Report Experiment Simple Harmonic Motion

Report for Experiment Motion in One & Two Dimensions Abstract The overall goal of our lab was to show the independent relationship between horizontal and vertical motion, in addition to studying the time dependence of displacement, velocity, and acceleration in one and two dimensions. Utilizing a hover puck and a glass table, we created an inclined plane for which we used to record the displacement of the puck and to compute its velocity in both vectors. Plotting these values of velocity with respect to the time allowed us to calculate the acceleration using the slope of the puck in our investigations. The acceleration allowed us to derive a formula to solve for gravity by multiplying it by 2 the inverse of the sin of our table's incline. Our average gravity came out to be 9. 50 𝑚/𝑠 , our value 2 2 being .30𝑚/𝑠 off from the accepted value of 9.80𝑚/𝑠 can be attributed to instrumental errors we encountered. Introduction When someone fires a cannonball, there's not just one single dimension being utilized, when the cannonball is in motion you have both a horizontal and vertical force acting upon the object. When you simply kick a ball on the ground, you don't have to worry about a vertical component as the ball never goes upwards or downwards vertically, its height remains the same. Galileo hypothesized that a ball dropped from a tower would hit the ground at the same time as a ball launched horizontally from a cannon, and through experiments, he was able to confirm this theory. His findings implied that the downwards motion when influenced by gravity was independent of the horizontal motion. This means that the overall movement of an object is not one motion or another but the vector sum of the motions. Some equations to explain motion and acceleration in the horizontal and vertical direction are shown below: Motion in the Horizontal Motion in the Vertical 𝑣𝑥 = 𝑣𝑥0 + 𝑎𝑥𝑡 𝑥 = 𝑥0 + 𝑣𝑥0𝑡 + 1 2 𝑣𝑦 = 𝑣𝑦0 + 𝑎𝑦𝑡 𝑎𝑥 𝑡 𝑦 = 𝑦0 + 𝑣𝑦0𝑡 + 1 2 𝑎𝑦 𝑡 We conducted this experiment for two reasons, one was to study the time dependence of displacement, velocity, and acceleration in one and two dimensions. Another reason was to show that motion in one direction is independent of motion in a perpendicular direction. In investigation 1 we dropped an air hockey puck on an elevated surface in order to track position vs time. Using this data we then were able to account for velocity and by plotting this on a velocity vs. time graph we could then solve for acceleration by using the slope. Investigation 2 had us execute the same general setup, yet this time the puck had a horizontal velocity applied to it by us. We used a graph of the position in both the x and y component vs time, in order to calculate the velocities. We calculated the velocity in both the horizontal and vertical components and graphed it in relation to time, to be able to solve for the acceleration with respect to time. Investigation3 and 4 had us calculate the acceleration with respect to gravity. Displacement is the change in distance over time of an object in motion. It is tracked front the object's initial position to its final at the end of the given time. Displacement allows us to calculate the average velocity which allows us to have an understanding of the relation between the rate and the overall direction of motion. The equation for displacement is as follows: 𝑉𝑦 = Δ𝑦 Δ𝑡 Acceleration measures objects' change in velocity in respect to time, which can be calculated in both the vertical and horizontal direction. The equation for acceleration is shown below: 𝐴𝑎𝑣𝑔 = Δ𝑉 Δ𝑡 Investigation 1 Investigation 1 was set up by using a glass table, a hover puck, a wooden block, and a webcam. We began the investigation set up by leveling the glass table so that when the hover puck was placed in the center of it, it would not move. Once the table was level we placed our wooden block under the back leg of the table so that we created an inclined plane. Once our physical components of the experiment were set up we began to set up the webcam which would track our data. We were asked to set up the IPL Tracker program located on our computers, once we did we were able to see the live feed being recorded by the overhead camera on our desktop. Next, we had to calibrate the camera settings, so that the table was fully in frame, and we had to play around with the focus and exposure settings but kept in mind that the frame rate had to remain at 30 frames/sec while achieving the clearest image we could. Our objective was to make the color of the hover puck appear as bright as possible on camera, since our tracking software, tracks color. Once that was calibrated we had to identify the outer corners of our glass table, while doing so we measured the outside length of our glass table and saved the data. The last two things we had to do for our setup were color masking and recording our experiment. The process of color masking was simple, as we just adjusted the settings until everything but the puck appeared black. To begin collecting data, I clicked on the start recording button by doing so the software began recording the time the X and the Y coordinates of the puck. I allowed the program to run for 10 seconds with the hover puck placed in the center of the table. Once my data was collected, I copy and paste the data table from the software program, following the lab instructions into my shared Microsoft excel spreadsheet with my partner. The table stored the data for the time in seconds, the x and the y- coordinate position of the puck, as well as the standard deviation of the x and y components and stored under the columns titled δx and δy. The standard deviation was calculated using the equation tool in excel by highlighting the column with the data we desired. We used the x and y positions in respect to time to create two histograms, one for the x-values and one for the y-values. The width of these histograms is equivalent to the respective standard deviation. The value for δx was 0.01705 and δy was 0.01277. While we did calculate the random error in the experiment there could have been other human errors, like systematic errors in regards to our position measurement or errors in our calculations. Investigation 2 Investigations 2’s purpose was to analyze the case of one-dimensional motion, and it followed the same setup in regards to equipment as Investigation 1. The table and the wooden block remained in the same position, and the camera remained calibrated to the same settings setup in Investigation 1. The only things required for the procedure were to make sure that our table corners were still properly identified, and then we had to turn on the puck and place it in the middle of the slanted glass table. Once it was placed on the slanted glass table, we started the data recording and held the puck stationary for a second at the top of the incline, before releasing it and letting it freely fall, making sure to try to avoid applying push in any directions. We took the data for a total of 8 seconds before we stopped. The data table was then copied and pasted from the software into our Excel sheet. Using our collected data we once again calculated the velocity of the puck in the y-direction using our position vs time data. We then also calculated the 𝑡 using our data for the 8 seconds. Both equations are 𝑎𝑣𝑔 shown below, and once computed were put into our data chart. 𝑉𝑦 = Δ𝑦 Δ𝑡 𝑡𝑎𝑣𝑔 = 1 2 (𝑡1 − 𝑡2) Next, we had to calculate the error in 𝑉𝑦 by using the error propaganda shown in Appendix A, this is done by utilizing the following equation: δ𝑉𝑦 = δ𝑉𝑦 𝑉𝑦 = δ𝑉 𝑦 Using the data in our chart for Investigation 2 we created a scatter plot to represent the relationship of Vy vs. tavg, with error bars being neglected due to how small they are. We created our chart so that it had 4 straight line portion data series. Once we had the data for each series we were able to right-click and add a 2 trendline so that we could separately fit the two straight parts of our data. We included our 𝑅 value in our 2 plot; 𝑅 which is the statistical measure that shows how close the data is to the fitted trendline. Ideally, 2 you would want your 𝑅 to be equal to 1, as that means the points are exactly on the trendline. Our R^2 value was equal to 0.99 for our 2nd through 4th series, and had a value of .96 for the 1st series, meaning our data was very close to the trendline. The slope of the scatter plot represents the acceleration of the hover puck as it falls from rest, we used the IPL Straight Line Fit Calculator to calculate the actual slope and its uncertainty. The slope of the scatter plot for the first was -55.85, the 2nd slope was -57.66, the 3rd was -60.26 and the last slope for series 4 was -62.74. During this trial you can see that the puck had its greatest average velocity between 1.7 and 3.2 I would say overall our slopes were close to what we had anticipated from this experiment. Investigation 3 Investigation 3, was set up in the same manner as Investigation 2 was, with the glass table being made into an inclined plane by use of the wooden block positioned under the back leg of the table. We were asked to take the hover puck and place it into the left-hand bottom corner of the glass table to begin. We clicked start recording and waited for a second before we launched the park in a way that it would create a parabola from the left-hand corner to the right-hand corner. We took extra care to make sure that the puck did not touch the top of the glass box. we stopped recording before the puck hits the edge of the table. This data was then copied and pasted from the program into our Microsoft Excel. We then repeated the calculations we had done for Investigation 2, using the new set of data. This meant that we calculated, 𝑡𝑎𝑣𝑔, 𝑉𝑦 , 𝑉𝑥, δvx, and δvy. We calculated for both the x and y component of movent this time as we were taking into account two dimensions as opposed to 1 since we are simulating movement in a parabola. The equation to calculate 𝑡𝑎𝑣𝑔 was shown in investigation 2, and is still solved the same way in investigation 3, this also applies for the equation of Vy and δVy. Vx and δVx are calculated similarly to Vy and δVy, but we are just using our x-value components instead of as seen below: 𝑉𝑥 = Δ𝑥 Δ𝑡 δ𝑉𝑥 = δ𝑉𝑥 𝑉𝑥 = δ𝑉 𝑥 Once all the data and calculations were recorded we created a scatter plot. The scatter plot for Investigation 3 showed the relationship between velocity and time. It includes both Vx and Vy as separate data series. The scatter plot was set up the same way it was in Investigation 2, but the steps to set up the Vy scatter plot must be executed twice due to the same steps needing to be executed for Vx as well. We 2 made sure to again include our 𝑅 values. The slopes are representative of the acceleration in both the x and y directions. The slope of horizontal 2 trajectory was equivalent to 6.04, while the slope of vertical trajectory was equal to -57.53 𝑚/𝑠 . These slopes of acceleration are what we would expect as the vertical acceleration will have more fluctuation since that is the axis that is changing, while the horizontal acceleration should have remained consistent which our data did for the most part. Investigation 4 The goal of investigation 4 was to solve mathematically for the acceleration due to gravity, using our data from Investigation 2 and 3. We were first asked to draw a diagram of our setup and to include vectors to illustrate the force of both gravity and acceleration acting on the object as well as the normal force. Length of Table 63 cm Height of Table and Block 14.9 cm Height of Block 3.9 cm Using our diagram we are able to come up with a formula to use to solve for gravity. We used the height of the back portion of the table and the length of the table to create a triangle, this allowed us to find the angle at which our glass table was tilted on. 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒(ℎ𝑒𝑖𝑔ℎ𝑡) ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑒𝑢𝑠𝑒(𝑙𝑒𝑛𝑔𝑡ℎ) 𝑠𝑖𝑛 (θ) = −1 (ℎ𝑒𝑖𝑔ℎ𝑡(14.9 𝑐𝑚) 𝑙𝑒𝑛𝑔𝑡ℎ(63 𝑐𝑚) θ = 𝑠𝑖𝑛 ( ) θ=16.1538462 We then used the angle to come up with a formula for g which was: 𝑔 = 𝑎 𝑠𝑖𝑛 θ Using this equation that we derived we can then solve for the acceleration free fall fro Investigation 2 and 3, in addition, we will also use this to make a formula to calculate the error of g fro both investigations using this equation: 𝛿𝑔 = 𝛿𝑎 2 ) 𝑎 +( 𝛿𝑙 2 ) 𝑙 +( 𝛿ℎ ℎ ) 2 Investigation 2 Data 𝑔 m(a) 55.85 𝑚/𝑠 57.67 𝑚/𝑠 60.16 𝑚/𝑠 62.75 𝑚/𝑠 2 2 2 2 𝛿g 2 9. 02 𝑚/𝑠 2 9. 32 𝑚/𝑠 2 9. 72 𝑚/𝑠 10. 14 𝑚/𝑠 2 0.0216 𝑚/𝑠 2 0.0216 𝑚/𝑠 0.0209 𝑚/𝑠 0.0204 𝑚/𝑠 2 2 2 Investigation 3 Data m(a) 57.539 𝑚/𝑠 𝛿g g 2 2 9. 29 𝑚/𝑠 0. 0216 𝑚/𝑠 2 gav can be calculated by averaging the acceleration calculations, which happens when you add up the acceleration values and divide it by the quantity of data. The equation for the average acceleration can be seen below as well as the error average: 𝑔𝑎𝑣 = 𝑔1+ 𝑔2 + 𝑔3+𝑔4+ 𝑔5 # 𝑜𝑓 𝑣𝑎𝑙𝑢𝑒𝑠 𝑔𝑎𝑣 = 9.50𝑚/𝑠 δ𝑔𝑎𝑣 = 2 δ𝑔1+ δ𝑔2 +δ 𝑔3+δ𝑔4+δ 𝑔5 # 𝑜𝑓 𝑣𝑎𝑙𝑢𝑒𝑠 δ𝑔𝑎𝑣𝑔 = 0. 0212 2 2 The accepted value for g is 9. 81 𝑚/𝑠 our value for g came out to be 9. 50 𝑚/𝑠 which is only 0.29𝑚/𝑠 off from the accepted value. This means that our value is outside of our error range and does not agree with our experimental uncertainty, and the slight difference can be accounted for by systematic and random errors in Investigation 2 and 3. 2 Conclusion Experiment 2 consisted of 4 experiments that proved Galileo’s ideas about motion in one and two dimensions. He was able to come to the conclusion that when motion was under the influence of gravity it was independent of the horizontal motion. This implies that the overall motion of an object is actually equal to its vector sum of motions. Experiment 2 allowed us to test Galileo's ideas by analyzing the one and two dimension movement of a hover puck on an inclined plane and computing its acceleration, velocity, and other types of movement. In the first experiment, we placed the hover puck in the center of the table and recorded its motion for a set period of time, this investigation allowed us to document the random errors that can occur. Investigation 2 had us analyze the one-dimensional movement of the hover puck. We did this by dropping the puck from the top of the inclined surface and seeing how the velocity changed in relation to gravity, and acceleration. We tested for the same values in investigation 3 but this time in both dimensions as we launched the puck from the corner in a parabola. After all of these sets of data were collected, in Investigation 4 we calculated the acceleration in relation to gravity-based upon the data. 2 We found that our average gravity (g) came out to be 9.50𝑚/𝑠 which is only .29 off from the 2 accepted value of 9.80 𝑚/𝑠 . Our error was calculated to be 0.0212, and our value is outside of this error range, even though our number was close to the actual value. This can be accounted for most likely in instrumental errors as we were not able to get our table properly level due to the legs being broken. In order to improve this experiment, I would recommend trying to find better lab equipment, as there were no fundamental issues with the experiment but the poor quality of our glass table ultimately ended up skewing our data. Possibly finding a way to have the puck moved without human involvement could also help as you can have a lot of random errors due to how we may launch the puck. Questions 1. Find the time for the puck to reach the bottom of the air table starting from the top for Investigations 2 and 3. Are these times all the same, as shown in the picture of the cannon (Fig. 3.2)? Why or why not? a. Because of the way we set up our experiment for investigation 3, the puck did not fall from the top of the table to the bottom. This means that we calculated the time it took from our highest y-point which was 20.156 to our lowest y which was 16.85. The time that it took to travel this distance was .37s. Investigation 2, had a time of 12s to travel the same distance as experiment 3. We can attribute this to the fact that in investigation 2 by the time the hover puck is at 20.15, there has already been acceleration and downward movement applied in the same direction as gravity. With investigation 3, the peak of the parabola is at y equals 20.156 and the puck only begins moving in the same direction of gravity at that point. 2. If the acceleration of a puck falling down an incline of angle θ (where θ<45∘ ) is aθ , what is a2θ? a. If the acceleration of a puck falling down an incline on an angle where θ is < 45, then a2θ is equal to 2𝑎θ𝑐𝑜𝑠θ. 3. You determined the acceleration from the slope of your plot of v vs. t. What is the meaning of the intercept of your line with the v-axis? Do you expect this value to be close to the origin? a. The intercept of our line with the v-axis is representative of the initial velocity of the hover puck. This value is expected to be close to the origin because the initial velocity in Investigation 1 and 2 in the y-direction should have been zero due to the hover puck being dropped from rest. 4. If you performed Investigations 2 and 3 again with an incline of θ=0, what would the trajectories look like in each case? a. If the investigations were to be performed again with that incline, the hover puck would not move from its initial position in investigation 2, and for investigation 3 it would only have a constant horizontal velocity, the vertical velocity would not exist. 5. In Investigation 3, you launched the puck so that it goes all the way up and comes back down in the right corner of the table. Since the table is a square, the puck goes up the same distance in the y-direction as it travels in the x-direction. Calculate the theoretical angle that is required for this. 2 a. 𝑑 = 2 𝑣 𝑠𝑖𝑛(2θ) 𝑔 2 𝑣 𝑠𝑖𝑛(2θ) 𝑔 ℎ = 2 = 2 𝑣 𝑠𝑖𝑛 (θ) 𝑔 2 𝑣 𝑠𝑖𝑛 (θ) 𝑔 2 →2𝑠𝑖𝑛(2θ)= 𝑠𝑖𝑛 (θ) The theoretical angle would come out to be 75 degrees. Honors Questions: 1. How would your results of Investigations 1 and 2 change if the masses of the pucks were doubled? a. There would be no effects on Investigation 1 and 2 as mass has no relation to velocity and or acceleration in this situation. 2. Imagine that two pucks in the first investigation are released 1 s apart. Will the distance between them remain the same during their “free fall” prior to the impact? a. No the distance between them would not remain the same because the distance between them depends on their velocities. As we know velocity is affected by the acceleration due to gravity in this situation. This means that the first puck will fall and its rate of 2 acceleration will increase by 9. 81 𝑚/𝑠 and puck twos velocity will also accelerate at the same rate. This means the distance will not remain the same as their velocities are changing. 3. If we imagine a “reverse” situation where the puck from your second investigation is launched backward from the point of impact with reversed vx and vy velocities (vx pointing left and vy pointing up), what will be the average acceleration and velocity in both x and y directions? What will be the travel time to the highest point? a. The average velocity along the y-axis will be zero and the average velocity in the horizontal component is the negative version of our data so the average velocity for Investigation 2 was 12.28 m/s, so the velocity in a reverse situation will be -12.28𝑚/𝑠. The average acceleration would be negative 9.50. The travel time to the highest point can be calculated by the equation shown below. 𝑡= 𝑉𝑎𝑦 𝑔