Copy of James Clear Final Physics Lab Report #1

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University of Alabama *

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520

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Physics

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Apr 3, 2024

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Title of experiment: Hooke’s Law & Simple Harmonic Motion Name: James Clear Date Performed: 01/23/24 Date Due: 01/30/24 Date Handed in: 01/30/24 Name of Partner(s): Alexia Luciano Introduction (Purpose/Hypothesis/Theory) The purpose of this experiment was to test a theory on Hooke’s law. Hooke's Law states that, “the force that is needed to extend and or shorten a spring by a proportional distance, is directly related to the proportion of that distance.” The first step of this experiment begins with determining the spring constant, k. A well developed understanding of Hooke's Law is gained by carefully examining the springs and the in response that is to follow from a known applied force. The second following phase of the experiment focuses on exploring simple harmonic motion. Simple harmonic motion is a repeated movement where a mass is attached to a spring, which in turn causes the spring to go through elongation or compression from the spring's original position length and then oscillate through its equilibrium position. There were five separate sections of data, and four different instances of data collection with necessary materials, which was completed with careful and precise experimentation. Data Part 1a: Determining k for a single spring Blue spring: Measured spring constant = 26 N/m Blue spring: Actual spring constant = 20 N/m Red spring: Measured spring constant = 12 N/m Red spring: Actual spring constant = 10 N/m

Green spring: Measured spring constant = 43 N/m Green spring: Actual spring constant = 40 N/m Part 1b: Parallel combinations of springs Equilibrium (starting) position of hook bar, 0.130 m = y 0 Mass hanging on springs Weight (force) hanging on springs Position of hook bar Displacement of hook bar from equilibrium 0.15 kg 1.47 0.137 m 0.003 m 0.20 kg 1.96 0.140 m 0.006 m 0.25 kg 2.45 0.143 m 0.009 m 0.30 kg 2.94 0.154 m 0.020 m 0.35 kg 3.43 0.162 m 0.028 m 0.40 kg 3.92 0.173 m 0.030 m Part 1c: Series combination of springs Equilibrium (starting) position of hook bar, 0.305 m = y 0 Mass hanging on springs Weight (force) hanging on springs Position of hook bar Displacement of hook bar from equilibrium 0.07 kg 0.686 0.318 m 0.013 m 0.10 kg 0.980 0.340 m 0.035 m 0.12 kg 1.17 0.370 m 0.065 m 0.15 kg 1.47 0.398 m 0.093 m 0.17 kg 1.66 0.418 m 0.105 m 0.20 kg 1.96 0.449 m 0.144 m

Part 2a: Amplitude variation for a spring A(m) △ t 1 (s) △ t 2 (s) △ t 3 (s) △ t (s) σ t (s) T (s) σ T (s) 0.18 6.19 6.26 6.33 6.26 0.070 0.626 0.007 0.19 6.22 6.54 6.26 6.34 0.541 0.634 0.054 0.20 7.02 7.54 7.22 7.24 0.262 0.726 0.026 0.22 7.21 7.04 7.11 7.12 0.085 0.712 0.0085 0.24 6.92 6.31 7.01 6.74 0.380 0.675 0.003 Part 2b: Mass variation for a spring M (kg) △ t 1 (s) △ t 2 (s) △ t 3 (s) △ t (s) σ t (s) T (s) T 2 (s 2 ) σ T (s) σ T 2 (s) 0.1 kg 5.53 5.52 5.77 5.603 0.144 0.5603 0.333 0.0144 0.00021 0.2 kg 7.56 7.31 7.65 7.506 0.176 0.7506 0.585 0.0176 0.00031 0.25 kg 7.75 8.01 8.01 7.923 0.150 0.7923 0.641 0.0150 0.00023 0.30 kg 7.88 7.99 8.15 8.067 0.136 0.8060 0.664 0.0136 0.00018 0.35 kg 8.85 8.54 8.77 8.721 0.161 0.8720 0.769 0.0161 0.00026 0.40 kg 9.31 9.55 9.63 9.497 0.167 0.9497 0.927 0.0167 0.00028 Sample Calculations: Part 1a: Determining k for a single spring Red spring: Measured spring constant of red spring = 12 Actual spring constant of red spring (instructor given value) = 10 Percent error equation = measured value − true value / (truevalue) × 100, equation; = 12 − 10 ÷ (10) ×100 = 20% Red spring percent error = 20% Part 2a: Amplitude variation for a spring 1. Calculate the mean △ t, and standard deviation, σ t of the three trials for each amplitude. - The formula for standard deviation (SD) is: σ = √(∑(x−¯x) ( x − x) 2 /n) - Mean sample; 6.19 + 6.26 + 6.33 = 6.26 = △ t

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2. Calculate the period T , from △ t/10 and record the data. - Sample; 6.26/10 = 0.626 = T 3. Calculate the uncertainty in the period, σ t , as σ t /10, and record the data. - Sample; 0.07/10 = 0.007 = σ T Part 2b: Mass variation for a spring 4. Calculate the mean △ t, and standard deviation, σ t of the three trials for each amplitude. - The formula for standard deviation (SD) is: σ = √(∑(x−¯x) ( x − x) 2 /n) - Mean sample; 5.53 + 5.52 + 5.77 = 5.603 = △ t 5. Calculate the period T, from △ t/10 and record the data. - Sample; 5.603/10 = 0.5603 = T 6. Calculate the uncertainty in the period, σ t , as σ t /10, and record the data. - Sample; 0.144/10 = 0.0144 = σ T 7. If both sides of Equation 2 are squared, the result is: T 2 = 4π 2 M / k. This states that T 2 is proportional to M with 4π 2 /k as the slope. Calculate T 2 and record in your data table. - Slope = 1.785 = 4π 2 /k = 39.44/1.785 = 22.1 8. Calculate the uncertainty in the square of the period, σ T 2 , and record the data. - Sample; 0.0144 2 = 0.00021 = σ T 2 Graphs/Charts: Graph 1: This graph illustrates the relationship between period and amplitude. However, the graph does not show any notable correlation between the two variables, period and amplitude. This falls aligns with the understanding that period is independent of the amplitude, because when amplitude is altered, period remains as similar values as shown above.

Graph 2: This graph above is illustrating the direct relationship between the period squared, T 2 and the mass of the object that is oscillating, which is M. The graph shows a linear trend, which is true because the period squared, T 2 has a proportional relationship with mass, M. Questions: Part 1a: Determining k for a spring 1. What is the spring constant in (N/m) of each of your springs? Ask your instructor for what the actual spring constant of each spring is and calculate your percent error for each spring. Do your results agree within the uncertainty parameters of the actual spring constant? - The measured spring constant values all varied, but were near the value of the actual spring constants. The blue spring had a measured spring constant value 26 N/m, the red spring had a constant of 12 N/m, and the green spring had a constant of 43 N/m. Using the actual spring constants from our instructor, the percent error equation was used to calculate the percent error for each spring constant. The calculated percent error for the blue spring is 30%, the percent error for red is

20%, and the percent error for green is 7.5%. Yes, the results also agree with uncertainty parameters as they are constant throughout all of the values. 2. Do your results support Hooke’s Law? - The result that came from our group's experiment findings supported Hooke's law. This is because the ideal spring exhibits an equilibrium length. When the spring is being compressed, a proportional force to the decrease in the length from the equilibrium position acts. This acts as the two pushing at each end away from one and other. 3. Suppose you hung a 5 kg mass from each of your springs (but do not actually do this). Do you think this would have any permanent effect on the springs? Do you think each spring would be affected equally? If not, which spring would be affected more? - Supposing that there was a 5 kg mass hung from each spring, unless this 5 kg mass caused a permanent damage to the spring, it would have to be anticipated the spring's behavior to agree to Hooke's Law. This is because of the simple addition of mass to the spring. Which entails the oscillations around its unstretched equilibrium length, and it should not entail a permanent effect. But since the adding of mass alters the amplitude of the motion, it does also influence the period. The period formula being; T=2π × √M/k, which shows the subsequent relationship between an increase in mass leads to a lengthened period, while the opposite, a decrease in mass results would result in a period getting shorter. This means the red spring, which has the lowest of the spring constants of 12 (N/m) that would be most significantly affected of the springs.

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4. There are a multitude of examples of Hooke’s law in action in our lives. Investigate and explain a particular example (biological preferred but choose something that interests you). - Hooke's law, a somewhat biological definition which states that the elastic behavior of a material comes about from small displacements of molecules. A biological example that illustrates Hooke’s law would be the concept that can be found in muscle tendons inside the human body. Essentially, the human muscle tendons, which link tissue to the bone, need to be able to stretch very easily and swiftly in response to any applied force but must be able provide a significantly higher returning force as the strain in the muscles increases, because if not this results in muscle tendon strains, pulls, and tears. Which is seen in athletes all over the world. Part 1b: Parallel combination of springs 5. What is the effective spring constant of your parallel spring combination (in N/m)? What is the effect of combining springs in parallel? Can you come up with a mathematical rule that explains what happens to the effective spring constant when springs are connected in parallel? - For our groups effective spring constant of the parallel spring combination in our data is 81 N/m. Then, combining the parallel springs, the springs are able to stretch less in comparison to when supporting those springs are stretched individually. As a result of this, the combined parallel springs are less vulnerable to stretching, and as a result the effective spring constant for the combination would end up two times the value that of a single spring, and that is because they

are combined and parallel. A mathematical equation of representation for this would be; F = -2kx. Part 1c: Series combination of springs 6. What is the effective spring constant of your series spring combination (in N/m)? What is the effect of combining springs in series? Can you come up with a mathematical rule that explains what happens to the effective spring constant when springs are connected in series? - For our groups effective series spring constant combination from our data is 12 N/m. Each spring tested in this part undergoes an identical force exerted by the weight of the mass it can sustain. As a result to that, each spring is able to extend by the same amount and the same as it would extend individually. Since the combined system is more susceptible to stretching, the effective spring constant is half of a single spring constant for two in series. Represented by a mathematical equation would be; F = -0.5kx. Part 2a: Amplitude variation for a spring 7. Use your data to plot a graph of T versus A. Include your standard deviation as error bars on your graph. Include this graph in the Graphs section of your lab report. How is the period T expected to depend upon the amplitude A? Does your graph confirm this expectation? - Based on our data and creating a graph with all the necessary elements, it illustrates that the period is not dependent on amplitude. This is assured by the following equation, T = 2π × √M/k. With the evidence from this equation and the graph, it agrees that period, T , is not related to amplitude. If period were to be

dependent on the amplitude, the graph would have reflected a direct correlation between the two, but the graph represents no major or notable change in period when amplitude changes. 8. If you allow your spring to oscillate for a long period of time, say one minute, the amplitude gradually decreases (this is known as damping ). If you waited for 20 seconds after you set your spring in motion and then began your data collection, would you expect your average measured period to differ from that found in your part 2a results? Explain. - If our spring was allowed to only oscillate for only nearly twenty seconds, no, there would not be an expectation for the average measured period to change drastically to what was found in part 2a. When the given amplitude is decreased, the oscillating object covers a shorter full distance traveled per cycle. If the spring were to oscillate at a shorter distance and the speed were to stay the same constant, this would lead to a decreased period. But this is not the case because the “damping” causes a decrease in the speed to cancel out the distance being shorter. Part 2b: Mass variation for a spring 9. Plot a graph of T 2 versus M. Include your standard deviation σ T 2 as error bars on your graph. Include this graph in your graphs section.What is the slope of your T 2 versus M graph? Because your slope should correspond to 4π 2 /k, take this equality and solve for k. What is this value of k? Calculate the percent difference between this value of k and the value of k determined from Part 1a for the same spring.

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- The slope of the graph for part 2b is 1.785. Setting this slope value equal to the corresponding equation; 4π 2 /k, gives that the k value is = 22.1 (N/m). Using the percent difference for this value related back to the k value from part 1a, the percent difference between these values would be 16.2%. 10. In Part 1c of the experiment, you measured the dependence of the period on mass of an oscillating spring, keeping the amplitude a constant (thus amplitude was a control in this part of the experiment). Explain how your results would be affected if you weren't careful about keeping the amplitude a control. - During this experiment, if there was not a careful and precise procedure followed with the control of amplitude, since period is not directly dependent on amplitude, it still would not have a major effect on the outcome of the data. Although the decreasing of the amplitude does reduce the full distance, as per the equation period = 2π × √M/k, a variation in amplitude does not impact the period. Conclusion - After carefully following the experimental procedure of this experiment to test the theory and collecting the provided data above, it demonstrates in agreement with Hooke’s law. The data illustrated a direct proportional relationship between force needed to stretch or compress and the distance the body moves. After analyzing the data further it was able to be determined that a mass that goes through a one dimensional motion is occupied in simple harmonic motion. The experiment was able to exhibit that the force that is applied on a specific mass is tied directly proportional to the displacement of that said mass from its equilibrium position and acts in the opposing direction. Overall, the data is able to

confirm and agree with Hooke’s Law is the recorded data sections, graphs, and questions answered.

Copy of James Clear Final Physics Lab Report #1

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