● for quantities that are added or subtracted, the absolute uncertainties are added
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● for quantities that are multiplied together or divided, the fractional (or percentage) uncertainties are added
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● for a quantity that is raised to a power, to calculate a final uncertainty the frac-tional uncertainty is multiplied by the power and the result is treated as a positive uncertainty
Worked example 1
The currents coming into a junction are I1 and I2. The current coming out of the junction is I. In an experiment the values of I1 and I2 are measured as 2.0 ± 0.1 A and 1.5 ± 0.2 A respectively.
Write down the value of I with its uncertainty.
Answer
I = I1 + I2 = (2.0 ± 0.1) + (1.5 ± 0.2)
The quantities are being added so to find the uncertainty the uncertainties of the original quantities are added.
Hence I = 3.5 ± 0.3 A
Worked example 2
The acceleration of free fall g is determined by measuring the period of oscil-lation T of a simple pendulum of length L. The reoscil-lationship between g, T and L is given by the formula g = 4π2(L/T2).
In the experiment, L was measured as 0.55 ± 0.02 m, and T as 1.50 ± 0.02 s.
Find the value of g and its uncertainty.
Answer
g = 4π2(L/T2) = 4π2(0.55/1.502) = 9.7 m s–2
To find the uncertainties, the second and third rules are applied.
Fractional uncertainty in L = 0.02/0.55 = 0.036 Fractional uncertainty in T = 0.02/1.50 = 0.013 Fractional uncertainty in T–2 = 2 × 0.013 = 0.026
AS Experimental Skills and Investigations
International AS and A Level Physics Revision Guide
Fractional uncertainty in g = fractional uncertainty in L + fractional uncer-tainty in T–2 = 0.036 + 0.026 = 0.062
The absolute uncertainty in g = 9.7 × 0.062 = 0.6 Thus g = 9.7 ± 0.6 m s–2
It is worth noting that the examiners are looking for the absolute uncertainty, not the percentage uncertainty. If you take the short cut and leave your answer as 9.7 ± 6.2%, you will lose credit. It is also worth noting that it is poor experimental practice to take only one reading and to try to find a value of g from that. You should take a series of readings of T for different lengths L, and then plot a graph of T2 against L. The gradient of this graph would be equal to 4π2/g.
To ascertain if an experiment supports or fails to support a hypothesis, your result should lie within the limits of the percentage uncertainties. To support the hypoth-esis in the absence of any uncertainty calculations, a good rule of thumb is that the calculated value should lie within 10% of any predicted value.
The following worked examples take you through some of the stages of evaluating evidence.
Worked example 1
In an initial investigation into the time it takes for an ice cube to melt (Investigation 3 in the table on page 103) in a beaker of water the following results are obtained.
Trial 1:
initial temperature of the water = 50 °C time taken (t) to melt = 85 s
Trial 2:
initial temperature of the water = 80 °C time taken to melt = 31s
(a) Explain why it is only justifiable to measure the time taken for the ice cube to melt to the nearest second.
(b) Estimate the percentage uncertainty in this measurement in trial 1.
(c) Estimate the percentage uncertainty in this measurement in trial 2.
(d) Why is it more important to calculate the uncertainty in the time rather than in the initial temperature of the water?
AS experimental skills & in vestigations
Answer
(a) Even though the stopwatch that was used may have measured to the nearest one-hundredth of a second, it was difficult to judge when the last bit of ice disappeared.
(b) Suppose that the absolute uncertainty = ±5 s percentage uncertainty = ± 855 × 100% = 6%
(c) absolute uncertainty = ±5 s
percentage uncertainty = ± 315 × 100% = 16%
(d) The percentage uncertainty in measuring the temperature of the water is much less than the uncertainty in measuring the time. (±1 °C, leading to ±1 to 2%).
This example shows the reasoning in estimating the uncertainty in a measured quantity and how to calculate percentage uncertainty. You might feel that 5 s is rather a large uncertainty in measuring the time. It is at the upper limit, and you might be justified in claiming the uncertainty to be as little as 1 s. Nevertheless, if you try the experiment for yourself, and repeat it two or three times (as you should do with something this subjective), you will find that an uncertainty of 5 s is not unreason-able. The measurement of the initial temperature of the water has a much lower percentage uncertainty as less judgement is needed to make the measurement.
The next stage is to look at how to test whether a hypothesis is justified or not.
Worked example 2
It is suggested that the time taken (t) to melt an ice cube is inversely propor-tional to θ 2, where θ is the initial temperature of the water in °C.
Explain whether or not your results from Worked example 1 support this theory.
Answer
If t ∝ 1/θ 2 then t × θ 2 = constant Trial 1: 85 × 502 = 213 000 Trial 2: 31 × 802 = 198 400
difference between the constants = 14 600
percentage difference = (14 600/198 400) × 100% = 7.4%
This is less than the calculated uncertainty in the measurement of t (= 16%, for trial 2) so the hypothesis is supported.
AS Experimental Skills and Investigations
International AS and A Level Physics Revision Guide
There are various ways of tackling this type of problem — this is probably the simplest.
Note that it is important to explain fully why the hypothesis is/is not supported.
At the simplest level, if the difference between the two calculated values for the constant is greater than the percentage uncertainties in the measured quantities, then the evidence would not support the hypothesis.
Hint