A-level Physics

1.2 Estimation and the limitations of measurement — Learn

Prerequisites: SI units and prefixes from Section 1.1; standard form; multiplying and dividing numbers in standard form; finding a mean.

By the end of this section you can: estimate a quantity to the nearest order of magnitude; decide whether an answer is reasonable; describe the resolution of an instrument; and explain the difference between accuracy, precision, systematic error and random error.

1. Estimation checks an answer

An estimate is a value found from rounded numbers, not from a careful measurement. Estimation has two uses.

To estimate, round every quantity to one significant figure and work with those numbers. The result is good to the nearest order of magnitude, which is the nearest power of ten. A value quoted to one significant figure, such as 2×1022 \times 10^2, could stand for a wide range of values. An order-of-magnitude estimate cares only that the power of ten is right.

Worked example 1. A classroom is 10 m long, 6.0 m wide and 2.5 m high. The density of air is 1.2 kg m−31.2\,\text{kg}\,\text{m}^{-3}. Estimate the mass of air in the room.

V=10×6.0×2.5=150 m3m=ρV=1.2×150=180 kg=2×102 kg(1 significant figure)\begin{aligned} V &= 10 \times 6.0 \times 2.5 = 150\,\text{m}^3 \\ m &= \rho V = 1.2 \times 150 = 180\,\text{kg} \\ &= 2 \times 10^2\,\text{kg} \quad (1\ \text{significant figure}) \end{aligned}

The estimate is 2×102 kg2 \times 10^2\,\text{kg}. The exact value depends on the temperature and the pressure of the air, so one significant figure is all the data support.

2. Every instrument has a resolution

The resolution of an instrument is the smallest change it can show. A metre rule marked in millimetres has a resolution of 1 mm. A digital balance that reads to the nearest 0.01 g has a resolution of 0.01 g.

A reading can never be more precise than the resolution. The true value may lie anywhere within half a division of the mark, so the reading uncertainty is half the smallest division.

Instrument Smallest division Uncertainty in one reading
Metre rule 1 mm 0.5 mm
Vernier caliper 0.1 mm 0.05 mm
Micrometer 0.01 mm 0.005 mm
Stopwatch 0.01 s 0.005 s from the display alone

Worked example 2. A stack of 100 sheets of paper is 12 mm thick. Find the thickness of one sheet.

thickness=12 mm100=0.12 mm\text{thickness} = \frac{12\,\text{mm}}{100} = 0.12\,\text{mm}

One sheet is thinner than the smallest division of the rule, so a direct reading of one sheet is impossible. Measuring 100 sheets divides the reading uncertainty by 100, and the same measurement on 10 000 sheets would divide it by 10 000. This is why a book of known page count measures paper better than a rule.

For a hand-held timing, the limit is usually the reaction time, about 0.2 s, not the resolution of the display. That limit is still divided when you time many oscillations and divide by the number of oscillations.

Worked example 3. A student can start and stop a stopwatch with an uncertainty of about 0.2 s. The student times 20 oscillations of a pendulum and divides by 20. Find the uncertainty in the period.

uncertainty in T=0.2 s20=0.010 s\text{uncertainty in } T = \frac{0.2\,\text{s}}{20} = 0.010\,\text{s}

3. Accuracy and precision

Two words describe the quality of a set of readings, and they are not the same.

A set of readings can be precise and inaccurate at the same time. If every reading is 0.20 mm too large, the readings agree with each other but all sit away from the true value. A set of readings can also be accurate on average but imprecise, when the readings scatter widely around the true value.

4. Systematic and random error

The difference between a reading and the true value is an error. Errors come in two kinds.

Worked example 4. Classify each error and say how to reduce it.

Error Kind How to reduce it
A balance reads 0.05 g when empty Systematic Zero the balance before use
Reaction time on a stopwatch Random Repeat the timing and take the mean
A thermometer reads 0.5 °C too high Systematic Check it at a known fixed point and correct

Watch out. A small resolution does not make a result accurate. A digital balance that reads to 0.01 g can still be 0.05 g wrong if it is not zeroed, and the display will look precise while every reading is shifted. Check by asking whether the same mistake could affect each reading.

5. The limits of a measurement

Three limits appear in every experiment.

  1. The resolution of the instrument. A measurement cannot be finer than the instrument allows.
  2. The scatter of repeated readings. Random errors spread the readings, and the spread is itself a measurement you can report.
  3. A systematic error that the method hides. No number of repeats will remove it, and the readings will look precise while being wrong.

Repeating a measurement is the main tool against the second limit. Choosing a better instrument is the main tool against the first.

6. Try it: which error does repeating remove?

Eight repeated readings of one length are plotted below. The mean is the result that is quoted, and the uncertainty shown covers the range the readings span.

Predict first. Do not move either slider yet. Write down your answers.

  1. If you increase the zero error, does the stated range get wider, narrower or stay the same width?
  2. Does the zero error move the mean?
  3. If the readings sit close together but all to one side of the true value, are they accurate?

Test. Move the zero-error slider first, then the spread slider. Watch the group of readings, the stated range and the true length.

Explain. Why does repeating a reading and taking the mean reduce a random error but not a systematic error? Write two or three sentences. Use the idea that a random error shifts each reading by a different amount, and a systematic error shifts every reading by the same amount.

Progress check

Return to the chapter problem. Before measuring, estimate the period of the 0.800 m pendulum. A 1.0 m pendulum has a period of about 2.0 s, and the period is proportional to the square root of the length.

  1. Is the period of the 0.800 m pendulum greater or less than 2.0 s?
  2. Estimate the period, then estimate the time for 20 oscillations.
  3. Compare your estimate with the recorded 35.90 s.

If question 2 goes wrong, remember that 0.800=0.894\sqrt{0.800} = 0.894. Multiply the 2.0 s period of the 1.0 m pendulum by 0.894.

You should now be able to

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