Period of an inertia balance 8/27/14

We are trying to find the mass of the tray by using through adding weight to the tray and calculating its period. There were nine different points in our table, the first of which no weight was added to the tray. We incremented the weight by 100 eight times, making the final weight on top of the tray 800 grams or .8 kg.

We set up a photo gate to record the period of the mass(es) on top of the tray and the tray. And we used Logger Pro to find the variables to the equation and also the missing mass of the tray.

After finding all recording down all our values, we used the equation: T = A(m + Mtray)^n

to solve for the mass of Mtray. If we took the natural log of both sides, the equation would become ln(T)=n*ln(m+Mtray) + ln(A) which looks like the equation y = mx +b. Through Logger Pro, we noted that b was -.435 and that m = .6784.

Through a power law fit, it showed that our mass was somewhere between 292.5 grams + or - 12.5 grams. 

In this lab, we learned how to use the period to solve for the mass of an unknown. By adding a logarithm to both sides of the equation, T = A(m + Mtray)^n, we come up with the equation of ln(T)=n*ln(m+Mtray) + ln(A)  which gives us a way to solve for the missing mass of the tray. We also learned that the period becomes bigger the heavier the weight on top of the tray. On average it increased around .05 seconds per 100 grams.