SHM 2 Lab Skeleton PHY121 Sp21

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William Rainey Harper College *

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Mechanical Engineering

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Apr 3, 2024

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Simple Harmonic Motion (SHM) 2 PHY121 Spring 2021 Using Error Bars to Draw Conclusions Measuring Instrument Uncertainty Q’s 1) Interactive ruler and interactive timer 2) 0.5cm for the ruler, 0.0167 seconds for the timer 3) 0.25cm for the ruler, 0.00835 seconds for the timer Measuring Technique Uncertainty (your standard deviation with units): 5.1000, 5.1000, 5.1167, 5.1333, 5.0833, 5.1000, 5.1500, 5.1333, 5.1500, 5.1667 STDV: 0.02745008 3 Excel Graphs with Error Bars. 0 0.2 0.4 0.6 0.8 1 1.2 1.4 0 20 40 60 80 100 120 140 Average time for 1 period (s) Spring Constant (N/m) The effect of the spring constant (N/m) on the average time to complete 1 period (s) 0.67 0.68 0.69 0.7 0.71 0.72 0.73 0.74 0.75 0.76 0.77 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 Average time for 1 period (s) Amplitude (m) The effect of the amplitude (m) on the average time to complete 1 period (s)
Conclusions Using Error Bars - There was no overlap in Graph 1 between the error bars and the data, this means that the data does overcome the error. - There was overlap in Graph 2 between the error bars and the data, this means that the data does NOT overcome the error. - There was no overlap in Graph 3 between the error bars and the data, this means that the data does overcome the error. Confidence Using Error Bars - In Graph 1, my confidence level is 96% because the data DID NOT overlap the error. As mentioned above in the conclusion, it means that the data DOES overcome the error. - In Graph 3, my confidence level is 72% because the data DID overlap the error. As mentioned above in the conclusion, it means that the data DOES NOT overcome the error. - In Graph 3, my confidence level is 92% because the data DID NOT overlap the error. It means that the data DOES overcome the error. Let’s Make a Cut: Amplitude which is Graph 2 will be cut because in that graph the I.V. doesn’t seem to correlate with the given D.V. Using Mathematical Models to Draw Conclusions Add Curves to your Excel Graph put graphs here: Spring constant Graph 1: 0 0.2 0.4 0.6 0.8 1 1.2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Average time for 1 period (s) Mass of the end (kg) The effect of the mass of the end (kg) on the average time to complete 1 period (s)
y = -0.0069x + 1.0063 R² = 0.7067 0 0.2 0.4 0.6 0.8 1 1.2 1.4 0 20 40 60 80 100 120 140 Average time for 1 period (s) Spring Constant (N/m) The effect of the spring constant (N/m) on the average time to complete 1 period (s) Linear y = -0.325ln(x) + 1.805 R² = 0.9701 0 0.2 0.4 0.6 0.8 1 1.2 1.4 0 20 40 60 80 100 120 140 Average time for 1 period (s) Spring constant (N/m) The effect of the spring constant (N/m) on the average time to complete 1 period (s) logarithmic y = 3.171x -0.471 R² = 0.9984 0 0.2 0.4 0.6 0.8 1 1.2 1.4 0 20 40 60 80 100 120 140 Average time for 1 period (s) Spring constant (N/m) The effect of the spring constant (N/m) on the average time to complete 1 period (s) Power
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Mass of the end Graph 2: y = 1.0344e -0.011x R² = 0.8365 0 0.2 0.4 0.6 0.8 1 1.2 1.4 0 20 40 60 80 100 120 140 Average time for 1 period (s) Spring constant (N/m) The effect of the spring constant (N/m) on the average time to complete 1 period (s) Exponential y = 0.6509x + 0.546 R² = 0.945 0 0.2 0.4 0.6 0.8 1 1.2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Average time for 1 perriod (s) Mass of the end (kg) The effect of the mass of the end (kg) on the average time to complete 1 period (s) Linear y = 0.2939ln(x) + 1.1055 R² = 0.9929 0 0.2 0.4 0.6 0.8 1 1.2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Average time for 1 period (s) Mass of the end (kg) The effect of the mass of the end (kg) on the average time to complete 1 period (s) Logarithmic
Best- Fit Curve for each variable description (i.e.”power fit for mass”): Exploring Your Best-Fit Curve Best-fit equation #1 in x,y form: y = 3.171x ^ -0.471 R^2 = 0.9984 Best-fit equation #2 in x,y form (if needed): y = 1.1437 x ^0.3569 r^2 = 0.987 Best fit equation #1 in actual variable form: 3.171k ^ -0.471 Best-fit equation #2 in actual variable form: y = 1.1437 m ^0.3569 Using the Best-Fit Curve to Draw Conclusions Conclusion for IV #1 : Here I would choose the spring constant as my IV #1. The graph showed a negative correlation between the spring constant and the time because as the spring constant increased (N/m) the average time decreased. Confidence Statement for IV #1: I am 95% sure that there is a negative correlation between the two variables (spring constant and time). It is negative correlation because as one variable increases, the other decreases. y = 1.1437x 0.3569 R² = 0.987 0 0.2 0.4 0.6 0.8 1 1.2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Average time for 1 period (s) Mass of the end (kg) The effect of the mass of the end (kg) on the average time to complete 1 period (s) Power y = 0.6509x + 0.546 R² = 0.945 0 0.2 0.4 0.6 0.8 1 1.2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Average tiime for 1 period (s) Mass of the end (kg) The effect of the mass of the end (kg) on the average time to complete 1 period (s) Exponential
Conclusion for IV #2 (if needed): Here I would choose mass of the end (kg) as my IV #2. The graph showed that there was a positive correlation between the mass at the end (kg) and the average time (s) because as the mass at the end (kg) increases the average time (s) also increases. Confidence Statement for IV #2 (if needed): I am 98% sure that there is a positive correlation between the two variables (the mass at the end and the average time). It is positive correlation because as one variable increases, the other increases as well. Comparing Mathematical Models With the Class 2 mathematical models that are similar to your model along with the names of the people who own them: How are these mathematical models similar to your own? Brian Roggers: Spring constant power t = 3.2004 k^-0.477 - Mass of the end power: t = 1.2837 m^0.4845 Ayush Kachhia: Spring constant power t = 3.1067 k^-0.459 - Mass of the end power: t = 1.2814 m^0.4438 2 mathematical models that are different from yours along with the names of the people who own them: How are these mathematical models different from your own? Raamix Hussain: Spring constant power t = 8.8527 k^-0.453 Nataly M: Spring constant logarithmic t = -0.211ln (k) Questions about BOTH models: 1) Few students in my class had similar mathematical model to mine. I saw that Brain Roggers’s and Ayush Kachhia’s mathematical model were closest mine. This helped me be more positive and confident with my model because it helps me to understand that some people ended up getting similar data to mine and that is why they have similar mathematical model like me. This helps to clarify that my answer is mostly common and correct. 2) This helps me improve my confidence level because as I double check my work with my classmates it allows me to understand where I might’ve or might not have gone wrong in terms of my data and answers. It also helps me to problem solve on my own before I could reach out to the professor because as I double check my work if my mathematical models match to some of the classmates then it is easy to predict that my model is 90% correct. If my model was completely off from most of the class, then it allows me to think where I could’ve went wrong and then going forward, I redo my work and double check with professor. Comparing to the Theoretical Model Equation for a mass-spring system:
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Questions 1) The amplitude isn’t inside the model because if we look back at the graphs, all three (mass, spring constant and amplitude) are dependent on the period but the spring constant and mass showed a clear correlation with period which was either positive or negative. In terms of amplitude, it ended up giving me high error bars and its correlation with period was neither positive nor negative. This is why the included in the amplitude isn’t theoretical model. 2a) T=2pi/sqrt (11.15) *m^.5 T=1.88m^0.5 2b) The mathematical model and the theoretical model have the same function because my model was a line and the theoretical equation was also a line, so they match. 2c) No, they don’t have the exact same coefficient because my mathematical model had a coeffic ient of 1.1437 whereas the theoretical model had a coefficient of 1.88. 2d) The theoretical model's exponent didn’t exactly match the mathematical model (0.3569) because the theoretical model was 0.5. 3a) T=2pi/sqrt (.3) *K^.5 T= 3.44K^-0.5 3b) Yes, the mathematical model and the theoretical model have the same function because my model was a line and the theoretical equation was also a line, so they match. 3c) The mathematical model for mass was had almost same coefficient as the theoretical model. My mathematical model was 3.171 and theoretical model was 3.44. 3d) My mathematical model for mass also had almost the same exponent as the theoretical model. My mathematical model was -0.471 and theoretical model was -0.5 4) 5 reasons why my mathematic al model wasn’t exactly same as the theoretical model are: - Incorrectly timing the 10 intervals with the watch - Not aligning the ruler properly - Not writing the exact time but rounding it up to one decimal place - Not graphing the data properly - Messing up with the graph equation and R^2 value.