Feb 25, 2015: Spark Free Fall and Experimental Uncertainty

Purpose:

 To test the theoretical statement that g= 9.80m/s^2; where the experiment has negligible air resistance and friction.

Procedure:

We sue a wooden cylinder as a free-fall body. It's hung at the top of the apparatus. Once secured, the sparker is turned on followed by turning off the electromagnet to release the freefall body. The sparker works at 60Hz, meaning it will leave a dot for every 1/60th of a second. The electromagnet is turned off so that the body could freely fall, without any magnetic interference. While free-fall happens, the sparker leaves dots on the white tape. Turn off the sparker and remove the tape from the apparatus and measure the displacement for each dot.

displacement is measured here

A white tape is placed behind  the metal wires

Data and Explanation:

In excel, the these rows are created:

• time: previous column + 1/60. Example, c2+1/60
• distance from origin: displacement
• delta x: change between each displacement
• Mid-interval time: midpoint of each time interval
• speed: midpoint of each speed interval; (delta x)/ (1/60)
• delta V: change between two speed midpoints.
• acceleration: (delta V)/(1/60)

final data

Two graphs are required for this lab:

• Position vs. Time: use polynomial fit,
• Velocity vs. Time: (or mid-interval speed vs. time) use Linear fit. The linear regression (R^2), is closed to 1.

• x(t) takes the form of 0.5mx^2 + bx+ c
• V(t)= mx+b
• a(t)=m
• Our average acceleration is the slope of velocity graph: a= 9.62m/sec^2

Questions 

1. The result of a straight line from V vs. t graph, shows that with constant acceleration, V-ave in a time interval is the same as the mid-interval speed in that same time interval.
2. From the velocity graph, we can look at the slope and that would be the average acceleration. The calculation of average of sum of (delta V)/(1/60) would give almost the same answer. We are (9.62-9.80)*100%/9.80= -1.94% off from the accepted value of g.
3. If v(t) takes the form of mx+b, then x(t)= 0.5mx^2 + bx+ c. Our value for "0.5m" = 4.815; making m= 9.63. Hence, we are (9.63-9.80)*100%/9.80= -1.94% off from the accepted value of g.

Conclusion:

_ We've made an assumption that there's no friction and the sparks are exactly 1/60th sec apart from one another.

_ From these assumptions, we see the pattern in our data as the following:

• a curve for position graph, and a straight line for velocity graph, an spread out data for acceleration graph.

_ Should we count in the factors of friction and other interference,

• acceleration graph would have had less deviation (variation) from one another.
• In our case, we choose the first 26 dots. Although this is a good sample size, more would have been better for a better standard deviation.

_Nonetheless, our uncertainty is quite small: Absolute difference= 9.63-9.80= +/- 0.17

Relative difference: (9.60-9.80)*100%/9.80= -2.04%

Errors and Uncertainty

Purpose: To find the uncertainty in value of g from the whole class using the techniques of mean and standard deviation.
Data:

mean= 9.56m/sec^2; std. dev= 0.20m/sec^2

_ We are 95% confident that the actual g-value is in (+/- 2std. deviation) range of 9.16 and 9.96m/sec^2.

_ Our collected values of g seem almost to be under the theoretical value. The standard deviation of each data from the average seem to vary every time.
_ Our average class value of g is also under the accepted value: (9.56-9.80)*100%/9.80 = -2.44%
_ Systematic error (bias) that account for difference between our value and other groups',

• small sample size to compare to (n=9)
• The friction during free fall vary each time resulting in different displacements
• Number of dots that each group decide to measure vary, hence the mean and deviation for each group are unique.

_One possible random error (noise) that account for difference is the uncertainty of the displacement measuring device.

_ Conceptually the less spread out the data, the more accurate and precise the experiment is. From this part of lab, we use the calculated means and deviations to statistically state our percent confidence in our experimental g-value compared to the actual g-value. Ideally, the data should give a 95% confidence that the actual g-value is in the range of 9.80+/- 0.02m/sec^2.