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What is linearization?
What is linearization?
After having done an experiment investigating 2 variables you may be asked to plot a straight-line graph even though the data might not have a linear relationship. Manipulating the data to create a linear relationship is called linearization.
Linear graph planning is a skill that can save time in advanced data analyses and may be able to offer some correction to produce better results when the data may not be ideal. For our course, it also helps you develop the ability to represent data in different ways, which is one of our course objectives.
How to approach linearization?
How to approach linearization?
Identify the equation relating the variables (and rearrange to solve for the dependent variable)
Modify the variables to match the form of the equation
for example, square a variable that is squared, take the inverse of a variable in the bottom of a fraction, etc)
you actually manipulate the data, creating a new column and you will square the numbers when the variable gets squared, and take the inverse when a variable is inverted.
Don't forget that units are also affected.
Factor out the constants to give meaning to the slopes (addition/subtraction may give meaning to the intercepts).
For example, if you have collected data of time (t) and distance (d) for uniformly accelerated motion then
step 1: the equation that relates them is d = ½ at2. It is already solved for d so if that is your dependent variable.
step 2: you need to square t to match the equation and graph d vs. t2. If your times in your table are 1 s, 2 s, 3 s you will square each to be 1 s2, 4 s2, 9 s2.
step 3: The constant value of 1/2a is multiplying t2 (your x axis) in the equation, so the slope will be equal to ½ a since y=mx+b shows whatever is multiplying the x axis is the slope.
There are always options in how you can linearize, for example with the same data you could
create a column of 1/2t2 (1 s, 2 s, 3 s become 0.5 s2, 2 s2 , 4.5 s2), then plot d on the y and 1/2 t2 on the x, which would make the slope equal to the acceleration.
rearrange to get t=2da in which case you would plot t vs d and 2/a would be the value of the slope
create a column of 1/2t2 (1 s, 2 s, 3 s become 0.5 s2, 2 s2 , 4.5 s2), then plot d on the y and 1/2 t2 on the x, which would make the slope equal to the acceleration.
Practice
Practice
Read the descriptions and test to see if you can find the following for each equation show below.
What two quantities should be graphed to make a linear graph?
How you will modify the data in your table to create that graph?
What will the value of the slope of that line would be?
You should be able to do any of these using the principles above without knowing the Physics.
Students measure the electric force FE as a result of changing the amount of charge q in a constant electric field E.
Recommended: rearrange to F = Eq, so F on y axis, q on x axis.
No modification needed, because there are no modifications on F or q.
The slope will equal the electric field strength E.
Students measure the electrical resistance R as a result of changing the area A of a wire of constant length and resistivity.
Recommended: rearrange to R = (ρl)(1/A) so plot R on the y axis and A on the x axis.
Modify the area data by inverting each value [for example if trial 1 area is 10 cm^2, then modify it to 1/(10cm^2) = .1 cm^-2, repeat for each value].
The slope would be equal to the resistivity times length, ρl.
Students measure the electrical force FE between two charged particles as a result of changing the distance between the particles r. (4πε are all constants and the prompt implies the q values do not change either).
Recommended: rearrange to FE = (q1q2/4πε)(1/r2) so plot FE on the y axis and 1/r2 on the x axis.
Modify the separation distance r in the data by inverting and squaring each value [for example if trial 1 separation is 0.1 m, then modify it to (1/0.1 m)^2 = 10 m-2, repeat for each value].
The slope would be equal to the product of the charges over the constants, q1q2/(4πε0).
Students measure the how far a spring is squashed x as a result of changing the speed v of a cart that slams into the spring.
Assuming the cart and spring are the same, which is implied, m and k are constants.
recommended: rearrange to x = sqrt(m/k)(v), so plot x on the y axis and v on the x axis.
no modification needed since both variables are squared.
slope equals the square root of mass divided by the spring constant.
Students measure the distance a cart travels before reaching a set speed (starting from rest) as a result of changing the acceleration of the box.
Recommended: rearrange by substituting 0 in for the initial velocity and initial position, then solve for x = (v2/2)(1/a) so put x on the y axis and 1/a on the x axis.
Modify the acceleration data by inverting each acceleration [for example if trial 1 acceleration is 0.1 m/s2, then modify it to 1/(0.1 m/s2) = 10 0.1 s2/m. repeat for each value].
The slope would be equal to the square of the speed, v2/2.
Students measure the angle an object rotates through θ (from an initial angular position of 0 and angular speed ω of 0) as a function of the time it accelerates (constant angular acceleration).
Recommended: rearrange by substituting 0 in for the initial angular vleocity and initial angular position, then solve for θ = (α/2)(t2) so put θ on the y axis and t2 on the x axis.
Modify the acceleration data by squaring the time [for example if trial 1 time is 0.1s, then modify it to (0.1 s)2 = 0.01 s2. repeat for each value].
The slope would be equal to half of the angular acceleration, α/2.
Practice actually following through the process and see your data get linear with this worksheet we sometimes use in class. This one starts with recognizing the shape of the original data and comparing it with the standard graphs found on 2.D Analyzing Functional Dependence step 2, where there are 6 graphs shown. You should memorize those shapes and relationships.
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