Lesson Explainer: Measurement Uncertainty and Resolution | Nagwa Lesson Explainer: Measurement Uncertainty and Resolution | Nagwa

Lesson Explainer: Measurement Uncertainty and Resolution Physics • First Year of Secondary School

In this explainer, we will learn how to define resolution-based and random measurement uncertainties, and show how they affect the values of measurements.

In physics, we are often required to make measurements. These could be of an object’s size, of an amount of time, or of the brightness of a star. Before we can infer anything from the quantities we measure, we have to understand the limitations of the measurement.

All measurements are limited by the devices we use to make them. For example, imagine we wish to measure the length of an object using the ruler below, which has markings in centimetres.

The degree of fineness to which an instrument can be read is known as the resolution. In this case, the ruler has a resolution of 1 cm.

We can see that the object is closer to the 5 cm marking than the 6 cm mark, so we would record the length as 5 cm. However, it is clearly not exactly 5 cm.

Using this ruler, we would record any object that is closer to the 5 cm mark than to any other as measuring 5 cm. This means an object could be as short as 4.5 cm, or anywhere up to 5.5 cm, and we would record its length as 5 cm.

We call this the uncertainty in the measurement. There are many sources of uncertainty, but here it is the uncertainty due to the resolution of the ruler.

The uncertainty on that measurement is equal to half of the range of likely values. In this case, the range is 5.54.5=1cmcmcm, and half of the range is 0.5×1=0.5cmcm. We write this uncertainty as ±0.5 cm to indicate that the true value may be as low as 50.5=4.5cmcmcm or as high as 5+0.5=5.5cmcmcm.

Note that this is equal to half of the resolution of the ruler. When calculating uncertainty due to the resolution of an instrument, the range of likely values is equal to the resolution. We can therefore say that the uncertainty is equal to half of the resolution.

We could reduce the uncertainty in the measurement of our object by using a different ruler, say, one that has markings every millimetre instead of every centimetre. This ruler has a resolution of 1 mm. When an instrument can be read more finely, we say that it has higher resolution.

With the higher resolution of this ruler, we can now say that our object is closest to the 5.3 cm mark. In reality, when we report this measurement, this means it could lie anywhere between 5.25 cm and 5.35 cm, so we would write the measurement as 5.3±0.05 cm. By increasing the resolution of our measuring device, we have therefore reduced the uncertainty on the resulting measurement. A measurement with lower uncertainty is said to be more precise.

We could use an even higher resolution instrument to measure this object, and this would reduce the uncertainty further still and result in an even more precise measurement. However, some amount of uncertainty will always exist. All instruments that we use to make measurements have some limited resolution, and therefore all measurements have some amount of uncertainty.

Definition: Uncertainty and Resolution

The resolution of a measuring device is the “fineness” to which the instrument can be read. An instrument that can measure a quantity more finely is said to have higher resolution.

The uncertainty of a measurement is the interval over which the “true” value of a measured quantity is likely to fall. It is equal to half of the range of likely values.

A measurement with a smaller uncertainty is said to be more precise.

Let’s look at an example of comparing the precision of two instruments.

Example 1: Identifying Instrument Resolution

The diagram shows two digital timers that have different resolutions. Both timers display time in seconds.

  1. Which of the two digital timers has the higher resolution?
  2. Which of the two digital timers can make more precise measurements?

Answer

Part 1

In this example, we are shown two digital timers. Timer (a) shows a time of 25.56 seconds, and timer (b) shows a time of 16.9 seconds. The first part of the question asks us to decide which timer has the higher resolution.

Recall that resolution is the degree of fineness to which an instrument can be read. An instrument with higher resolution can be read more finely than one with lower resolution.

Looking at our two timers, we have timer (a) that records the time to the nearest 0.01 seconds and timer (b) that records time to the nearest 0.1 seconds. Timer (a) can be read more finely. Therefore, the digital timer with the highest resolution is timer (a).

Part 2

In the next part of the question, we are asked which of the two digital timers can make more precise measurements.

A more precise measurement is one with lower uncertainty, so let’s consider the uncertainty in the two readings.

Timer (a) shows a reading of 25.56 s. The true value could be anywhere between 25.555 s and 25.565 s. This is a range of likely values of 25.56525.555=0.01sss.

Timer (b) shows a reading of 16.9 s, which could indicate a true value anywhere between 16.85 s and 16.95 s. The range of likely values is 16.9516.85=0.1sss.

The timer with the smallest interval in which the true value could lie has the lowest uncertainty, and hence the highest precision. Therefore, the timer that can make more precise measurements is timer (a).

In the next example, we will calculate the range and uncertainty of a measured value.

Example 2: Understanding Measurement Uncertainties

A small object is measured using a measuring stick with marks 1 cm apart, as shown in the diagram. The left-hand end of the object is closer to the first mark (zero cm) than it is to the 1 cm mark, and the right-hand end of the object is closer to the 2 cm mark than it is to the 3 cm mark.

  1. What is the maximum length that the object could have?
  2. What is the minimum length that the object could have?
  3. What is the measured length of the object?
  4. What is the uncertainty in the measured length of the object?

Answer

In this example, we are trying to measure the length of a small object using a ruler with a resolution of 1 cm.

We are told that the left-hand end is somewhere between the 0 cm and 1 cm marks but is closer to 0 cm. The right-hand end lies somewhere between the 2 cm and 3 cm marks but is closer to 2 cm. To measure the length of the object, we take the reading at the right-hand end and subtract the reading at the left-hand end.

Part 1

We first need to determine the maximum length that the object could have. This is the measurement we would read if the right-hand end was the furthest to the right it can be and the left-hand end is the furthest to the left. This is the range marked in blue on the diagram. The furthest to the right that the right-hand end can be is 2.5 cm; any further and it would be read as 3 cm. Similarly, the furthest left that the left-hand end can be is at 0 cm. So the maximum length that the object could have is 2.50=2.5cmcmcm.

Part 2

Next, we need to find the minimum length the object could have. This range is indicated in red on the diagram; it covers the range from the furthest right that the left-hand end could be to the furthest left that the right-hand end could be. We know that the left-hand end is closer to 0 cm than to 1 cm, so the largest value it could have is 0.5 cm. Similarly, we know that the right-hand end lies somewhere between 2 cm and 3 cm, so the lowest measurement it could have is 2 cm. Therefore, the minimum length the object could have is 20.5=1.5cmcmcm.

Part 3

For the next part, we need to determine the measured length of the object. Here, we take the closest marks on either end. On the right-hand end, that is 2 cm, and on the left-hand end, it is 0 cm. The measured value is therefore 20=2cmcmcm.

Part 4

Finally, we need to determine the uncertainty in the measured length of the object. The uncertainty is defined as half of the range of likely values.

We have already found the maximum value, 2.5 cm, and the minimum value, 1.5 cm. The range of possible measurements is therefore 2.51.5=1cmcmcm. Half of the range is 0.5×1.5=0.5cmcm.

The uncertainty in the measured length of the object is therefore ±0.5 cm.

Instrument resolution is a source of uncertainty that applies to every measurement we make, but it is not the only source of uncertainty. Another form of uncertainty that we encounter regularly is random uncertainty due to changes in the quantity being measured.

For example, say we are trying to measure the length of a metal pipe. The metal expands when it is warm and contracts when it is cold, so we might obtain different measurements depending on the temperature on the day we make the measurement.

The way we reduce random uncertainty is to make many repeated measurements. We can then take the mean of the set of values as a best estimate of the true value. To estimate the uncertainty on that measurement, we can then give the range of values recorded, and the uncertainty is again half of the range of likely values, so that randomuncertaintymaximumvalueminimumvalue=2.

We will see this in practice in the following example.

Example 3: Understanding Random Uncertainty

The length of a metal pipe is measured, and the length varies slightly for different measurements. The measurements are shown in the table.

Measurement12345
Length (cm)100.6100.3100.2100.2100.2
  1. Find the mean length of the pipe.
  2. Find the uncertainty in the length of the pipe due to its length changes.
  3. The pipe lengths are measured to a resolution of 0.1 cm. Is the uncertainty in the pipe length due to the precision of the measurements greater than, less than, or equal to the uncertainty due to the changes in the length?

Answer

Part 1

We have a metal pipe that we are trying to measure the length of. In the table, we see five measurements that indicate the length is changing between measurements. The first part of the question asks us to find the mean length of the pipe. To do this, we need to recall that meansumofmeasurementsnumberofmeasurements=.

In this case, the number of measurements is 5, so we can substitute that and the measurements themselves in and we find meancmcmcmcmcmcm=100.6+100.3+100.2+100.2+100.25=100.3.

So, the mean length of the pipe is 100.3 cm.

Part 2

The uncertainty in this measurement is the random uncertainty due to the changes in the length. We can find this uncertainty by taking randomuncertaintymaximumvalueminimumvalue=2.

Here, the maximum value measured is 100.6 cm, and the minimum value is 100.2 cm, so we have randomuncertaintycmcmcmcm=100.6100.22=0.42=0.2.

The uncertainty in the length of the pipe due to its length changes is therefore ±0.2 cm.

Part 3

Finally, we are told that the resolution of the instrument used to measure the pipe is 0.1 cm. Recall that uncertainty due to resolution is equal to half of the resolution of the instrument. Therefore, the uncertainty due to the precision of the measurement is uncertaintyresolutioncmcm=2=0.12=0.05.

So, we have a random uncertainty due to length changes of ±0.2 cm and uncertainty due to the precision of the measurement of ±0.05 cm. So, the uncertainty due to the precision of the measurement is less than the uncertainty due to changes in the length.

Another type of uncertainty we may encounter is systematic uncertainty. This occurs when there is some flaw in the experimental design: perhaps a ruler that been warped, a scale that has not been correctly calibrated, or a repeated error in reading the measurement. Systematic uncertainties result in measurements being consistently read as too high or too low. Unlike random uncertainties, we cannot reduce systematic effects by taking repeated measurements, as the error is present in every measurement. The best way to reduce systematic uncertainties is to take a measurement of a known quantity and check that we obtain the expected result.

When we state a measurement as some value ± some uncertainty, this is known as the absolute uncertainty. We might also express the uncertainty as a percent uncertainty.

To calculate the percent uncertainty, we use percentuncertaintyabsoluteuncertaintymeasuredvalue=×100%.

For our measurement of 5±0.5 cm, we would calculate the percent uncertainty as percentuncertaintycmcm=0.55×100%=10%.

If we measured a length of 50 cm for another object with the same ruler, we would obtain the same absolute uncertainty of ±0.5 cm. The percent uncertainty in this case would be percentuncertaintycmcm=0.550×100%=1%.

So, two measurements with the same absolute uncertainty can have different percent uncertainties. The percent uncertainty is useful to see how significant the uncertainty is. In the example here, we had two measurements with the same absolute uncertainty of ±0.5 cm but different measured lengths of 5 cm and 50 cm. The uncertainty is much more significant when measuring smaller lengths, and we can see this more clearly when we look at the percent uncertainties of 10% and 1%.

Let’s work through an example of converting absolute to percent uncertainties.

Example 4: Converting from Absolute to Percent Uncertainty

Find the difference in the percent uncertainties of the two following measurements: 10±0.5 s and 5±0.1 s.

Answer

In this example, we are given two measurements with their absolute uncertainties, and we are asked to find the difference in the percent uncertainties.

To find the percent uncertainty of each quantity, we need to recall that percentuncertaintyabsoluteuncertaintymeasuredvalue=×100%.

For the first quantity, we have a measured value of 10 s and an absolute uncertainty of ±0.5 s, which gives percentuncertaintyss=0.510×100%=5%.

And for the second measurement, we have a measured value of 5 s and an absolute uncertainty of 0.1 s, so percentuncertaintyss=0.15×100%=2%.

To find the difference in the percent uncertainty, we take the first result minus the second result, so dierenceinpercentuncertainty=5%2%=3%.

Returning to our two rulers, we were able to obtain two measurements for the length of an object: a measurement of 5 cm from the ruler marked out in centimetres and a measurement of 5.3 cm from the ruler marked in millimetres.

One way of looking at these two measurements is that we can say there is more information contained in the measurement of 5.3 cm than in the measurement of 5 cm. This is because it has more significant figures.

Significant figures are digits that carry meaning. In this case, the measurement of 5.3 cm has two significant figures, whereas the measurement of 5 cm has only one significant figure.

The number of significant figures in a measured quantity indicates the resolution of the instrument used to make the measurement. Here, the measurement of 5.3 cm has more significant figures, so we know that it was measured with an instrument that has higher resolution than that used to take the measurement of 5 cm.

When counting the significant figures in a quantity, we do not include any leading or trailing zeros that are used as placeholders. For example, we might record the length of the object we measured above as 0.053 m. Here, the leading zeros are placeholders so they do not contribute to the number of significant figures, which is still two. Reporting a value in different units does not change the number of significant figures.

Similarly, we might have a map on which the scale can be read to the nearest 1‎ ‎000 m and on which we measure a distance of 5‎ ‎000 m. This value has only one significant figure, as we do not include the trailing zeros. We could equally write this value as 5 km, measured to the nearest 1 km. This also has one significant figure.

Note that trailing zeros are not always placeholders, though. If the scale on the map had high enough resolution that we could read it to the nearest metre, we might still obtain a measurement of 5‎ ‎000 m, but here the value has four significant figures. It is important to take account of the resolution of the instrument used to make the measurement when counting how many significant figures a value has.

If we are given a value of 5‎ ‎000 m, we might be told that this is stated to four significant figures, or equivalently that the instrument used to make the measurement has a resolution of 1 m. This tells us that the true value lies between 4‎ ‎999.5 m and 5‎ ‎000.5 m, whereas a value of 5‎ ‎000 m reported to one significant figure implies a true value of anywhere between 4‎ ‎500 m and 5‎ ‎500 m.

Trailing zeros after a decimal point (such as the last zero in 0.0530 m) are always significant, so 0.0530 m has 3 significant figures. If a value is written this way, we know the measurement was made with a resolution of 0.0001 m.

When we do calculations, we need to be sure to only write trailing zeros after a decimal point if they are significant.

As this example suggests, the number of significant figures a value is quoted to can tell us about the resolution of the measurement and the range of likely true values.

In the following example, we will practice counting the number of significant figures in measured quantities.

Example 5: Understanding Significant Figures

A 1-milligram-resolution digital scale measures the masses shown in the table.

Measurement12345
Mass (g)0.0800.2421.40110.08412.440
  1. How many significant figures are in the first measurement?
  2. How many significant figures are in the second measurement?
  3. How many significant figures are in the third measurement?
  4. How many significant figures are in the fourth measurement?
  5. How many significant figures are in the fifth measurement?

Answer

In this example, we have a digital scale that we are told has a resolution of 1 milligram, and we are asked to determine the number of significant figures in each of five different measurements made with the scale.

The first thing to note is that the mass measurements are listed in grams, and the resolution of the scale is given as 1 milligram. Recall that 1=1000gmg, so 1=11000=0.001mggg. Therefore, a digital scale with 1 milligram resolution can measure mass to the nearest 0.001 g.

Part 1

Looking at the first measurement, then, we see that it is recorded as 0.080 g. The first two digits are leading zeros, which are placeholders and therefore do not count toward the number of significant figures. The last zero, however, is significant because we always include trailing zeros after a decimal point. In this case, we also know that the scale has sufficiently high resolution to record this digit accurately. The number of significant figures in the first measurement is therefore two.

Part 2

In the second measurement of 0.242 g, we can ignore the leading zero, and that leaves us with three significant figures.

Part 3

The third measurement is 1.401 g. We cannot ignore zeroes that are in between nonzero digits, so the number of significant figures here is four.

Part 4

Similarly, in the fourth measurement of 10.084 g, we need to count all of the digits before and after the decimal point for a total of five significant figures.

Part 5

Finally, in the fifth measurement of 12.440 g, we include all of the digits, including the zero because it is a trailing zero after a decimal point. This measurement therefore has five significant figures.

As demonstrated here, the same instrument can provide measurements with different numbers of significant figures depending on the size of the quantity being measured. If the measurement is much larger than the resolution of the instrument, we can record a measurement with more significant figures.

Since we can never make a completely precise measurement in physics, is it important to understand how to work with significant figures to be able to state measurements to the appropriate level of precision.

We frequently encounter situations in which we need to use two measured quantities to calculate a third derived value. For example, we might want to know the speed of a car. To calculate the speed, we use the formula speeddistancetime=.

The distance the car has covered might be the one we measured above as 5‎ ‎300 m to two significant figures. To record the time it took for the car to cover that distance, we used a digital timer with a resolution of 0.1 s, which records the time as 166.7 s. This measurement has four significant figures.

We can calculate speed as speeddistancetimemsms==5300166.7=31.79/.

Here, we combined two quantities, where one has two significant figures and the other has four. When we calculate the speed, we always quote the result to the least number of significant figures of the quantities we used in the calculation. In this case, it is two significant figures. So we need to quote this result to two significant figures. We do this by taking the first two digits (31) and then looking at the next digit. If it is 4 or lower, we round down and keep the first two digits as they are. If it is 5 or higher, we round the last digit up by one. In this case, the next digit is 7, so we round up. That makes the final value speedmstotwosignicantgures=32/.

It is important to note that the number of significant figures is not necessarily equal to the number of decimal places. The time value we used above, 166.7 s, has four significant figures but only one decimal place. A value of 0.05 m has two decimal places, but only one significant figure.

Definition: Significant Figures

The number of significant figures in a measured quantity is the number of digits that carry meaning. This excludes leading and trailing zeros when they are used as placeholders.

When combining two or more values with different numbers of significant figures, the result should always be stated to the least number of significant figures of any of the contributing quantities.

Let’s now work through a couple of examples of working with significant figures.

Example 6: Working with Significant Figures

The sides of a rectangular tile are measured to the nearest centimetre, and they are found to be 6 cm and 8 cm. Rounding to the same number of significant figures that the side lengths were measured to, what is the area of the tile?

Answer

Here, we need to calculate the area of a rectangle given the measured lengths of its two sides.

Recall that to find the area of a rectangle, we multiply the lengths of the two sides. So, given the side lengths of 6 cm and 8 cm, we have areacmcmcm=6×8=48.

Now, we are asked to give the result to the same number of significant figures as the side lengths were measured to. Both side lengths are given to 1 significant figure, so we should also give the answer to 1 significant figure. To do this, we keep the first digit (40 cm2), and then look at the second one to decide whether to round up or down. In this case, the second digit is 8, so we want to round up. This gives us a final answer of areacmtosignicantgure=501.

Example 7: Combining Measurements with Different Numbers of Significant Figures

A distance of 115 metres is measured to the nearest metre. The distance is run in a time of 12 seconds, measured to the nearest second. Rounding to an appropriate number of significant figures, what was the average running speed?

Answer

In this example, we need to calculate the speed of a runner given the distance and time. First, recall that speeddistancetime=.

Substituting the values in the question in, we find speedmsms=11512=9.58̇3/.

Now, we need to determine the appropriate number of significant figures to round this result to. We started with a distance of 115 m, which has 3 significant figures, and a time of 12 s, which has 2 significant figures. When combining values with different numbers of significant figures, we always state the result with the least number of significant figures of the quantities used to calculate it. In this case, the time has the least significant figures at 2, so we should state the resultant speed to 2 significant figures.

To do this, we start with the first two digits (9.5), and since the third digit is 8, we round up to 9.6 m/s. The final answer is therefore speedmstosignicantgures=9.6/2.

Key Points

  • The resolution of a measuring device is the “fineness” to which the instrument can be read.
  • The uncertainty of a measurement is the interval in which the true value of a measured quantity is likely to fall and is stated as half of the range of likely values.
  • The number of significant figures is the number of digits in a value that carry meaning, excluding leading and trailing zeros used as placeholders.
  • When combining measurements with different numbers of significant figures, we should always state the result to the lowest number of significant figures of any of the measurements used in the calculation.

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