Chapter 1: Review

The chapter problem, solved

The pendulum in the church tower has a length of 0.800 m from the pivot to the centre of the bob. The time for 20 oscillations was measured three times as 35.90 s, 35.85 s and 35.95 s. The metre rule reads to 1 mm, so each reading carries an uncertainty of 0.0005 m, and the length carries $0.0005 + 0.0005 = 0.001\,\text{m}$ .

1. The mean time for 20 oscillations.

\begin{aligned} \text{mean} &= \frac{35.90 + 35.85 + 35.95}{3} = 35.90\,\text{s} \\ \text{range} &= 35.95 - 35.85 = 0.10\,\text{s} \\ \text{uncertainty} &= \frac{0.10}{2} = 0.05\,\text{s} \end{aligned}

The time is $35.90 \pm 0.05\,\text{s}$ .

2. The period and its absolute uncertainty. Dividing by 20 divides both the time and its uncertainty.

\begin{aligned} T &= \frac{35.90}{20} = 1.795\,\text{s} \\ \Delta T &= \frac{0.05}{20} = 0.0025\,\text{s} \end{aligned}

3. The percentage uncertainty in the period.

\frac{0.05}{35.90} \times 100\% = 0.139\%

4. The percentage uncertainty in the length. The length is $0.800 \pm 0.001\,\text{m}$ .

\frac{0.001}{0.800} \times 100\% = 0.125\%

5. The acceleration due to gravity.

\begin{aligned} g &= \frac{4\pi^2 L}{T^2} \\ &= \frac{4\pi^2 \times 0.800}{1.795^2} \\ &= 9.802\,\text{m}\,\text{s}^{-2} \end{aligned}

6. The percentage uncertainty in $g$ . The period is squared, so its percentage uncertainty is doubled, and the exact constant $4\pi^2$ adds nothing.

\begin{aligned} \text{percentage uncertainty in } g &= 0.125\% + 2 \times 0.139\% \\ &= 0.40\% \end{aligned}

7. The absolute uncertainty and the quoted result.

\begin{aligned} \Delta g &= 9.802 \times \frac{0.40}{100} = 0.04\,\text{m}\,\text{s}^{-2} \\ g &= 9.80 \pm 0.04\,\text{m}\,\text{s}^{-2} \end{aligned}

8. Does the experiment agree with the accepted value? The measured range is 9.76 to 9.84 m s⁻². The accepted value, 9.81 m s⁻², lies inside the range, so the experiment agrees with it within the stated uncertainty.

9. The length the clock needs. A clock keeps time when its pendulum has a period of exactly 2.000 s.

\begin{aligned} T &= 2\pi\sqrt{\frac{L}{g}} \\ L &= \frac{gT^2}{4\pi^2} \\ &= \frac{9.81 \times 4.000}{4\pi^2} \\ &= 0.994\,\text{m} \end{aligned}

The clock pendulum must be about 0.994 m long. The 0.800 m pendulum has a period of 1.795 s, which is shorter than 2.000 s, so the clock would run fast. The technician must lengthen the pendulum from 0.800 m to about 0.994 m.

10. The judgement. The measurement is consistent with the accepted value, so it gives no reason to doubt the accepted value. It does not prove that the value is right. The uncertainty allows for random scatter, and a systematic error in the length, at the centre of the bob, would not appear in it. This final judgement is a reasoned answer. It is not the kind of claim a computer can check, so compare your reasoning with the evidence above.

Mixed practice — choose the method first

For each question, say which rule or equation you will use before you calculate.

The 20-oscillation timings are $35.90\,\text{s}$ , $35.85\,\text{s}$ and $35.95\,\text{s}$ . Find the mean time and its uncertainty using the half-range convention.
The time for 20 oscillations is $35.90\,\text{s}$ . Find the period $T$ of one oscillation in seconds.
Use $g = \dfrac{4\pi^2 L}{T^2}$ with $L = 0.800\,\text{m}$ and $T = 1.795\,\text{s}$ to find $g$ .
The percentage uncertainty in the length is $0.125\%$ and the percentage uncertainty in the period is $0.139\%$ . Find the percentage uncertainty in $g$ , using $g = \dfrac{4\pi^2 L}{T^2}$ .
The value of $g$ is $9.802\,\text{m}\,\text{s}^{-2}$ and its percentage uncertainty is $0.40\%$ . Find the absolute uncertainty in $g$ .
The value of $g$ is $9.802\,\text{m}\,\text{s}^{-2}$ and the absolute uncertainty is $0.04\,\text{m}\,\text{s}^{-2}$ . Write the result in the correct form.
The experiment gives $g = 9.80 \pm 0.04\,\text{m}\,\text{s}^{-2}$ . The accepted value is $9.81\,\text{m}\,\text{s}^{-2}$ . Decide whether the experiment agrees with the accepted value, and give a reason.
A clock keeps time when its pendulum has a period of exactly $2.000\,\text{s}$ . Use $T = 2\pi\sqrt{\dfrac{L}{g}}$ with $g = 9.81\,\text{m}\,\text{s}^{-2}$ to find the length the pendulum needs.
Show that the period equation $T = 2\pi\sqrt{\dfrac{L}{g}}$ is homogeneous.
A pressure is quoted as $2.5\,\text{MPa}$ . Write it in pascals and in base units.
Estimate the number of seconds in a year to one significant figure.
A metal block has mass $84.6 \pm 0.1\,\text{g}$ and volume $30.0 \pm 0.5\,\text{cm}^3$ . Find the density and its absolute uncertainty.
A student must find the acceleration of a trolley down a ramp. Say which quantity the student should measure directly, which equation relates the measurements, and which uncertainty is likely to dominate the result.

Answers and concise solutions

1. The mean time and its uncertainty

Answer: The mean is $\dfrac{35.90 + 35.85 + 35.95}{3} = 35.90\,\text{s}$ . The range is $0.10\,\text{s}$ , so the uncertainty is half the range, $0.05\,\text{s}$ . The time for 20 oscillations is $35.90 \pm 0.05\,\text{s}$ .

2. The period of one oscillation

Answer: $T = \dfrac{35.90}{20} = 1.795\,\text{s}$ .

3. The value of g

Answer: $g = \dfrac{4\pi^2 \times 0.800}{1.795^2} = 9.80\,\text{m}\,\text{s}^{-2}$ to three significant figures.

4. The percentage uncertainty in g

Answer: The period appears squared, so its percentage uncertainty is doubled: $2 \times 0.139\% = 0.279\%$ . The constant $4\pi^2$ contributes nothing. Add the length: $0.125\% + 0.279\% = 0.40\%$ to two significant figures.

5. The absolute uncertainty in g

Answer: $0.40\%$ of $9.802$ is $\dfrac{0.40}{100} \times 9.802 = 0.039\,\text{m}\,\text{s}^{-2}$ , which is $0.04\,\text{m}\,\text{s}^{-2}$ to one significant figure.

6. The result in the correct form

Answer: The uncertainty is $0.04\,\text{m}\,\text{s}^{-2}$ to one significant figure, so the value is quoted to the same decimal place: $g = 9.80 \pm 0.04\,\text{m}\,\text{s}^{-2}$ .

7. Whether the experiment agrees with the accepted value

The experimental value lies between 9.76 and 9.84 m s^-2.
The accepted value 9.81 m s^-2 falls inside that range.
The experiment agrees with the accepted value within its stated uncertainty. A disagreement would need a gap larger than the uncertainty, not merely a different value.

8. The length the pendulum needs

Answer: Rearranging $T = 2\pi\sqrt{\dfrac{L}{g}}$ gives $L = \dfrac{gT^2}{4\pi^2} = \dfrac{9.81 \times 4.000}{4\pi^2} = \dfrac{9.81}{\pi^2} = 0.994\,\text{m}$ . The 0.800 m pendulum is too short, so its period is less than 2.000 s and the clock would run fast. The technician must lengthen the pendulum to about 0.994 m.

9. A homogeneity check

Answer: Squaring both sides gives $T^2 = \dfrac{4\pi^2 L}{g}$ , and $4\pi^2$ has no unit. The left side has base units $\text{s}^2$ . On the right, $\dfrac{\text{m}}{\text{m}\,\text{s}^{-2}} = \text{s}^2$ . Both sides match, so the equation is homogeneous.

10. The pressure in pascals

Answer: $2.5\,\text{MPa} = 2.5 \times 10^6\,\text{Pa}$ . A pascal is $\text{kg}\,\text{m}^{-1}\,\text{s}^{-2}$ , so the pressure in base units is $2.5 \times 10^6\,\text{kg}\,\text{m}^{-1}\,\text{s}^{-2}$ .

11. The number of seconds in a year

Answer: $365 \times 24 \times 3600 = 3.15 \times 10^7$ , so about $3 \times 10^7\,\text{s}$ to one significant figure.

12. The density and its absolute uncertainty

Answer: The density is $\dfrac{84.6}{30.0} = 2.82\,\text{g}\,\text{cm}^{-3}$ . The percentage uncertainty in the mass is $0.12\%$ and in the volume is $1.7\%$ , so in the density it is $0.12\% + 1.7\% = 1.8\%$ . The absolute uncertainty is $2.82 \times 0.018 = 0.05\,\text{g}\,\text{cm}^{-3}$ , so the density is $2.82 \pm 0.05\,\text{g}\,\text{cm}^{-3}$ .

13. The method and the dominant uncertainty

Measure the time for the trolley to travel a measured distance, using light gates or by timing several runs.
Measure the distance along the ramp, and take account of the ramp angle if the acceleration along the ramp is wanted.
Starting from rest, use s = 1/2 a t^2 to find a = 2s / t^2.
The time appears squared, so its percentage uncertainty is doubled and usually dominates the uncertainty in the acceleration.

What you can now do

name the SI base units, and write a derived unit in base units;
change a prefix, including for a squared or a cubed unit;
check that an equation is homogeneous;
estimate a quantity to the nearest order of magnitude;
state the resolution and the reading uncertainty of an instrument;
tell accuracy from precision, and a systematic error from a random error;
find the mean and the half-range uncertainty of repeated readings;
convert between absolute and percentage uncertainty;
combine uncertainties through a sum, a difference, a product, a quotient and a power;
quote a result with its uncertainty, and decide whether it agrees with an accepted value.

Your relevance reflection

Return to the prompt from the chapter opener: think of a measurement you rely on in a normal week, and what could make the reading wrong. Write one more example now that you have finished the chapter. Has the chapter changed what you notice about a reading, a quoted value or an uncertainty?