Lab#1 Finding a relationship between mass and period for an inertial balance

Purpose 
  The purpose of this lab is to come up with an equation to describe the period of an inertial balance when different masses are set on it.

MaterialsC-ClampInertial BalanceMasking Tape USB cable, photogateComputer Multiple Different Masses

Set up    1. use a C-clamp to secure the inertial balance to the tabletop, Pur a thein piede of jmasking taoe one the end the inertial balance;    2. Set up a photogate so the when the balance is oscillating the tape completely passes through the beam of the photogate.
Procedure    With no extra mass on the tray, pull back the inertial balance. Take notice of how much force you are using to pull back the balance, for you must use the same force for each trial in order to obtain accurate results. Record the period of the balance.For each additional trial, add an extra 100 grams to the balance and record the period. In this experiment we repeated the trail until 800 grams. Multiple trials will allow us to create a more accurate curve. Always use the same amount of force when pulling back the inertial balance.               
    3. we were given the initial equation:T=A(m+Mtray)^n. With this equation, we had three unknowns: A, Mtray, and n. To make a more reasonable looking equation, we took the natural logarithm of each side, which gave us: lnT=nln(m+Mtray)+lnA. This equation mimicked that of y=mx+b, in that lnT is y, n is m, ln(m+Mtray) is x, and lnA is b. To begin graphing our data, we first opened a blank LoggerPro document, which allowed us to plot data in an "x" and "y" table. In this table, we labeled the "x" column "Mass (kg)" and the "y" column "Period (sec)." We then created three new columns for (m+Mtray), lnT, and ln(m+Mtray) by first clicking the "Data" tab, and then "New Calculated Column"  Finally, we had to create a parameter Mtray, which would be our initial guess for the mass of the inertial balance by itself. To do this, we clicked the "Data" tab, and then "User Parameters." 
T=0.6489(m+0.369)^0.6489 

                                                

    When making the graph, we set the vertical axis as lnT and the horizontal axis as ln(m+Mtray). This gave us a linear graph with a slope "m" (which is n) and a y-intercept "b" (which is lnA). We then had to adjust the parameter Mtray several times to get a correlation coefficient that was as close to 1 as possible. 

Deriving The Final Equation

    Once we had our originally unknown values for n, lnA, and Mtray, we were now able to complete our equation of T=A(m+Mtray)^n using algebra  When testing the equations, we found that we were roughly 0.005 seconds off from the actual period for most of the masses. When evaluating the mass of the tape dispenser , using Mtray equal to 0.369kg