Lab 5 Springs Worksheet (1)

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Researcher Name: Reagyn Beckwith DA Name: Sierra Butler PI Name: Zach Scott Lab 5 Worksheet Use the data you have organized to answer the following questions. LAB DATA – PART 1 LAB DATA – PART 2

Researcher 1: Include a description of how force and stretch were determined as they are not directly measured (a Free Body Diagram is required). This experiment included finding the exerted force by using Newton’s second law of f=mg. The mass for this equation was found and recorded during the experiment. The gravitational force is a known value of 9.8 𝑚 𝑠 2 . The values of mass and gravitational force were then multiplied to find the exerted force. To find the stretch, measurements from the length of each spring were recorded before mass was added and then measured again with the added mass. The next step was to find the total measurement of the actual stretch experienced by the spring. This was done by subtracting the length of the stretched spring from the length of the unstretched spring. DA 1: Create a “ Force vs Stretch Data table” for at least one spring (like Prelab question 2 but with all the raw data included as well) as well as a scatter plot (Force on the y axis and stretch on the x axis) showing the data for all three springs (with trend lines and equations for all three springs). If you can’t achieve clarity with careful labelling, you can have multiple scatter plots. RED SPRING Stretch Force 0.64255 2.153942 0.62055 2.643942 0.60855 3.133942 0.60055 3.623942 0.57855 4.113942

y= 25.263x + 0.21979 BLUE SPRING y= 40.734x + 0.26962 0 1 2 3 4 5 0.56 0.58 0.6 0.62 0.64 0.66 Force (N) Stretch (m) Force (N) vs Stretch (m) 0 1 2 3 4 5 0.58 0.6 0.62 0.64 0.66 Force (N) Stretch (m) Force (N) vs. Stretch (m)

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GREEN SPRING y= 48.562x + 0.27743 Researcher 2: Should the mass be moving while you measure the spring’s distance? Explain why or why not using Newton’s Law. Yes, it would be accurate for the mass to move as the distance of the spring is measured. If the mass were to remain the same and not move while the spring is in an unstretched position, the forces acting on the object would be equal. An object where its mass is multiplied by its acceleration is a contingent force from Newton’s second law. If there were to be an acceleration of zero, the object would be immobile, it would not move because the forces acting on the object would also be equal to zero. To find the stretch of each spring, distance had to be measured. For the spring to exhibit stretch, enough mass must be added to the spring. When mass was added, the net force was disturbed because of the acceleration that the spring experienced. During this time, the spring stretched and bounced from the added mass, allowing for the distance to be determined. Ultimately, this proves that when the distance of the spring is measured, there will be movement from the mass. PI 1: What is the meaning of the y intercept found on the Force vs distance graph generated by the DA. Also, please report the measured k value for all three springs. Make certain that your k value has units of N/m. The y-intercept on a Force vs. Distance graph from the DA signifies the initial force applied to a spring or object when it's in its equilibrium position or at rest. This intercept represents the point where no displacement has occurred, indicating the force required to maintain the spring at its natural state. By applying Hooke's law (F = -kx), the spring constant can be calculated from the gradient of the graph, providing a quantitative measure of each spring's stiffness. Red spring: 25.263N/m. Blue spring: 40.734N/m. Green spring: 48.562N/m Researcher 3: Take a screen shot of one of the position vs time graphs to explain how to measure period and amplitude in the procedure section. Make sure you also explain how you estimate the uncertainty in both the period and amplitude measurements. 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 0.62 0.625 0.63 0.635 0.64 0.645 0.65 0.655 0.66 Force (N) Stretch (m) Force (N) vs Stretch (m)

The position of one wave at the peak was subtracted by the trough to find the amplitude for the graph of position versus time. The second peak must be left of the first peak rather than to the right. The obtained amplitude value was then divided by two. Between the wave peak and the trough, a halfway point was revealed as a result of dividing by two. The time value from one peak is subtracted by its corresponding peak to the left to determine the period. Uncertainty is estimated for the period and amplitude by finding the standard deviation of each. Therefore, the standard deviation of the period determines the value of uncertainty for the period and the standard deviation of the amplitude determines the uncertainty value for amplitude. DA 2: Provide a table of Period vs Amplitude for one spring. Create a scatter plot to determine how the period depends on the amplitude. DA 3: From g. and h. (in step 3 of the lab manual) the DA should construct a data table (and a scatter plot) of Period vs k value. Spring St.Dev period K-value Red 0.010954 25.263 Blue 0.007071 40.734 Green 0.016432 48.562 0 0.2 0.4 0.6 0.8 1 0 2 4 6 AMPLITUDE PERIOD PERIOD VS AMPLITUDE RED SPRING PERIOD AMPLITUDE 0.88 0.1493 0.9 0.146 0.88 0.129 0.88 0.136 0.9 0.1331

PI 2: What determines how long it takes a mass to bounce on a spring? Make sure you reference using the data tables/graphs provided by the DA. One reason that determines how long it takes for a mass to bounce on a spring is the stiffness of the springs or the k value. A stiffer spring spring has a higher k-value and it causes the mass to bounce back faster because the spring exerts a stronger force to bring back equilibrium. This can be seen with the graph above with the red and blue data. However, due to human error the green data is flawed. By comparing the blue and red data, the blue spring recorded a higher k value and had a shorter period. 0 10 20 30 40 50 60 0 0.005 0.01 0.015 0.02 K-value Period Period vs. K-value

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