Why Timing Ten Oscillations Reduces Percentage Uncertainty

A pendulum takes about 2 seconds to complete one oscillation. You start a stopwatch as it passes a reference point, then stop it when it returns. But your reaction time might introduce an uncertainty of 0.4 seconds across those two button presses.

That’s a substantial fraction of a 2-second measurement.

Now imagine timing ten complete oscillations instead. The stopwatch runs for about 20 seconds, but you still only press start once and stop once. Has the uncertainty increased tenfold?

No. The absolute timing uncertainty stays approximately the same, while the measured time becomes ten times larger. When you divide the total time by ten to calculate the average period, the absolute uncertainty in the period also becomes ten times smaller.

This is why timing ten oscillations reduces percentage uncertainty: it spreads approximately the same start-and-stop timing uncertainty across a much longer measurement interval.

Let’s work through the reasoning, including when this technique does not work as expected.

01The measurement problem: why time more than one oscillation?

Consider a simple pendulum consisting of a small bob attached to a string. The bob swings backwards and forwards around an equilibrium position.

The time taken for one complete oscillation is called the period, represented by \(T\).

Suppose you want to measure the period using a handheld stopwatch.

A sensible first attempt is:

1. Start the stopwatch when the bob passes the equilibrium position.
2. Stop it when the bob returns to that position after one complete oscillation.
3. Record the elapsed time.

There is a problem. You must react to two separate events: the starting crossing and the finishing crossing.

Even if the stopwatch displays times to two decimal places, your reaction time might be much larger than the display resolution.

For example, suppose your uncertainty in each button press is approximately \(0.20\text{ s}\). Using a conservative maximum-uncertainty convention, the uncertainty in the elapsed time is:

\[ \Delta t = 0.20+0.20 \]

\[ \Delta t = \pm 0.40\text{ s} \]

If the measured period is \(2.00\text{ s}\), this is a significant uncertainty.

The obvious improvement is to time more oscillations.

The revised experimental method

Instead of stopping the stopwatch after one oscillation:

1. Start the stopwatch when the bob passes the equilibrium position in a chosen direction.
2. Count ten complete oscillations.
3. Stop the stopwatch when the bob passes the same position in the same direction for the tenth time after starting.
4. Divide the total elapsed time by ten.

Why use the same position and direction?

Because passing the equilibrium position in the opposite direction is only half an oscillation later. You need ten complete cycles, not ten crossings.

The revised method uses exactly the same two stopwatch actions, but the measured interval is approximately ten times longer.

02Why the percentage uncertainty decreases

The key is understanding the difference between absolute uncertainty and percentage uncertainty.

Absolute uncertainty describes the estimated uncertainty in a measurement, using the same units as the measurement.

For example:

\[ t = 20.0 \pm 0.4\text{ s} \]

The absolute uncertainty is \(0.4\text{ s}\).

Percentage uncertainty compares that absolute uncertainty with the size of the measurement.

\[ \%\text{ uncertainty} = \frac{\text{absolute uncertainty}} {\text{measured value}} \times 100 \]

A fixed absolute uncertainty becomes less significant when the measured value increases.

What happens when you time ten oscillations?

Suppose the period of a pendulum is approximately \(2.00\text{ s}\).

We will use an illustrative stopwatch uncertainty of \(\pm 0.40\text{ s}\) for the complete elapsed-time interval.

Notice something important.

The stopwatch uncertainty remains \(\pm 0.40\text{ s}\) because you still have one start action and one stop action.

However, when you divide the ten-oscillation measurement by ten, you also divide its absolute uncertainty by ten.

The mathematical explanation

Let:

• \(t_N\) = total time for \(N\) complete oscillations, in seconds
• \(N\) = number of complete oscillations
• \(T\) = average period, in seconds
• \(\Delta t_N\) = absolute uncertainty in the total measured time, in seconds
• \(\Delta T\) = absolute uncertainty in the calculated period, in seconds

The average period is:

\[ T=\frac{t_N}{N} \]

Since \(N\) is an exact count, dividing the measured time by \(N\) also divides its absolute uncertainty by \(N\).

Therefore:

\[ \Delta T=\frac{\Delta t_N}{N} \]

Substituting these expressions into the percentage uncertainty equation:

\[ \%\text{ uncertainty} = \frac{\Delta T}{T}\times100 \]

\[ = \frac{\Delta t_N/N}{t_N/N}\times100 \]

The factors of \(N\) cancel:

\[ \boxed{ \%\text{ uncertainty} = \frac{\Delta t_N}{t_N}\times100 } \]

This is the important result.

If the absolute uncertainty in the total elapsed time remains approximately constant, increasing the number of oscillations increases \(t_N\), reducing the percentage uncertainty.

For a constant period and a fixed endpoint uncertainty, timing ten oscillations gives approximately one-tenth of the percentage uncertainty of timing one oscillation.

03Be precise about your uncertainty convention

There is a common source of confusion in HSC Physics: students sometimes calculate an uncertainty without explaining where the number came from.

You should always identify the convention being used.

1. Reaction-time uncertainty

For a handheld stopwatch, there are two important human actions:

• Starting the stopwatch.
• Stopping the stopwatch.

Suppose each action has a reaction-time uncertainty of \(\pm 0.20\text{ s}\).

For a conservative maximum-uncertainty estimate, add the absolute uncertainties:

\[ \Delta t = 0.20+0.20 \]

\[ \boxed{\Delta t=\pm0.40\text{ s}} \]

This is the convention used in our examples.

It is not a universal stopwatch uncertainty. Your teacher, practical instructions, or experimental procedure might specify a different value, such as \(\pm0.20\text{ s}\) for the entire interval.

Use the uncertainty actually specified, and do not add the start-and-stop uncertainties again if they have already been included.

2. Instrument resolution

Resolution is the smallest change an instrument can display.

For example, a digital stopwatch displaying hundredths of a second has a resolution of \(0.01\text{ s}\).

However, this does not mean the complete measurement uncertainty is automatically \(0.01\text{ s}\).

If you are using a handheld stopwatch, human reaction time may be much more significant than the display resolution.

A common school convention for an analogue instrument is to use half the smallest scale division. For digital instruments, the uncertainty convention should be specified or stated explicitly.

The important distinction is that resolution describes the instrument’s display, while measurement uncertainty also depends on how the measurement is performed.

3. Uncertainty is not the same as error

These terms are related, but they describe different things.

For a signed error, we can define:

\[ e=x_{\text{measured}}-x_{\text{accepted}} \]

A positive error means the measured value is higher than the accepted value. A negative error means it is lower.

Uncertainty does not tell you the actual error. You need an accepted or reference value to calculate that.

Standard deviation is different again. It tells you how spread out your repeated measurements are.

For example, you might repeatedly time ten oscillations and obtain:

\[ 19.8,\quad20.0,\quad20.2\text{ s} \]

These values have a mean of \(20.0\text{ s}\), with a sample standard deviation of \(0.2\text{ s}\).

That standard deviation describes the variation between trials. It is not automatically the same as the stopwatch’s reaction-time uncertainty.

A systematic timing bias could affect every trial in a similar way while leaving the standard deviation small.

For a deeper explanation of how repeated measurements contribute to percentage uncertainty, see How to Find Percentage Uncertainty from Repeated Measurements.

05The important contrast: timing every oscillation separately

There is a subtle mistake worth avoiding.

Suppose a student times ten oscillations individually, rather than starting the stopwatch once and stopping it after ten cycles.

They then calculate the mean of the ten measured periods.

Surely this is equivalent?

Not necessarily.

Each individual timing requires its own start and stop actions. If each interval has an absolute uncertainty of \(\pm0.40\text{ s}\), the student has introduced many more opportunities for reaction-time variation.

Compare the two methods:

Method A: Ten separate measurements

• Start and stop the stopwatch for the first oscillation.
• Repeat for each of the remaining nine oscillations.
• Calculate the mean period.

Method B: One continuous measurement

• Start the stopwatch once.
• Time ten complete oscillations.
• Stop the stopwatch once.
• Divide the total time by ten.

Under a conservative maximum-uncertainty convention, Method B has a major advantage.

Its total timing uncertainty is approximately \(\pm0.40\text{ s}\), which becomes \(\pm0.04\text{ s}\) when calculating the average period.

For Method A, each individual period has an uncertainty of approximately \(\pm0.40\text{ s}\). The conservative maximum uncertainty in the mean remains \(\pm0.40\text{ s}\) if the individual timing errors can all contribute in the same direction.

This does not mean repeated measurements are useless. If the errors are independent random variations, averaging can reduce their contribution to the uncertainty of the mean. However, the reduction follows a statistical relationship, not automatically the tenfold improvement obtained from one continuous interval.

The practical lesson is simple: when using a handheld stopwatch, time several complete oscillations with one start and one stop.

07When timing ten oscillations does not solve the problem

Timing more oscillations is useful, but it relies on an important assumption:

The period must remain approximately constant throughout the measurement interval.

Consider the simple pendulum again.

For small angular displacements, its period is approximated by:

\[ T=2\pi\sqrt{\frac{L}{g}} \]

where:

• \(T\) is the period, in seconds.
• \(L\) is the pendulum length, measured from the pivot to the centre of the bob, in metres.
• \(g\) is gravitational acceleration, in \(\text{m s}^{-2}\).

Under the small-angle approximation, the period is independent of the initial amplitude.

However, this is an approximation. At larger angular displacements, the period becomes slightly dependent on amplitude.

A real pendulum also loses energy due to air resistance and friction. Its amplitude gradually decreases.

Consequently, its period can change during a long measurement.

Imagine that the first five oscillations have an average period of \(2.03\text{ s}\), while the final five have an average period of \(2.00\text{ s}\).

Timing all ten oscillations gives:

\[ t_{10}=5(2.03)+5(2.00) \]

\[ t_{10}=20.15\text{ s} \]

The calculated average period is:

\[ T=\frac{20.15}{10} \]

\[ T=2.015\text{ s} \]

This is a good estimate of the average period across those ten oscillations.

It is not an estimate of the period at the initial amplitude alone.

A longer measurement interval has reduced the relative influence of stopwatch reaction time, but it has also averaged over a changing physical system.

There are other limitations worth remembering:

• A missed or incorrectly counted oscillation is a counting error, not a small stopwatch uncertainty.
• A systematic calibration error in the stopwatch is not eliminated by timing more oscillations.
• A changing pendulum length or amplitude can affect the measured period.
• Timing more cycles does not automatically remove systematic errors.

If the experiment is intended to measure how the period changes with amplitude, a photogate or suitable electronic timing system is more useful. It can record individual cycles without requiring a person to react to every crossing.

The choice of measurement interval should therefore balance two things: reducing endpoint timing uncertainty and keeping the physical system sufficiently constant.

08Questions and solutions

The following questions progress from routine period calculations to experimental design and the limitations of long measurement intervals.

Question 1

A student measures eight complete oscillations of a pendulum in \(12.80\text{ s}\).

The absolute uncertainty in the complete elapsed-time measurement is \(\pm0.20\text{ s}\).

Calculate:

(a) The average period of the pendulum.

(b) The absolute uncertainty in the calculated period.

(c) The percentage uncertainty in the calculated period.

Solution 1

The average period is \(1.600\text{ s}\), with an absolute uncertainty of \(\pm0.025\text{ s}\) and a percentage uncertainty of approximately \(1.56\%\).

Step 1: Calculate the average period.

\[ T=\frac{t_N}{N} \]

\[ T=\frac{12.80}{8} \]

\[ \boxed{T=1.600\text{ s}} \]

Step 2: Calculate the absolute uncertainty.

Since the oscillation count is exact:

\[ \Delta T=\frac{\Delta t_N}{N} \]

\[ \Delta T=\frac{0.20}{8} \]

\[ \boxed{\Delta T=\pm0.025\text{ s}} \]

Step 3: Calculate the percentage uncertainty.

\[ \%\text{ uncertainty} = \frac{0.025}{1.600}\times100 \]

\[ \boxed{1.56\%} \]

The complete result is:

\[ \boxed{T=(1.600\pm0.025)\text{ s}} \]

The important point is that the stopwatch uncertainty applies to the entire eight-oscillation interval, not separately to each oscillation.

Question 2

An independent reference measurement establishes that the period of a pendulum is \(2.00\text{ s}\). Treat this reference value as exact for this question.

A student obtains two results using a handheld stopwatch:

• Method A: One oscillation takes \(1.92\text{ s}\).
• Method B: Ten oscillations take \(19.70\text{ s}\).

Assume an absolute uncertainty of \(\pm0.40\text{ s}\) for either complete stopwatch interval.

For each method:

(a) Calculate the period.

(b) Calculate the percentage uncertainty.

(c) Calculate the signed percentage error relative to the reference value.

(d) Explain whether the lower percentage uncertainty proves that the method also eliminates measurement error.

Solution 2

Method B gives a more precise result, with a percentage uncertainty of approximately \(2.03\%\), compared with \(20.8\%\) for Method A. Its measured period is also closer to the reference value in this particular example.

Step 1: Calculate the period using Method A.

\[ T_A=1.92\text{ s} \]

Its absolute uncertainty is:

\[ \Delta T_A=0.40\text{ s} \]

Therefore:

\[ \%\text{ uncertainty}_A = \frac{0.40}{1.92}\times100 \]

\[ \boxed{20.8\%} \]

Step 2: Calculate the period using Method B.

\[ T_B=\frac{19.70}{10} \]

\[ \boxed{T_B=1.970\text{ s}} \]

The absolute uncertainty becomes:

\[ \Delta T_B=\frac{0.40}{10} \]

\[ \Delta T_B=0.040\text{ s} \]

Therefore:

\[ \%\text{ uncertainty}_B = \frac{0.040}{1.970}\times100 \]

\[ \boxed{2.03\%} \]

Step 3: Calculate the signed percentage errors.

Recall that:

\[ e=T_{\text{measured}}-T_{\text{accepted}} \]

For Method A:

\[ e_A=1.92-2.00=-0.08\text{ s} \]

The signed percentage error is:

\[ \frac{-0.08}{2.00}\times100 = \boxed{-4.0\%} \]

For Method B:

\[ e_B=1.970-2.00=-0.030\text{ s} \]

The signed percentage error is:

\[ \frac{-0.030}{2.00}\times100 = \boxed{-1.5\%} \]

Step 4: Explain the difference between uncertainty and error.

Method B has a smaller percentage uncertainty because the same endpoint timing uncertainty is spread across a longer measurement interval.

It also happens to have a smaller error in this example.

However, timing ten oscillations does not guarantee a smaller error in every experiment.

For example, a stopwatch with a consistent calibration bias could produce a result that is very precise but systematically too high or too low.

Reduced percentage uncertainty means improved precision under the stated assumptions. It does not automatically mean systematic errors have been eliminated.

Question 3

A student uses a digital stopwatch with a display resolution of \(0.01\text{ s}\).

For this experiment, the following uncertainty conventions are specified:

• Start reaction-time uncertainty: \(\pm0.15\text{ s}\).
• Stop reaction-time uncertainty: \(\pm0.15\text{ s}\).
• Uncertainty contribution from the displayed elapsed-time reading: \(\pm0.01\text{ s}\).

The student measures twelve complete oscillations in \(24.36\text{ s}\).

Using a conservative maximum-uncertainty convention:

(a) Calculate the absolute uncertainty in the total elapsed time.

(b) Calculate the average period and its absolute uncertainty.

(c) Calculate the percentage uncertainty.

(d) Explain why using only the stopwatch’s \(0.01\text{ s}\) display resolution would give a misleading result.

Solution 3

The average period is \(2.030\text{ s}\), with an absolute uncertainty of approximately \(\pm0.026\text{ s}\) and a percentage uncertainty of \(1.3\%\).

Step 1: Calculate the absolute uncertainty in the total elapsed time.

Under the specified conservative convention, add the three uncertainty contributions:

\[ \Delta t=0.15+0.15+0.01 \]

\[ \boxed{\Delta t=\pm0.31\text{ s}} \]

Step 2: Calculate the average period.

\[ T=\frac{24.36}{12} \]

\[ \boxed{T=2.030\text{ s}} \]

Step 3: Calculate the absolute uncertainty in the period.

\[ \Delta T=\frac{0.31}{12} \]

\[ \boxed{\Delta T=0.0258\text{ s}} \]

To an appropriate number of significant figures:

\[ \Delta T\approx0.026\text{ s} \]

Step 4: Calculate the percentage uncertainty.

\[ \%\text{ uncertainty} = \frac{0.0258}{2.030}\times100 \]

\[ \boxed{1.27\%\approx1.3\%} \]

Step 5: Explain the display-resolution misconception.

The stopwatch’s display resolution is only one contribution to the measurement uncertainty.

The student must also account for the uncertainty associated with starting and stopping the stopwatch.

Using only \(0.01\text{ s}\) would ignore the much larger reaction-time contributions.

This is why the displayed number of decimal places does not necessarily tell you how accurately a measurement has been made.

Question 4

Two groups are asked to measure the average period of a pendulum over six complete oscillations.

They use different methods.

Group A:

The students time each oscillation separately, stopping and restarting the stopwatch for every cycle. They then calculate the mean of their six period measurements.

Group B:

The students start the stopwatch once, time six complete oscillations continuously, and divide the total time by six.

Each start or stop reaction has a maximum uncertainty of \(\pm0.20\text{ s}\).

Ignore instrument resolution and use conservative maximum-uncertainty addition.

(a) Determine the maximum absolute uncertainty in the mean period reported by Group A.

(b) Determine the maximum absolute uncertainty in the mean period reported by Group B.

(c) Calculate how many times smaller Group B’s maximum uncertainty is.

(d) Explain why the result does not mean that repeated measurements are always less useful.

Solution 4

Group B has a smaller conservative maximum uncertainty in the mean period: approximately \(\pm0.067\text{ s}\), compared with \(\pm0.40\text{ s}\) for Group A.

Step 1: Calculate the uncertainty in one individual timing interval.

Each interval requires one start action and one stop action.

\[ \Delta t=0.20+0.20 \]

\[ \Delta t=\pm0.40\text{ s} \]

Step 2: Calculate Group A’s maximum uncertainty.

Each of the six individual period measurements has an uncertainty of \(\pm0.40\text{ s}\).

The mean period is:

\[ \bar T_A=\frac{T_1+T_2+\cdots+T_6}{6} \]

Using conservative maximum limits, the maximum uncertainty in the mean is:

\[ \Delta\bar T_A = \frac{6(0.40)}{6} \]

\[ \boxed{\Delta\bar T_A=\pm0.40\text{ s}} \]

This assumes the individual timing errors can all contribute in the same direction.

Step 3: Calculate Group B’s maximum uncertainty.

Group B uses one start and one stop action for the entire six-oscillation interval.

Therefore, the total elapsed-time uncertainty is:

\[ \Delta t_6=\pm0.40\text{ s} \]

The average period uncertainty is:

\[ \Delta T_B=\frac{0.40}{6} \]

\[ \boxed{\Delta T_B=\pm0.0667\text{ s}} \]

Step 4: Compare the two methods.

The ratio of their maximum uncertainties is:

\[ \frac{0.40}{0.0667}\approx6 \]

\[ \boxed{\text{Group B has a sixfold smaller maximum uncertainty.}} \]

Step 5: Explain the role of repeated measurements.

Group A’s repeated measurements can still be useful.

If the timing variations are independent random errors, averaging the results can reduce the random contribution to uncertainty in the mean.

However, Group B avoids repeatedly introducing new start-and-stop reaction events. Under the conservative maximum-uncertainty convention specified here, this gives it a clear advantage.

The two methods are not equivalent simply because they both involve six oscillations.

Question 5

A student wants to measure the period of a pendulum immediately after it is released from an initial angular displacement of \(30^\circ\) from the vertical.

The pendulum is damped, so its amplitude gradually decreases.

For this question, use the following illustrative model:

• The average period during the first five oscillations is \(2.03\text{ s}\).
• The average period during the next five oscillations is \(2.00\text{ s}\).

The student’s handheld stopwatch has an absolute uncertainty of \(\pm0.40\text{ s}\) for the complete elapsed-time interval.

The student times all ten oscillations in one measurement and obtains \(20.15\text{ s}\).

They calculate:

\[ T=\frac{20.15}{10}=2.015\text{ s} \]

\[ \Delta T=\frac{0.40}{10}=0.040\text{ s} \]

They conclude:

“The period at the instant of release is \(2.015\text{ s}\), with a percentage uncertainty of approximately \(2\%\). Timing even more oscillations would give an even more accurate measurement of the initial period.”

Evaluate this conclusion.

Your response should:

(a) Calculate the percentage uncertainty in the ten-oscillation result.

(b) Determine what physical quantity the ten-oscillation result actually represents.

(c) Explain why the smaller percentage uncertainty does not necessarily make this a better measurement of the initial period.

(d) Propose an improved experimental method for measuring the initial period while reducing human reaction-time uncertainty.

Solution 5

The student’s percentage uncertainty is approximately \(1.99\%\), but their conclusion about the initial period is incorrect. The result represents the average period across ten oscillations, during which the pendulum’s amplitude and period change.

Step 1: Calculate the percentage uncertainty.

The calculated average period is:

\[ T=2.015\text{ s} \]

The absolute uncertainty is:

\[ \Delta T=0.040\text{ s} \]

Therefore:

\[ \%\text{ uncertainty} = \frac{0.040}{2.015}\times100 \]

\[ \boxed{1.99\%\approx2.0\%} \]

The student’s calculation is mathematically correct.

The problem is the interpretation of the result.

Step 2: Determine what the ten-oscillation measurement represents.

The first five oscillations have an average period of \(2.03\text{ s}\).

Their combined time is:

\[ t_1=5(2.03)=10.15\text{ s} \]

The final five oscillations have an average period of \(2.00\text{ s}\).

Their combined time is:

\[ t_2=5(2.00)=10.00\text{ s} \]

The total time is:

\[ t_{\text{total}}=10.15+10.00 \]

\[ t_{\text{total}}=20.15\text{ s} \]

Dividing by ten gives:

\[ T_{\text{average}}=\frac{20.15}{10} \]

\[ \boxed{T_{\text{average}}=2.015\text{ s}} \]

This is the average period across the ten oscillations.

It is not the period at the initial release amplitude.

Step 3: Explain why the physical system changes the interpretation.

For a small angular displacement, the simple pendulum model predicts:

\[ T\approx2\pi\sqrt{\frac{L}{g}} \]

This approximation treats the period as independent of amplitude.

However, at larger angular displacements, the period is slightly longer. As the pendulum loses energy, its amplitude decreases and its period approaches the small-angle value.

The ten-cycle measurement therefore averages periods from two different stages of the pendulum’s motion.

The lower percentage uncertainty tells us that the average period has been measured with relatively little endpoint timing uncertainty.

It does not tell us that the average period is the same as the initial period.

In fact, the difference between the early-cycle average and the ten-cycle average is:

\[ 2.03-2.015=0.015\text{ s} \]

This is smaller than the calculated absolute uncertainty of \(0.040\text{ s}\).

Consequently, the handheld stopwatch measurement cannot reliably distinguish these two values.

Step 4: Propose an improved experimental method.

A photogate connected to an electronic timer or data logger would be more appropriate.

Place the photogate at a suitable reference position so the bob’s crossings can be recorded electronically.

Record successive crossings in the same direction to determine individual oscillation periods, rather than relying on human reaction time to start and stop a stopwatch.

Then:

1. Calculate the period of each early oscillation.
2. Record how the period changes as the amplitude decreases.
3. Use the earliest measured cycles to estimate the initial period.
4. If necessary, examine the trend in period against cycle number or amplitude and extrapolate towards the initial release condition.

A photogate reduces the uncertainty associated with human reaction time, but it does not prevent the pendulum’s amplitude from changing. Recording individual periods is what allows you to distinguish the initial behaviour from the later behaviour.

The deeper lesson: timing more oscillations improves the precision of an average-period measurement when the period is approximately constant. It does not guarantee a more accurate measurement of a quantity that changes during the experiment.

09Where to take this next

Once you can calculate the uncertainty in a period, the next step is understanding how uncertainty affects comparisons between experimental results.

In particular, learn how to distinguish overlapping uncertainty intervals from genuinely agreeing measurements, and how to use uncertainty when calculating gradients.

Two useful continuations are:

• Percentage Error vs Percentage Uncertainty: HSC Physics Guide
• Error Bars and Maximum-Minimum Gradients: HSC Physics Guide

The central decision to remember is this: use a longer measurement interval when the timing uncertainty is dominated by fixed start-and-stop effects, but make sure the physical quantity you are measuring stays approximately constant throughout that interval.