Question

Spring Constants & Periods By varying the mass placed on a spring, the period of oscillation can be changed. When multiple data sets are collected and plotted, the spring constant can be determined. Before data was collected, the minimum and maximum masses were determined by visually watching the harmonic motion. For the minimum, a mass was chosen such that the resulting simple harmonic motion had an appreciable amplitude without significant dampening over ten periods. For the maximum, the mass was selected such that the harmonic motion did not appear to exceed the linear elasticity of the spring. The suitable range of mass was determined to be between 25 grams and 45 grams. The same spring was used throughout the experiment. Several different masses were hung from the spring, and the spring was stretched approximately 2 cm from its equilibrium point and allowed to oscillate with each mass. A stopwatch was used to determine the total time for ten (10) periods. Plot #1 -- Period vs Mass Although the constraints of the experimental set-up did not allow collecting data using masses smaller than 25 grams, a data point near (0,0) can be approximated. If the spring was an ideal spring with zero mass, the data point (0,0) would be a valid data point. Using the provided data set, and the data point(0,0), create a scatter plot of Period vs Mass using Excel. If plotted correctly, the resulting plot will not be linear. If there is any doubt that the plot is not linear, fit the plot with a linear fit. Nonlinear plots such as this can be difficult to fit properly. Even if a proper fit can be applied using software, interpreting the resulting fit coefficients can be challenging. Manipulating the parameters placed on each axis such that a plot results in a linear relationship can be of great value. This can be accomplished even if an equation does not appear to be linear. Linear

Answer 1

0.8 0.7 total mass hung from spring sqrt(m) total time for 10 periods T=t/10 T^2 411?m 0.6 0.5 o 0 0 0 0.025 0.501 5.01 5.54 time period 0.4 0.554 0.03 0.035 -Series 1 0.3 0.15811388 0.17320508 0.18708287 0.2 0.21213203 5.92 6.36 6.75 0.592 0.636 0.251001 0.98596 0.306916 1.183152 0.350464 1.380344 0.404496 1.577536 0.455625 1.774728 0.04 0.2 0.045 0.675 0.1 0 0 0.01 0.02 0.03 0.04 0.05 mass

k = 41 slope plot 2 2 0.8 0.7 slope of the 1/vk k= 3.1894 3.877049 N/m y = 3.1894x -0.002 R2 = 0.9988 0.6 0.5 time period T 0.4 -Series 1 0.3 Linear (Series 1) 0.2 0.1 0 0 0.05 0.1 0.15 0.2 0.25 vm

plot 3 y=0.257x-0.0015 R=0.9989 slope of the graph from equation y = 0.257x -0.0011 is 0.257 1/k = 0.257 k= 3.89 N/m 0.5 0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 -Series 1 Linear (Series 1) 0 0.5 1.5 2 4ném