Comparing a Model with Experimental Data - Physics Primer

If we're testing a claimed equation relating physical variables, once we have the experimental data, a scatter plot, and a best fit equation for that data, how do we compare the best fit equation with the claim? Here we discuss how to make this comparison and identify what the best fit parameters should physically correspond with (if the claim is true). This gives us a method for identifying whether a set of experimental data supports or refutes the claims of a proposed physical model.
Full details of Newton's 2nd Law experiment: • Newton's 2nd Law - Tes...
Full details of projectile motion experiment: • Measuring g From Proje...
Full details of Mass Spring Oscillation experiment: • Springs in Motion - Te...
Link to Radioactive Decay Simulator: academo.org/demos/radioactive...
0:00 Introduction
2:30 Types of Fit Equations
5:40 Linear Fit Example (Newton's 2nd Law)
8:48 Making the Comparison (Linear Fit)
13:45 Quadratic Fit Example (Projectile Motion)
15:55 Making the Comparison (Quadratic Fit)
20:12 Power Fit Example (Mass Spring System)
23:11 Making the Comparison (Power Fit)
26:15 Exponential Fit Example (Simulated Radioactivity)
28:53 Making the Comparison (Exponential Fit)
31:48 Concluding Remarks
Opening Image Credit: NASA, ESA, CSA, STScI, Klaus Pontoppidan (STScI), Image Processing: Alyssa Pagan (STScI)
science.nasa.gov/missions/jam...

[@PhysicistMichael]

A quick example of using those % error and % difference equations that I mention at the end. I'll apply these to the mass spring system case (25:27 to see where I get the numbers from)
For the exponent in the power fit, we were expecting exactly 0.5 and got 0.48 in the best fit. So, since we know what we should have precisely gotten for the exponent (the reliably accepted value), we'd use the % error equation and get:
% error = |accepted - experimental|/accepted x 100%
= |0.5-0.48|/0.5 x 100%
= 4%
For the coefficient in the power fit, we found the expected value to be 2.69 and got 2.54 in the best fit. Since that expected value was based on the measurement of the spring constant, which could have been off itself, we'd go with the % difference equation and get:
% difference = |value 1 - value 2|/[1/2(value 1 + value 2)] x 100%
= (2.69 - 2.54)/[1/2(2.69+2.54)] x 100%
= (0.15 / 2.615) x 100%
= 5.7%

[@ukoni8667]

This is great.