January 2002
Number 28
Graphs in Physics
This Factsheet will cover the graphical techniques required for AS and A2 Physics:
• plotting graphs from data and choice of the best straight line;
• gradient and intercept;
• linear form, including logarithmic plots;
• standard curves;
• area under a curve;
• tangent to a curve.
Before starting on this Factsheet, you should make sure you are familiar with exponentials and logarithms (Factsheet 10)
Plotting graphs
Axes
• When you are told to plot “A against B”, the first named quantity is always plotted on the y (vertical) axis. So a graph of “distance against time” has distance on the y-axis and time on the x-axis.
• If you are not told which quantity to plot on which axis, then you
should put the variable you control on the x-axis. For example:
! If you are measuring temperature at specified time intervals, then time goes on the x-axis, since you are choosing the time intervals
! If you are measuring extension of a spring for specific loads, then you are choosing the load, so load goes on the x-axis.
• Label your axes, and include units. If you don’t you will lose
marks!
Scales
• Choose your scale so that it is easy to use – for example, 5 units to every large square is fairly easy, but 7 units to a large square is not.
• Your scale should be chosen so that your graph uses at least half the graph paper.
• You do not generally have to start at 0 on the scale – for example, if you had the values 10.02, 10.04, 10.07, 10.09, 10.12, it would be more sensible to start at 10. However:
! If you are plotting a graph to show two quantities are directly proportional, you must start at 0 on both axes, since you will need to show that the graph goes through the origin.
! If you are using your graph to find an intercept, you must make sure the intercept is on the graph paper!
Points, lines and curves
• Accuracy is important – use a sharp, hard (H) pencil. If you use a cross, where the two lines cross must be the point required.
• If the points look roughly like a straight line, you should draw the best straight line. This is not necessarily the line that passes through the most points; you should aim to have roughly the same number of points above the line and below the line. If you have one obviously strange result, ignore it when drawing the line.
• If the points look like a curve, then draw a smooth curve. This may not go through every single point (like the best straight line)
! Do not draw a “wiggly” curve. ! Do not “join the dots” using a ruler.
Straight line graphs
You need to be able to recognise the equation of a straight line graph: x and y are variables (the values you plot) y = mx + c m and c are constants (numbers whose value is known or is to be determined)
m is the gradient of the graph; m =
x y in change in change
c is the y-intercept - where the straight line crosses the y-axis
Examples of this include:
A speed against time graph for a body moving in a straight line with constant acceleration
v = u + at
• speed (v) takes the place of y – it is on the y-axis
• time (t) takes the place of x – it is on the x-axis
• acceleration (a) takes the place of m (it is the constant in front of t)
• initial speed (u) takes the place of c (it is the constant on its own) So the gradient gives the acceleration, and the y-intercept gives the initial speed.
A graph of pressure against temperature (K) for a constant volume of gas.
pV = nRT where p = pressure, V = volume, n = no. of moles R = gas constant and T = temperature (K) So p =
V nR
T
• p is on the y-axis, T on the x-axis
• the gradient is V nR
• the y-intercept is zero – the graph should pass through the origin A graph of acceleration against displacement for a body performing
SHM
a = -ω2x a = acceleration, x = displacement
• a is on the y-axis, x on the x-axis
• the gradient is -ω2 and the y-intercept is zero
A graph of total length of a spring against applied force for a spring obeying Hooke’s law
We know T = kx T = tension, x = extension, k = spring constant Also L = Lo + x L = total length, Lo = unstretched length
So: L = Lo +
k 1
T
• L is on the y-axis, T on the x-axis
• The gradient is k 1
and the y-intercept is Lo
Graphs in Physics
Physics Factsheet
Conversion to linear form
Straight line graphs are more useful for many purposes than curves because their gradients and intercepts can be used to find physical quantities. It is useful, therefore, to be able to convert other graphs to a straight-line – or linear – form. The following examples illustrate some of the commonest cases:
Distance and time for body moving from rest in a straight line with constant acceleration
Since the body is moving from rest, we have s = ½ at2
If we let X = t2, the equation would then become s = ½ aX – which looks like a straight line equation
So if we plot s on the y-axis, and t2 on the x-axis, the gradient will be ½a, and there will be no intercept.
Period and length for a simple pendulum
T = 2π g
l
where g is the acceleration due to gravity
By squaring each side we have T2 = g l 2 4π Or T2 = l g 2 4π
∴ plotting T against l we will get a straight line, graph of gradient g
2
4π
, passing through the origin.
Gravitational force and distance
The gravitational force (F) between two bodies of masses M and m at a distance of r from each other is given by:
F = 2 r GMm
, where G is the universal gravitation constant
This can be rewritten as F = GMm 12
r - so a plot of F against 2 1 r will produce a straight line with gradient GMm
Displacement and speed in SHM
In simple harmonic motion, speed and displacement are related by v2 = ω2(r2 – x2) where r is the amplitude and ω is the angular
speed of the associated circular motion.
Expanding the brackets, this becomes v2 = ω2r2−ω2x2
So a plot of v2 against x2 would give a straight line graph with gradient −ω2 and y-intercept ω2r2
Exam Hint: When using a straight line graph to determine physical quantities, you must:
• Use the correct units. If you are not sure what they are, go back to the original equation and work them out.
• Use an appropriate degree of accuracy – the gradient and intercept cannot be given to a greater degree of accuracy than the data used to plot your graph. Two significant figures is usually the greatest degree of accuracy possible from a graph.
Conversion to linear form for exponentials
Some processes in physics are represented by an exponential decay law; two examples are radioactive decay and charge decay for a capacitor: Radioactive decay: N = Noe
-λt
N = no. of radioactive atoms at time t No = initial no. of radioactive atoms
λ = decay constant (s-1) t = time (s) Charge decay: Q = Qo RC t − e Q = charge (C) at time t Qo = initial charge (C) t = time (s) R = resistance (Ω ) C = capacitance (F) RC = time constant (ΩF (=s))
Production of a linear graph enables the value of the decay constant or the time constant to be determined. To get a linear graph, logarithms must be used (Factsheet 10).
The strategy is to take logarithms of both sides of the equation, then simplify using the above laws. Taking logarithms “brings down” the power and gets rid of the e. The resulting equation can then be plotted as a linear graph, in the same way as the previous examples.
Specimen Question
The following data were obtained for a body moving with constant acceleration:
Distance (m) 2.55 6.41 11.37 17.61 25.00
Time (s) 2 4 6 8 10
(a) Plot a graph of (distance/time) against time;
(b) Hence determine the acceleration and initial speed of the body. s/t values are 1.275, 1.603, 1.895, 2.201, 2.500 3s.f. So plot: (2,1.275) (4, 1.603) (6,1.895) (8, 2.201), (10, 2.500) 2/3 d.p. We know s = ut + ½ at2 So s/t = u + ½ at So gradient = ½ a = (2.5 – 1.0)/(10 – 0) = 0.15 ms So a = 0.30 ms-2 Intercept = u = 1.0 ms-1 Time/s Laws of logarithms ln(ab) = lna + lnb ln(a/b) = lna – lnb ln(an) = nlna ln(ea) = a Disa tnc e/tim e (m s -1 ) 0 0.5 1 1.5 2 2.5 3 0 2 4 6 8 10
Graphs in Physics
Physics Factsheet
Conversion to linear form for power laws
If an equation is known to be in the form of a power law – in other words, it looks like y = Axn, where A and n are (unknown) constants – then the power can be determined from a suitable linear graph. Again, this requires the use of logarithms; the procedure is very similar to the one for exponential equations:
Original equation y = Axn 1. Take logarithms lny =ln (Axn) 2. Simplify lny = lnA + lnx
n
lny = lnA + nlnx 3. Plot graph of lny against lnx 4. The gradient is n
5. The intercept is lnA
Specimen Question
The following data are from an experiment to investigate the period of a pendulum (T/s) with its length (L/m)
L/m 0.10 0.20 0.30 0.40 0.50 0.60
T/s 0.63 0.89 1.09 1.26 1.40 1.54
It is known that T = kLn, where k and L are constants to be determined.
Plot a suitable graph and hence: (a) determine the value of n;
(b) determine the value and units of k.
T = kLn⇒ lnT = lnk + lnLn⇒ lnT = lnk + nlnL So plot lnT against lnL
Gradient = n = 0.50 (2SF)
Intercept = lnk = 0.69 (2SF) ⇒ k = e0.69 = 2.0 (2SF) Units: k = T/L½ , so units are sm-1/2
Specimen Question
The following data relate to eight asteroids
Asteroid Period (years) Length of
semi-major axis (AU)
Achilles 11.77 5.18 Amor 2.66 1.92 Ceres 4.60 2.77 Eros 1.76 1.46 Hidalgo 14.14 5.76 Icarus 1.12 1.08 Juno 4.36 2.67 Vesta 3.63 2.36
Kepler’s third law for planetary motion states that the square of the period is proportional to the cube of the semi-major axis. Draw an appropriate logarithmic graph to test the accuracy of Kepler’s law for these asteroids.
Kepler’s law: T2∝ L3 (T = period, L = length of semi-major axis) This can be written as T =kL3/2
Taking logs gives lnT = lnk + 2 3
lnL
So if the law applies, a graph of lnT against lnL will be a straight line with gradient
2 3 .
Gradient = (1.5 – 0)/(1.0 – 0) = 1.5, as required
Exam Hint: You may use log (logarithm to the base 10) instead of ln (logarithm to the base e) for power laws. The only difference is that you use the “LOG” and “INV LOG”or “10x” buttons on your calculator where you would have used the “LN” and “ex” buttons. Table 1. Conversion of exponential equations to linear form
Process Radioactive decay Charge decay
Original equation N = Noe -λt Q = Qo RC t e− 1. Take logarithms ln(N) = ln(Noe-λt) ln(Q) = ln(Qo RC t − e ) ln(N) = ln(No) + lne-λt ln(Q) = ln(Qo) + ln( RC t e− ) 2. Simplify ln(N) = ln(No) - λt ln(Q) = ln(Qo) − RC 1 t
3. Plot graph of ln(N) against t ln(Q) against t
4. The gradient is -λ (s-1) − RC
1
(Ω-1F-1[= s-1] 5. The intercept is ln(No) (no units) ln(Qo) (C) (no units)
Table 1 (right) shows the stages in converting exponential equations to a linear form. Note that the simplification must be done in two stages in the order shown. In particular: ln(Noe−λ
t
) IS NOT THE SAME AS -λtln(No).
This is because only the e is to the power −λt. Note that these graphs also allow the determination of the half life for radioactive decay, and the capacitance (if the resistance is known) for charge decay.
The gradient of the graph is −λ = − ln2/T1/2 for
radioactive decay, and –1/RC for charge decay The same approach can be taken for similar equations – such as those for activity in radioactivity and current and voltage in capacitor discharge. -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 -2.5 -2 -1.5 -1 -0.5 0 lnL (AU) ln T (y rs) 0 0.5 1 1.5 2 2.5 3 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
Graphs in Physics
Physics Factsheet
x y
Equation y = kx
Description Straight line graph through the origin
Points to note k is the gradient, so the larger k, the steeper the line Example extension against applied force for a spring.
Equation y = kx2
Description parabola
Points to note the larger k, the steeper the curve
curve is symmetrical about y-axis
curve never goes below y-axis
Example potential energy against displacement in SHM x y
Standard curves
You need to be able to sketch and recognise the following common graphs:
Equation y = mx + c
Description straight line graph
Points to note m = gradient, so the higher m, the steeper the line
c is the y-intercept.
Example pressure against Celsius temperature for a gas at constant volume. y x c A Equation y = Ae-kx
Description exponential decay curve
Points to note A is the y-intercept – the “starting value”.
the higher the value of k, the faster the decay. the curve never touches the x-axis
Example activity against time for a radioactive sample. x y
Equation y =
x k
Description hyperbola; graph for inverse proportion
Points to note curve approaches axes, but does not reach them. Example potential against distance (for charge or mass)
(positive part of graph only)
x y
Equation y = 2
x k
Description graph for inverse square law
Points to note curve approaches both coordinate axes, but does
not reach them.
curve is always above x-axis
Example gravitational attraction between bodies (positive x only) x y
Equation y = cosx
Description cosine graph
Points to note curve repeats every 360o (= 2π radians)
Example displacement against time for SHM
(timing starts at one of the extremes) 1 -1 360o x y Equation y = sinx
Description sine graph
Points to note curve repeats every 360o (= 2π radians)
Example displacement against time for SHM
(timing starts at equilibrium position) x y 1 -1 360o 0 0 0 0 A 0 0 0 0
Graphs in Physics
Physics Factsheet
Exam Hint: When finding the gradient:
• use a large “triangle” to increase your accuracy
• try to choose two points whose coordinates are exact numbers Area under a graph
The area under a graph often represents a physical quantity. It can be determined in various ways:
• Calculation
If the graph consists of straight lines only, then the area under it can be determined by splitting it into triangles, rectangles and/or trapeziums and using the appropriate formulae for their area.
e.g. two ways of splitting a graph are shown below:
• Counting squares
You are unlikely to be required to do this for large numbers of squares! Take care in rounding squares up or down.
Scale
Whatever method for finding the area is used, you must ensure you take account of the scale of the graph.
• When using calculation, make sure the sides of your trapeziums, triangles and rectangles – and hence their areas – are calculated using the scales.
• When counting squares, multiply the total number of squares by what each square represents on the x-axis and on the y axis. For example, if the total area is 73 small squares, and on the x-axis each small square represents 0.2 units, and on the y-axis it represents 4 units, then the final answer is 73 × 0.2 × 4 = 58.4
Units
The units of the area are obtained by multiplying the units for the y-axis by the units for the x-axis.
Graph of y-axis
units
x-axis units
Area units
current against voltage A V A × V = W speed against time ms-1 s ms-1× s = m force against extension m N m × N = J voltage against charge V C V × C = J
pressure against volume Nm-2 m3 Nm-2× m3 =Nm = J
What does the area represent?
The idea of an area under a graph representing a physical quantity is common throughout physics. The units of the area should give you an idea of what is represented. Be aware that some areas do not represent anything physical – for example, the area under a displacement-time graph does not represent anything. The table below gives some common examples of areas which represent physical quantities.
Graph of Quantity represented by
area
speed against time distance
acceleration against time speed
force against time impulse
force against displacement work done
force against extension for a spring energy stored in the spring power against time total work done
voltage against charge for a circuit containing a capacitor
energy stored in the capacitor
current against time total charge passed pressure against volume for a gas work done on or by the gas magnetisation against current
(magnetic hysteresis loop)
area inside loop = energy needed to take material round one cycle of magnetisation stress against strain (elastic
hysteresis loop)
area inside loop = energy lost per unit volume in one cycle
Tangents to curves
A tangent to a curve is used to find its gradient at a particular point. It is a straight line which must just touch the curve at the specified point. The diagram below shows a tangent drawn to the curve at the point (2,4)
By using the straight line, the gradient of the curve at this point
is 4 1 5 0 16 = − −
Accuracy is very important when drawing tangents – use as fine a
pencil as possible for both the original curve and the tangent, and spend some time ensuring the tangent is just touching the curve, rather than missing it altogether or crossing it twice.
a b
h Trapezium Area = ½ h(a + b)
triangle + rectangle + trapezium two triangles + two rectangles
Triangle Area = ½ h(a + b) b h 25 20 15 10 5 0 0 1 2 3 4 5
Graphs in Physics
Physics Factsheet
What are tangents for?
Finding the gradient of a curve is useful in many areas of physics. Examples include:
• finding the velocity from a displacement-time graph
• finding the acceleration from a velocity-time graph
• finding the current from a charge-time graph
It is always important to check the units of any gradient found – it will be the units of what’s on the y-axis divided by the units of what’s on the x-axis. It is worthwhile checking that these are the units you’d expect for the quantity you are finding:
eg: Gradient of a charge-time graph has units of C ÷ s = Cs-1 = A These are the units of current, as we would expect.
Exam Hint: - Some candidates become confused about which quantities are found from the gradient of a graph and which from the area under a graph. Working out the units provides a ready check :-
• area under a velocity- time graph has units ms-1 × s = m (displacement)
• gradient of a velocity-time graph has units ms-1 ÷ s = ms-2 (acceleration)
Questions
1. The equation of a straight line is y = mx + c. Explain the meaning of the constants m and c. 2. Sketch the following graphs
(a) y = x2 (b) y = 2x + 3 (c) y = 2e−x (d) y = sinx
(e) y = 1/x (f) y = 1/x2
3. Give five examples of cases when the area under a graph represents a physical quantity. In each case, state the physical quantity represented and the graph to be plotted.
4. Give three examples of cases when the tangent to a graph represents a physical quantity. In each case, state the physical quantity represented and the graph to be plotted.
5. For each of the following situations: (i) write down an appropriate equation
(ii) suggest a suitable straight line graphs to plot.
(iii) state how the required quantity is to be determined from the graph.
(a) A body is moving in a straight line with constant acceleration. Its acceleration is to be obtained from data on its speed at various times.
(b) The acceleration due to gravity is to be obtained from data on the displacement at various times for a particle falling from rest under gravity.
(c) The decay constant of a radioactive isotope is to be obtained from data on its activity at various times.
6. In an experiment on black-body radiation, the following data were obtained:
Temperature (K) 300 350 400 450 500 Power radiated/ m2 (W) 460 850 1450 2330 3540 The temperature and power radiated are thought to be related by an equation of the form P = σTn, where σ and n are constants.
Plot a suitable graph to determine the value of n.
Answers
Answers for questions 1. – 4. can be found in the text.
5. (a) (i) v = u + at (ii) v against t (iii) gradient (b) (i) s = ½ gt2 (ii) s against t2 (iii) gradient (c) (i) A = Aoe−λt (ii) lnA against t (iii) − gradient
6. Plot lnP against lnT Gradient = n ≈4 5 6 7 8 9 5.6 5.7 5.8 5.9 6 6.1 6.2 6.3 ln(P/W) Ln(T/K)
Acknowledgements: This Factsheet was researched and written by Cath Brown Curriculum Press, Unit 305B The Big Peg, 120 Vyse Street, Birmingham B18 6NF.
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