Analysing experimental data
A student in a school in Kildare measures $g$ with a pendulum and gets $9.4\ \text{m s}^{-2}$. The accepted value is $9.81\ \text{m s}^{-2}$. Has the experiment failed? Or is the result good enough? In Leaving Certificate Physics you are expected to answer that question with numbers, not with a shrug. By the end of this page you will be able to.
Errors, accuracy and precision
Every measurement is slightly wrong. Your job is to know how wrong, and why.
Two words that sound alike but are different:
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Accuracy: how close a result is to the true (accepted) value.
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Precision: how close repeated readings are to each other.
Random errors are reduced by repeating readings and taking the average. Systematic errors are not reduced by repeating: you have to find and remove the cause (zero the balance, compensate for friction, read the scale at eye level).
Why time many oscillations?
Your reaction time when starting and stopping a stopwatch is about $\pm 0.2\ \text{s}$, no matter how long the event is. So it matters far less on a long time than on a short one.
A student measures the period $T$ of a simple pendulum of length $l = 1.00\ \text{m}$ by timing $20$ complete oscillations, which take $40.4\ \text{s}$. a) Calculate the…
A student measures the period $T$ of a simple pendulum of length $l = 1.00\ \text{m}$ by timing $20$ complete oscillations, which take $40.4\ \text{s}$.
- a) Calculate the period of the pendulum.
- b) Using $T=2\pi\sqrt{l/g}$, calculate the value of $g$.
- c) Explain why timing $20$ oscillations is better than timing one.
Tables, ratios and percentage difference
Most experiments give you pairs of readings. The first step is to turn them into the quantities you actually need, and the second is to ask whether a pattern holds.
Take a spring. In Hooke's law, $F = ks$, the force should be proportional to the extension, so $F/s$ should be the same for every reading. Calculate the ratio for each row and see.
| Load $F$ / N | Extension $s$ / m | $F/s$ / N m⁻¹ |
|---|---|---|
| 1.0 | 0.025 | 40.0 |
| 2.0 | 0.051 | 39.2 |
| 3.0 | 0.074 | 40.5 |
| 4.0 | 0.101 | 39.6 |
| 5.0 | 0.124 | 40.3 |
The ratios are close but not identical. That is what random error looks like. If instead the ratio drifted steadily (say from $40$ to $30$), the spring would not be obeying Hooke's law, or you would have passed its elastic limit.
Be careful to distinguish this from percentage uncertainty, which compares the size of the possible error with the measurement itself:
$$\text{percentage uncertainty} = \frac{\text{absolute uncertainty}}{\text{measured value}} \times 100$$The two together let you judge a result. If the percentage difference is no bigger than the percentage uncertainty, the result agrees with the accepted value within experimental error.
Back to the Kildare student: $g = 9.4$ against $9.81$ gives $\dfrac{0.41}{9.81}\times 100 = 4.2\%$. If the estimated uncertainty in the experiment is about $5\%$ (reading $l$ and timing $T$), the result is consistent with $9.81$ within experimental error. The experiment has not failed, though a more careful one would do better.
Significant figures
A result cannot be more precise than the readings it comes from. Give your answer to the same number of significant figures as the least precise reading (usually $2$ or $3$). Keep extra digits in the middle of a calculation and round only at the end.
Density from measurements
To find density, $\rho = \dfrac{m}{V}$, measure the mass with a balance and get the volume from the dimensions. Convert to SI units first: $1\ \text{cm}^3 = 10^{-6}\ \text{m}^3$. For example, a block $5.0\ \text{cm}\times 2.0\ \text{cm}\times 4.0\ \text{cm}$ has $V = 40\ \text{cm}^3 = 4.0\times 10^{-5}\ \text{m}^3$. To identify the material, compare your density with the table of densities in your data booklet and pick the closest, quoting the property you compared (density) and how close it is.
A student in a school in Kildare verifies Hooke's law by hanging masses from a spring and measuring its extension. | Mass $m$ / kg | Extension $x$ / m | |---|---| | 0.10…
A student in a school in Kildare verifies Hooke's law by hanging masses from a spring and measuring its extension.
| Mass $m$ / kg | Extension $x$ / m |
|---|---|
| 0.10 | 0.020 |
| 0.20 | 0.041 |
| 0.30 | 0.059 |
| 0.40 | 0.081 |
- a) Calculate the force on the spring for each mass, using $g = 9.8\ \text{m s}^{-2}$.
- b) Calculate the ratio $F/x$ for each reading.
- c) State what the results suggest about Hooke's law.
A student determines the density of a metal block. Its mass is $0.2160\ \text{kg}$, and its dimensions are $6.0\ \text{cm}\times 3.0\ \text{cm}\times 4.0\ \text{cm}$. a)…
A student determines the density of a metal block. Its mass is $0.2160\ \text{kg}$, and its dimensions are $6.0\ \text{cm}\times 3.0\ \text{cm}\times 4.0\ \text{cm}$.
- a) Calculate the volume of the block in $\text{m}^3$.
- b) Calculate the density in $\text{kg m}^{-3}$.
- c) Suggest, from the table of densities of metals in the data booklet, a metal that this block could be made from, and state the property you compared.
Graphs, gradients and linearising
A single ratio uses only one reading. A graph uses them all, and a line of best fit averages out random errors. The aim is nearly always to arrange the variables so that the graph is a straight line, because the gradient then gives the quantity you want.
Why square the time? Because $T$ against $l$ is a curve, and you cannot easily read a constant from a curve. Squaring $T$ gives a straight line through the origin.
Rules for good graphs:
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Put the variable you control (independent) on the $x$-axis and the one you measure on the $y$-axis (unless the formula says otherwise).
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Label each axis with the quantity and its unit, and choose a scale that uses most of the page.
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Draw the line of best fit so that points lie evenly above and below it. A good straight line with points scattered around it is not joined dot to dot.
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For the gradient, take two points on the line, far apart, and show the triangle. (When a question tells you to use the first and last data points, do that.)
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A point far from the trend is an anomalous value: check it, repeat it if you can, and say why you left it out.
A line that does not pass through the origin when theory says it should shows a systematic error. If the pendulum graph of $T^2$ against $l$ has an intercept on the $T^2$ axis, you may have measured $l$ only to the top of the bob instead of to its centre.
A student verifies $F_{net}=ma$ using a trolley of total mass $0.80\ \text{kg}$ on a runway with friction compensated. The trolley is pulled by different forces and the…
A student verifies $F_{net}=ma$ using a trolley of total mass $0.80\ \text{kg}$ on a runway with friction compensated. The trolley is pulled by different forces and the acceleration is found using light gates.
| Force $F$ / N | Acceleration $a$ / $\text{m s}^{-2}$ |
|---|---|
| 0.20 | 0.26 |
| 0.40 | 0.49 |
| 0.60 | 0.77 |
| 0.80 | 1.02 |
| 1.00 | 1.24 |
- a) Describe the graph that should be plotted and the shape expected.
- b) Calculate the gradient from the first and last data points.
- c) Use the gradient to calculate the mass of the trolley, and calculate the percentage difference from $0.80\ \text{kg}$.
- d) Give one reason for the difference.
Students investigate the period $T$ of a pendulum for different lengths $l$. | $l$ / m | $T$ / s | |---|---| | 0.20 | 0.90 | | 0.40 | 1.27 | | 0.60 | 1.55 | | 0.80 |…
Students investigate the period $T$ of a pendulum for different lengths $l$.
| $l$ / m | $T$ / s |
|---|---|
| 0.20 | 0.90 |
| 0.40 | 1.27 |
| 0.60 | 1.55 |
| 0.80 | 1.79 |
| 1.00 | 2.01 |
- a) Calculate $T^2$ for each length.
- b) Calculate the gradient of a graph of $T^2$ against $l$ using the first and last points.
- c) Use the gradient to calculate $g$, given that $T^2=\dfrac{4\pi^2}{g}l$.
- d) State why the amplitude of swing should be kept small.
A student uses a ball dropped from different heights $h$ and a light-gate timer to measure the time $t$ to fall, and then finds $g$. | $h$ / m | $t$ / s | |---|---| |…
A student uses a ball dropped from different heights $h$ and a light-gate timer to measure the time $t$ to fall, and then finds $g$.
| $h$ / m | $t$ / s |
|---|---|
| 0.20 | 0.203 |
| 0.40 | 0.286 |
| 0.60 | 0.350 |
| 0.80 | 0.404 |
| 1.00 | 0.452 |
- a) Show that $h=\tfrac12 g t^2$ means a plot of $h$ against $t^2$ is a straight line through the origin, and give the gradient in terms of $g$.
- b) Calculate $t^2$ for each point and the gradient from the first and last points.
- c) Calculate $g$ and the percentage difference from $9.81\ \text{m s}^{-2}$.
- d) Discuss the main source of error and how it would affect the intercept of the graph.
Testing conservation laws and judging results
Some experiments do not use a graph. They compare a quantity before and after, and check whether it is conserved within experimental error.
The same approach works for energy. If a trolley falls through a height $h$, the loss in $E_p = mgh$ should equal the gain in $E_k = \tfrac12 mv^2$. A gap means energy went elsewhere (friction, air resistance, rotation of the wheels) or there is a measurement error. State at least two specific causes, not just "human error".
Speeds from light gates
A light gate measures the time $t$ for which a card of length $d$ blocks the beam. This gives the average speed $v = d/t$ over the card, not the instantaneous speed at one point.
If the trolley is accelerating, the speed at the middle of the card is a better value than at its front edge, and a shorter card gives a better estimate of the speed at a point (but a smaller time to measure). It is a trade-off.
Resonance of an air column
In a resonance tube, a tuning fork of frequency $f$ is held over a tube partly filled with water. Resonance (a loud note) occurs at tube lengths $L_1$, $L_2$ … The air column behaves like a tube closed at the water end, so:
$$L_1 + e = \frac{\lambda}{4}, \qquad L_2 + e = \frac{3\lambda}{4}$$where $e$ is the end correction (the antinode forms slightly beyond the open end). Subtracting, $L_2 - L_1 = \lambda/2$, and $e$ cancels. So the difference between two resonances gives $\lambda$ without needing to know $e$, which is why using two resonances beats using one. Then $v = f\lambda$.
To estimate the percentage uncertainty, divide the reading uncertainty (for example $\pm 0.5\ \text{cm}$ on each length) by the quantity it affects. To combine uncertainties, add the absolute uncertainties when quantities are added or subtracted (so $L_2 - L_1$ carries $\pm 1.0\ \text{cm}$), and add the percentage uncertainties when quantities are multiplied or divided (a square such as $T^2$ doubles the percentage uncertainty of $T$).
Writing the conclusion
A good conclusion has three parts:
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The result, with units and sensible significant figures.
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The comparison with the accepted value (percentage difference) and with the uncertainty.
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The main source of error and a specific improvement (for example "use a shorter card", "time $20$ oscillations", "repeat three times and average", "keep the amplitude below about $10^\circ$").
A student investigates conservation of momentum with two trolleys on a runway. Trolley A has mass $0.50\ \text{kg}$ and moves at $0.80\ \text{m s}^{-1}$ towards trolley…
A student investigates conservation of momentum with two trolleys on a runway. Trolley A has mass $0.50\ \text{kg}$ and moves at $0.80\ \text{m s}^{-1}$ towards trolley B, of mass $0.50\ \text{kg}$, which is stationary. They stick together on impact. The student measures the velocity after the collision as $0.37\ \text{m s}^{-1}$.
- a) Calculate the total momentum before the collision.
- b) Calculate the total momentum after the collision using the measured velocity.
- c) Calculate the percentage difference and state whether momentum is conserved within experimental error.
- d) Calculate the kinetic energy before and after, and explain the difference.
A student verifies conservation of energy for a trolley rolling down a low-friction ramp. The trolley has mass $0.60\ \text{kg}$. It starts from rest and descends a…
A student verifies conservation of energy for a trolley rolling down a low-friction ramp. The trolley has mass $0.60\ \text{kg}$. It starts from rest and descends a vertical height of $0.15\ \text{m}$. A light gate measures its speed at the bottom as $1.62\ \text{m s}^{-1}$.
- a) Calculate the loss in gravitational potential energy.
- b) Calculate the gain in kinetic energy.
- c) Calculate the percentage of energy lost and suggest two possible sources of this loss.
- d) The light gate measures the speed using a card of length $5.0\ \text{cm}$ and a time of $0.0309\ \text{s}$. Explain why this gives an average speed rather than the speed at the bottom, and how this affects the result.
A student measures the speed of sound in air by the resonance of a tuning fork of frequency $512\ \text{Hz}$ above a tube partly filled with water. The first resonance…
A student measures the speed of sound in air by the resonance of a tuning fork of frequency $512\ \text{Hz}$ above a tube partly filled with water. The first resonance is found when the air column is $16.0\ \text{cm}$ long, and the second resonance at $49.5\ \text{cm}$.
- a) Calculate the wavelength using the difference between the two resonance lengths.
- b) Calculate the speed of sound.
- c) Use the first resonance length to find the end correction $e$, given that $L_1+e=\lambda/4$.
- d) Explain why the use of two resonances gives a better value than the use of one, and estimate the percentage uncertainty if each length has an uncertainty of $\pm 0.2\ \text{cm}$.
Back to the start: $9.4\ \text{m s}^{-2}$ against $9.81\ \text{m s}^{-2}$ is a $4.2\%$ difference. With an uncertainty of around $5\%$ the result agrees with the accepted value within experimental error, so the experiment has worked, and the report should say what to improve.
Summary
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Random errors scatter readings and are reduced by repeating and averaging; systematic errors shift all readings the same way and must be removed at source. Accuracy is closeness to the true value, precision is closeness of repeated readings.
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Timing many oscillations reduces the percentage uncertainty in $T$ because the reaction-time error is shared out.
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Percentage difference $= \dfrac{|\text{measured}-\text{accepted}|}{\text{accepted}}\times 100$; percentage uncertainty $= \dfrac{\text{absolute uncertainty}}{\text{measured value}}\times 100$. A result agrees if its difference is within the uncertainty.
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Rearrange the formula to $y = mx + c$ (for example $T^2$ against $l$, $h$ against $t^2$) so the gradient gives the quantity wanted. Label axes with units, draw a best-fit line and use a large gradient triangle.
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To test conservation of momentum or energy, compare before and after, calculate the percentage difference and name specific sources of loss.
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Give answers to sensible significant figures and always finish with a conclusion that includes one improvement.