Lab 8.docx - Lab#8 Simple Harmonic Motion Data Table#1...

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Lab #8: Simple Harmonic Motion Data Table #1: Hooke’s Law Initial position of hanger, x 0 = 0.1550 m Added Mass [g] Equivalent Force, F [N] Final Position, x f [m] Displacement x= x f – x 0 [m] 25 0.245 0.2100 0.0550 50 0.490 0.2855 0.1305 75 0.735 0.3565 0.2015 100 0.980 0.3925 0.2375 125 1.225 0.4670 0.3120 150 1.470 0.5200 0.3650 Displacement of unknown mass: 0.1105 m .04533 kg Data Table #2: Simple Harmonic Motion Mass of hanger pan, m 0 = 24.10 g Added Mass, m i [g] Total mass, m= m i +m 0 [kg] Time for 20 cycles [s] Period, T [s] Period Squared, T 2 [s] 25 g 0.0491 12.837 0.64185 0.41197 50 g 0.0741 15.569 0.77845 0.60598 75 g 0.0991 18.953 0.94765 0.89804 100 g 0.1241 21.052 1.0526 1.1079 125 g 0.1491 22.996 1.1498 1.3220
150 g 0.1741 25.290 1.2645 1.5989 Insert graphs here, remember you need to have three graphs. Make sure you are using the variables in the axis that is asked in the manual.
Spring constant work Table 1 1/0.2485=4.024=k Table 2 For t^2 REFLECTION Hooke’s Law
1. Is zero included within the range of uncertainty for your y-intercept? Consider the significance of your answer. Should it be zero and how does your answer predict the elongation of the spring with no mass attached?
Simple Harmonic Motion 2. Why do you think it was better to measure the time for 20 periods then divide by 20? Why not just measure a single period directly with a stopwatch?
3. Why would you not use a linear fit for the plot of Period vs. Mass?

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