Sparker Lab 9/24/14

In this lab, we tried the find the value of g through a spark generator. Marks were made on the spark sensitive tape when an object fell after the electromagnet was turned off. We took note of the distances where there was a mark through a two meter stick, and used excel to get a graph.

We used a meter stick to count the distance (column b) and we set it up in meters. Column A was the time which was incremented by (1/60) per second as that was how fast the spark generator made a dot on the tape. Delta X was found to be the distance between two points in column B. The mid-interval speed was calculated through time + (1/120). Mid-interval speed was found through delta x multiplied by 60. Using Excel our graph came to look like below. 

In our testing, the constant g was 9.61 m/s our about .2m/s off 9.81. This could be due to the fact this experiment was not conducted in a vacuum or that we did not take into account air resistance. There also could have been things that we might have messed up during the experiment, as the data points we took down for the distance at a certain point in time could be a bit off as we did eyeball the distance. 

Questions:
1. Velocity is the integral of acceleration and if acceleration is constant, throughout the interval which the interval is constant the velocity should be the same.
2. We found the constant g through the slope of our graph which came out to be 9.61 m/s. It was slightly off compared to 9.81. Our percent error was (9.81-9.61)/9.81*100, which was roughly off by two percent.
3. You can get acceleration by deriving the position graph, which would give you the velocity graph. And by taking the derivative of the velocity graph, you would get your acceleration graph. Our value was 9.61 m/s which is slightly off compared to 9.81.

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.