Springpractice

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).

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

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

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

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

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

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

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

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.

y2 = mx2 + b

y1 = mx1 + b

Subtracting one equation from the other gives

y2 - y1 = mx2 - mx1 = m(x2 – x1)

Therefore12

runrise

We’ll call this

Weight (Newtons)

0 10 20 30 40 50 60

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.

X2 =X1 =

Point 2

Point 1

Weight (Newtons)

0 10 20 30 40 50 60

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

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

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

m x + b•=

Weight (Newtons)

0 10 20 30 40 50 60

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 would be about 10.2 Newtons

Weight (Newtons)

0 10 20 30 40 50 60

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.

x0 10 20 30 40 50 60

Suppose you have some x and y data related to each other in the following way.

x0 10 20 30 40 50 60

A replot of this data might straighten this line to give a linear relationship.

Let’s try y2 versus x.

This relationship would be

Springpractice

mass stretch

level stretch

corresponding stretch

weight newtons s

weight run

title weight

meters yintercept

independent variable

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