate of \(0.0225\ \mathrm{N\,cm^{-1}}\).

The difference comes from treating a shared measurement as though it were independently uncertain in every trial.

This is a more advanced experimental-analysis issue, but it reveals something important: error bars describe individual measurement intervals, while the relationship between measurements can also depend on whether their uncertainties are independent or correlated.

08What should you remember?

When interpreting an experimental graph, keep three questions separate:

1. What is the best estimate of the gradient?
2. How much can the gradient vary within the stated measurement uncertainties?
3. Does the experimental relationship actually support the physical model being tested?

Error bars help answer the second question. Maximum and minimum gradient lines turn those intervals into an estimated uncertainty in the gradient.

But neither technique fixes a systematic error, proves a linear relationship, or guarantees that the true value lies within the calculated interval.

Once you’ve established a reliable gradient and its uncertainty, the next decision is how that gradient connects to the physical quantity you’re investigating. Sometimes it represents the quantity directly, as with the spring constant. Sometimes it represents its reciprocal, as with the resistance example.

Getting that final connection right is what turns graph analysis into a meaningful physics result.