# Physics: Graph Drawing

- This applies to both AS and A Level exams, with the exception of the blue subheadings, which are A Level only.

You can find more examples in Appendix 10 in the Edexcel specification.

## What to plot:

- You'll need to rearrange to the form $y = mx + c$
y = mx + c where your two variables are $y$y and $x$x - 'A graph of $a$
a against $b$b ' means that the first variable mentioned goes on the $y$y -axis and the second goes on the $x$x -axis - 'directly proportional to' can be written as $=k$
=k where $k$k is a constant - For example, to test if the measurement $a$
a is directly proportional to $b$b , we need to find if it satisfies the equation $a = kb$a = kb - This can be rearranged to $y = mx + c$
y = mx + c form: $a = kb + c$a = kb + c - Therefore, we need to plot $a$
a against $b$b . It should form a straight line with a $y$y -intercept of 0 (because $c = 0$c = 0 ) and gradient $k$k - Use this method to determine whether measurements satisfy a relationship
- 'inversely proportional' means $a = k\frac{1}{x}$
a = k\frac{1}{x}

## Axes:

- Axes must be labelled with the quantity that it represents, and the units
- For example, if you're plotting distance squared on the $x$
x -axis, you could label it '$d^2\ (\mathrm{m}^2)$d^2\ (\mathrm{m}^2) ' - Although writing $d^2 /\mathrm{m}^2$
d^2 /\mathrm{m}^2 ' is preferred by SI

## Units in log axes:

- If you're plotting the log of distance, you would write $\log(d/\mathrm{m})$
\log(d/\mathrm{m}) '

## Error bars:

- Usually, you would pick a measurement in the middle, calculate the absolute error in that value, and use that for every error bar

## Example 1:

- Let's say you're taking measurements of the time of something
- Pick the middle reading
- Let's say the measurements of time for the middle reading are $1.42\ \mathrm{s}$
1.42\ \mathrm{s} , $1.46\ \mathrm{s}$1.46\ \mathrm{s} , $1.67\ \mathrm{s}$1.67\ \mathrm{s} , $1.49\ \mathrm{s}$1.49\ \mathrm{s} and $1.43\ \mathrm{s}$1.43\ \mathrm{s} - $1.67\ \mathrm{s}$
1.67\ \mathrm{s} should be discarded because it's an anomaly - This gives a mean of $1.45\ \mathrm{s}$
1.45\ \mathrm{s} . Mark this as a data point - Our minimum value is $1.42\ \mathrm{s}$
1.42\ \mathrm{s} ($0.03\ \mathrm{s}$0.03\ \mathrm{s} away from the mean) and maximum is $1.49\ \mathrm{s}$1.49\ \mathrm{s} ($0.04\ \mathrm{s}$0.04\ \mathrm{s} away from the mean) - Therefore, pick the largest distance from the mean ($0.04\ \mathrm{s}$
0.04\ \mathrm{s} ). This will be the length of our error bar in each direction - So draw a bar of width $0.04 \times 2 =$
0.04 \times 2 = $0.08\ \mathrm{s}$0.08\ \mathrm{s} on each data point

## Example 2:

- We'll use the same values as above
- Except this time, we need to plot $\ln(t/\mathrm{s})$
\ln(t/\mathrm{s}) - The mean value is $1.45\ \mathrm{s}$
1.45\ \mathrm{s} , so plot the natural log of this ($0.372$0.372 ) - The largest distance from the mean is $0.04\ \mathrm{s}$
0.04\ \mathrm{s} - This makes the maximum possible value $1.45 + 0.04 = 1.49$
1.45 + 0.04 = 1.49 $\mathrm{s}$\mathrm{s} and the minimum $1.45 - 0.04 = 1.41$1.45 - 0.04 = 1.41 $\mathrm{s}$\mathrm{s} - The length of the error bar is therefore $\ln(1.49) - \ln(1.41) = 0.0552$
\ln(1.49) - \ln(1.41) = 0.0552

## Line of best fit:

- First, draw the best fit line through the data points
- Now draw the error boxes and draw the steepest and shallowest possible line which passes through all of them
- Points which are clearly anomolous can be ignored when drawing a line of best fit