Springpractice
Jun 20, 2015
This is adapted from www.physics.sfasu.edu/downing/Graphing.ppt in order to use it with a 3º ESO group of students
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Transcript
GraphingGraphing
When weight is added to a spring hanging from the ceiling, the spring stretches.
How much it stretches depends on how much weight is added.
You can see it in the next slide.
Starting level
Add a mass
Starting level
Add a massStretch is now here
Starting levelStretch is now here
Add another mass
Add a mass
Starting level
We control the mass that is added.It is the independent variable.The stretch depends on the mass added.It is the dependent variable.
Stretch is now here
Add another massStretch is now here
Add a mass
The following data were obtained by adding different weights to a spring and measuring the corresponding stretch.
Weight (Newtons)
Stretch (meters
)6.0 0.1240
14.0 0.147522.0 0.177530.0 0.195038.0 0.219540.0 0.230047.0 0.252554.0 0.267558.0 0.2875
A graph of this experimental data is shown on the next slides.
Each graph should be identified with a title
Weight, the independent variable, will be plotted along the horizontal axis (the abscissa).
Weight (Newtons)
0 10 20 30 40 50 60
Stretch, the dependent variable, will be plotted along the vertical axis (the ordinate).
Str
etc
h (
mete
rs)
0.10
0.14
0.18
0.22
0.26
0.30Stretch Versus Weight
The independent variable is always plotted on the horizontal axis, the abscissa.
The dependent variable is plotted on the vertical axis, the ordinate.
Notice that each axis is labeled as to what is plotted on it and also the units in which the variable is measured.
Units are important.
REMEMBER!
Weight (Newtons)
0 10 20 30 40 50 60
Str
etc
h (
mete
rs)
0.10
0.14
0.18
0.22
0.26
0.30Stretch Versus Weight
Let’s plot the data.
(6.0, 0.1240)
(14.0, 0.1475)
(22.0, 0.1775)
(38.0, 0.2195)(40.0, 0.2300)
(47.0, 0.2525)
(54.0, 0.2675)
(58.0, 0.2875)
Each data point should be circled, so that it can be easily found and distinguished from other dots on the paper.
(30.0, 0.1950)
Weight (Newtons)
0 10 20 30 40 50 60
Str
etc
h (
mete
rs)
0.10
0.14
0.18
0.22
0.26
0.30Stretch Versus Weight
This is not a connect-the-dot exercise.
The data appear to fit a straight line similar to this one.
Weight (Newtons)
0 10 20 30 40 50 60
Str
etc
h (
mete
rs)
0.10
0.14
0.18
0.22
0.26
0.30Stretch Versus Weight
If there is a general trend to the data, then a best-fit curve describing this trend can be drawn.
In this example the data points approximately fall along a straight line.
This implies a linear relationship between the stretch and the weight.
Weight (Newtons)
0 10 20 30 40 50 60
Str
etc
h (
mete
rs)
0.10
0.14
0.18
0.22
0.26
0.30Stretch Versus Weight
Important information can be obtained if we know the equation that describes the data.
With an equation we are able to predict the values of the variables beyond the boundaries of the graph.
Weight (Newtons)
0 10 20 30 40 50 60
Str
etc
h (
mete
rs)
0.10
0.14
0.18
0.22
0.26
0.30Stretch Versus Weight
If data points follow a linear relationship (straight line), the equation describing this line is of the formy = mx + b where y represents the
dependent variable
(in this case, stretch), and
x represents the independent variable (weight).
Weight (Newtons)
0 10 20 30 40 50 60
Str
etc
h (
mete
rs)
0.10
0.14
0.18
0.22
0.26
0.30Stretch Versus Weight
The value of the dependent variable when x = 0
is given by b and is known as the y-intercept.
The y-intercept is found graphically by finding
the intersection of the y-axis (x = 0) and the line
through the data points.
(From the equation y = mx + b, if we set x = 0
then y = b.)
In this case b = 0.11 meters.
Weight (Newtons)
0 10 20 30 40 50 60
Str
etc
h (
mete
rs)
0.10
0.14
0.18
0.22
0.26
0.30Stretch Versus Weight
m is the slope of the best-fit line.
It is found by taking any two points,
for instance (x2, y2) and (x1, y1), on the straight
line and subtracting their respective x and y values.
Note
y2 = mx2 + b
y1 = mx1 + b
Subtracting one equation from the other gives
y2 - y1 = mx2 - mx1 = m(x2 – x1)
Therefore12
12
xxyy
m
runrise
We’ll call this
slope
Weight (Newtons)
0 10 20 30 40 50 60
Str
etc
h (
mete
rs)
0.10
0.14
0.18
0.22
0.26
0.30
y-intercept = 0.11 meters
Stretch Versus Weight
run = (34.0 – 17.0) Newtons = 17.0 Newtons
rise = (0.21 – 0.16) meters = 0.05 meters
runrise
To find the slope of this line take two separate points on the line.
17.0
0.16
34.0
0.21
Y1 =
Y2 =
X2 =X1 =
Point 2
Point 1
slope
Weight (Newtons)
0 10 20 30 40 50 60
Str
etc
h (
mete
rs)
0.10
0.14
0.18
0.22
0.26
0.30
y-intercept = 0.11 meters
Stretch Versus Weight
run = (34.0 – 17.0) Newtons = 17.0 Newtons
rise = (0.21 – 0.16) meters = 0.05 meters
Newtons0.17meters05.0 Newton/meters00294.0
runrise
17.0
0.16
34.0
0.21
Y1 =
Y2 =
X2 =X1 =
Point 2
Point 1
To find the slope of this line take two separate points on the line.
)meters11.0(
Weight (Newtons)
0 10 20 30 40 50 60
Str
etc
h (
mete
rs)
0.10
0.14
0.18
0.22
0.26
0.30Stretch Versus Weight
slopeNewtons0.17meters05.0 Newton/meters00294.0
runrise
At this point everything needed to write the
equation that describes the data has been
found.
Recall that this equation has the form
yStretch )Newton/meters00294.0( Weight
y-intercept = 0.11 meters
m x + b•=
Weight (Newtons)
0 10 20 30 40 50 60
Str
etc
h (
mete
rs)
0.10
0.14
0.18
0.22
0.26
0.30Stretch Versus Weight
Here are two ways we can get useful
information
How much weight will
produce a 0.14 m
stretch? From the graph:
we could read it
from the plot like
this.
This would be about 10.2 Newtons
Weight (Newtons)
0 10 20 30 40 50 60
Str
etc
h (
mete
rs)
0.10
0.14
0.18
0.22
0.26
0.30Stretch Versus Weight
Here are two ways we can gain useful
information
How much weight will
produce a 0.14 m
stretch?
Solving the
equation
gives x (Weight)=10.2
Newtons.
Stretch = 0.00294 · Weight +
0.11for x (Weight) when y (Stretch)=0.14 m
A curve through this data is not straight and x and y are not linearly related. Their relationship could be complicated.
y
x0 10 20 30 40 50 60
0
1
2
3
4
5
6
Suppose you have some x and y data related to each other in the following way.
y2y
x0 10 20 30 40 50 60
0
1
2
3
4
5
6
A replot of this data might straighten this line to give a linear relationship.
Let’s try y2 versus x.
This relationship would be
bmxy2