# 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
• 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