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21:32:43 `q001. The frequency of a pendulum is how frequently it oscillates back and forth. A very short pendulum oscillates much more quickly than a very long pendulum.
A cycle is a complete oscillation, from one extreme point to the other and back. Frequency is usually measured in cycles/second. However, it could be measured in cycles/minute, cycles/millisecond, cycles/year, or cycles/(time unit), where (time unit) is any unit of time. In this experiment you will observe and measure frequency in cycles/minute. If you hang your arm loosely at your side and nudge it a bit, it should oscillate back and forth a few times. Try it. Estimate how many complete oscillations it will undergo in a minute if you keep nudging it slightly to keep it moving.......!!!!!!!!...................................
RESPONSE --> I got an estimated 52 oscillations in a minute when I allowed my arm to swing at my side.
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21:33:02 If you managed to keep your arm relaxed it probably took a bit over a second to complete one back-and-forth oscillation. For most people a relaxed arm will oscillate about 40 to 50 times in a minute.
If your arm is tense, then you probably had to force the oscillations in you might have ended up with anything from about 30 to 100 times a minute.......!!!!!!!!...................................
RESPONSE --> ok
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21:37:29 `q002. In this activity you will:
1. Observe the frequency f of a pendulum vs. its length L. 2. Make a table of f vs. L. 3. Make a graph of f vs. L. Note the following conventions for determining which variable is independent and which dependent, and for placement of the dependent and independent variables on tables and graphs. f depends on L. We control L by holding the pendulum string at different lengths and observe its effect on f, so 1. f is the dependent variable, L the independent. 2. When we make a table, the independent variable goes in the first column, the dependent in the second. 3. When we make a graph, the independent variable goes horizontally across the page, and the dependent variable up and down (dependent is vertical vs. independent, which is horizontal). Note how this relates to the traditional way of graphing y vs. x, with y vertical and x horizontal. Summarize these conventions in your own words.......!!!!!!!!...................................
RESPONSE --> I will count the oscillations of a pendulum for a minute and then compare the number to the varying length of the string used. I'll then make one table and one graph comparing f (frequency) v/s L (length).
Since the frequency of the pendulum is dependent upon the length of the string, it is the dependent variable and L is the independent variable. When making the table, L is the first column and f is in the second and when making a graph, f is vertical (y axis) and L is the horizontal line (x axis)..................................................
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21:41:15 Predictions
Imagine a rock or some other mass hanging by a string over the rail of a high deck. As the rock swings back and forth, you gradually let more string over the rail, lowering the mass. Will the swings take longer and longer, or will they become more and more frequent? If it requires 2 seconds for the rock to swing back and forth from a certain length of string, how long do you think it would take if the string was twice as long? What if the string was half as long?......!!!!!!!!...................................
RESPONSE --> The swings will take longer and longer since the pendulum mass is lowered further from the railing and has more string to swing from.
I think it would take the rock 4 seconds or more to swing back and forth if the length of the string was doubled. I think it would take 1 second or less if the string was half as long since the mass of the pendulum is higher and there's a smaller amount of string for it to swing on..................................................
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21:50:03 `q003. Sketch a graph of the time required for a swing vs. pendulum length, as you would predict it. Describe your graph in some detail, using the conventions outlined in the 'Describing Graphs' exercise.
As with all graphs, be careful to use consistent units on each axis. If the marks on an axis are equally spaced, then they should represent the same change in the quantity represented by that axis. A common error would be, if for example the numbers 5, 10 and 20 were to be represented, would be to make three equally spaced marks representing the numbers 5, 10 and 20. The spacing between 5 and 10 should be half the spacing between 5 and 20. If you're not careful about your spacing, you will distort the shape of your graph.......!!!!!!!!...................................
RESPONSE --> Since Frequency, f, is the dependent variable, it is on the y-axis and Length, L, is independent and on the x-axis.
I am going to say that the length of the string will be tested at 5in, 10in, and 15in. The frequency of the pendulum should decrease with each test, therefore showing a graph that slopes downward, and I'm guessing at a pretty constant rate. This just shows that the longer the string is, the slower the pendulum will swing since the longer length allows for more movement and a wider swing. I'm not sure about the physics meaning as to why it actually swings slower, but that's my own observation. I'm sure I'll learn the rest soon..................................................
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21:52:49 If you said, as many people do, that a string twice as long to would imply a 4 second period (the period is the time per swing) and that a string half as long would imply a 1 second period, then you probably are assuming a linear relationship between period and length. In this case your graph should have been a straight line, a line which is decreasing at a constant rate.
If you said that a doubled length implies less than a doubled period, then to be consistent you probably said that half the length implies more than half the period. In this case your graph would be increasing but at a decreasing rate. Other responses are possible. The experiment will tell you whether your prediction was correct.......!!!!!!!!...................................
RESPONSE --> ok.
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22:07:50 `q004. If a pendulum of a certain length swings back and forth 35 times per minute, how many times per minute do you think it would take if the string was twice as long? What if the string was half as long?
Sketch a graph of the number of swings per minute vs. pendulum length, as you would predict it. Describe your graph. Be sure to describe how your graph behaves as pendulum length approaches zero.......!!!!!!!!...................................
RESPONSE --> If the string was twice as long, then the pendulum would swing slower so I divided 35 by 2 = 17.5 oscillations per minute. If the string was half the length, the pendulum would swing twice as fast so I did 35*2 = 70 oscillations per minute.
On a graph, the number of swings would be the y axis since it's the dependent variable and the length would be on the x-axis. As mentioned in my previous answer, my graph having the length go in ascending order (0, 10, 20, 40in for example), at zero, the pendulum wouldn't be able to swing at all so the graph line would start rather low and then peak at 10in, and then decline steadily at the 20in and 40in lengths. As the pendulum lenth approaches zero, the frequency should increase until there isn't enough string left for the pendulum to swing on. My description above doesn't really show this mainly because I'm not sure how to describe it..................................................
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22:09:32 The most common responses that the pendulum would swing back and forth 70 times per minute if the string was half is long, and 17.5 times per minute if the pendulum was twice as long. This would be consistent with the most common answer to the preceding question, which resulted in an increasing linear graph. The present graph would be decreasing, as we would expect, and would still be linear.
A prediction that half the length would imply less than 70 cycles per minute would be consistent with the idea that doubling the length will imply more than 17.5 cycles per minute. This would result in a graph that decreases but at a decreasing rate.......!!!!!!!!...................................
RESPONSE --> ok. I'm happy to know my answers are rather common.
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22:17:15 `q005. How are your answers to the first two questions related to your answers to the last two? Are they consistent?
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RESPONSE --> My answers do seem to be consistent and pretty in sync with the most common responses to the questions given. All my graphs were decreasing at a steady rate and seemed to agree with the explanations given.
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22:22:21 The Experiment
Frequency is measured in cycles/time units. Here we will measure frequency in cycles/minute, because it's easy to count the cycles in a minute. First make the pendulum: Tie a light string around a rock, a potato, or any relatively small, dense object. The length of the pendulum is measured from the fixed point of the swing to the center of gravity of the object. Describe how you constructed your pendulum.......!!!!!!!!...................................
RESPONSE --> I tied a string around the washer provided in the lab kit. the string is 33.5 inches long to the center of the object.
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22:37:46 `q006. Get the 'feel' of the pendulum.
Feel how quickly a pendulum 1 foot long swings back and forth. Don't time anything yet; just get the feel of the thing. Then get the feel of the rhythm of a two foot pendulum, then a three foot pendulum, then a4 foot pendulum. Try walking to the rhythm of each, stepping (either one or two steps to the cycle) the same distance on each 'beat' of the pendulum. Does the speed of your walk change by the same amount each time the length of the pendulum increases by a foot?......!!!!!!!!...................................
RESPONSE --> I think the speed of my walk should slow as the length of the pendulum gets longer but I couldn't get it to work too well. Every time I took a step, the swinging of the pendulum was thrown off and I couldn't really focus well on my walking speed while i was trying to get the pendulum to remain swinging at a steady pace.
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22:43:35 What would a graph of your speed vs. time look like? Would it be a straight line, would it curve upward, or would it go upward and tend to level off?
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RESPONSE --> I'm not quite sure how to vision a speed v/s time graph. I'm assuming that the question is referring to my walking along with the pendulum but since I couldn't get that to work out too accurately and I didn't time anything, I'm not sure what to do with a graph of these factors.
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23:17:22 `q007. Observe the number of cycles in a minute for different lengths:
Measure different lengths, from about a foot to the longest pendululum you can easily manage. Increase the length by the same amount each time, and time the pendulum for a minute at each length. Choose your length increment so you will end up with about six observations (i.e., so you will have six different lengths). Don't let the pendulum swing too far (no more than about 10 degrees from vertical). The frequency f, in cycles/minute, is the number of complete cycles in the minute. Write down in a table the frequency f and length L for each length. Let the first column be the length, the second the frequency. Label this table Data Set 1, and include the table in your response here.......!!!!!!!!...................................
RESPONSE --> Data Set #1
Length (inches) Freq. (cycles/minute) 12 55 15 49.5 18 45 21 40.5 24 39 27 37.................................................
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23:22:45 Analyzing The Experiment
Look at your data for Data Table 1 and see how changes in frequency are related to changes in length: Look at the numbers on the table. Do the frequencies change regularly with the lengths, or to they change faster and faster, or slower and slower?......!!!!!!!!...................................
RESPONSE --> The frequency, as expected, lowered with the increasing length of the rope. It did not, however, decrease by half when the length of the rope was doubled. The frequencies change drastically at first, decreasing by 5.5 oscillations when the rope was at 15 inches. Then they were even decreases between 15, 18, and 21 inches. After that, the oscillations declined at a lower rate, averaging around a decrease of 2 oscillations.
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23:24:11 `q008. As you look at the numbers, try to visualize the graph.
Sketch a graph of the general shape of your f vs. L data. Don't mark off a scale, don't plot points, just sketch the basic shape, from your examination of the numbers.......!!!!!!!!...................................
RESPONSE --> The general shape of the graph would start decreasing at a fast rate and then even out towards the end to where it's not declining as rapidly.
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23:31:23 Describe the graph.
How does the graph reflect the behavior of the numbers on the table? What is it about the graph that is related to the 'feel' of the walking exercise?......!!!!!!!!...................................
RESPONSE --> As before, the length is on the x axis and the frequency is on the y axis. The graph decreases more rapidly in the beginning and then slows in decline towards the longer lengths of the rope.
I'm not sure about how the graph relates to the feel of the walking exercise since I couldn't get it to turn out accurately. But as I mentioned earlier, my steps would end up slowing to match the pendulum swing, just as the number of oscillations decreased with the longer rope, the speed of my steps would also decrease so the two should match pretty accurately..................................................
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23:35:09 As the length increases, your table will show that the frequency decreases. That is, for a longer pendulum you will count fewer complete cycles in a minute.
Your table will also show that equal increases in the length of the pendulum result in smaller and smaller decreases in the frequency. For example, between a 1-foot pendulum and a 2-foot pendulum the frequency changes from about 55 cycles per minute to about 40 cycles per minute, the decrease of about 15 cycles per minute, while changing from a four-foot pendulum to a five-foot pendulum the decrease would be from about 28 cycles per minute to about 25 cycles per minute, a decrease of only about 3 cycles per minute. Note that the numbers given here are very approximate, so if your results differ by a few cycles per minute it is no cause for alarm. This behavior will cause the graph to decrease at a decreasing rate. With the walking exercise you should have noticed that each increase in length resulted in a decrease in walking speed, but that the decrease was less for the longer pendulums than for the shorter.......!!!!!!!!...................................
RESPONSE --> Ok.
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23:39:56 `q009. Sketch the graph by marking a scale and plotting points and compare with your rough sketch. How does the general shape of your graph compare with the shape of the rough sketch?
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RESPONSE --> Surprisingly, my graph matches my rough sketch relatively well. It's much neater and more accurate but the overall shape and curve is pretty much the same.
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23:44:55 Consider the period of the pendulum vs. length:
The period of a pendulum is the time required for 1 complete cycle. That is, if the pendulum requires 2 seconds to complete a cycle, the period is 2 seconds. Before doing any calculations to find the period, recall your direct experience of the pendulum. Does the period increase or decrease with length? What do you think a graph of the period vs. length would look like? Sketch a rough graph of period vs. length, and describe it in words.......!!!!!!!!...................................
RESPONSE --> The period should increase with the length of the rope because each cycle gets longer and slower to complete.
A graph of period v/s length would have the period on the y axis since it's dependent upon the length. The slope of the line would increase at a rapid rate at first and then slowing down as the distance in oscillations decreases and as the length increases. In short, it would increase and an increasing rate and then towards the longer lengths, the slope would increase at a decreasing rate..................................................
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23:48:25 `q010. From your data, figure out the period associated with each length. Make a table of period vs. length, label it clearly, and include the table in your response here.
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RESPONSE --> Length (inches) Period (seconds/cycle) 12 1.09 15 1.21 18 1.33 21 1.48 24 1.54 27 1.62
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23:53:40 The period is the number of seconds required for cycle. For example, if there are 30 cycles in a minute, then it takes 2 seconds for each cycle and the period is 2 seconds. If there are 40 cycles in a minute, 40 cycles require 60 seconds and a single cycle takes 60/40 seconds = 1.5 seconds.
If you miscalculated your periods, recalculate them and show the corrected table. Look at the numbers, asking yourself the same sort of questions as before. Visualize a graph of the numbers in this table. Then draw the graph. Can you look at the graph in such a way as to invoke the 'feel' of the things you have observed? Describe how the graph you made from the table is like, and how it is different than the graph you sketched before you made the table.......!!!!!!!!...................................
RESPONSE --> My sketch showed that the period increased at a higher rate in the beginning and then evened off some towards the longer lengths of rope.
My actual graph showed that the increase in the periods was even for the first three calculations, then the line increased even more between 18 and 21 inches and finally slowing down for the remainder of the lengths. It doesn't look like a gently sloping curve like my sketch does. It looks more like an S shape almost since the middle lengths increased more so than the shorter or longer lengths..................................................
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23:59:45 `q011. Let T stand for the period of the pendulum (the time required for a single cycle). You have a table of T vs. L. Now create a table for log(T) vs. log(L):
Use the log key on your calculator to obtain the required values. For example, if the period T is 3.1, log(T) will be approximately .5. For every T and every L on your table, find log(T) and log(L) and place these values in your new table. Show your table here.......!!!!!!!!...................................
RESPONSE --> Log (T) Log (L) .037 1.08 .083 1.12 .124 1.26 .170 1.32 .188 1.38 .209 1.43
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00:08:41 Next plot your new table, obtaining a graph of y = log(T) vs. x = log(L).
If you have taken your data carefully, and if you have used your calculator and plotted your points correctly, your graph should form a straight line. If not, go back and fix anything that needs to be fixed and replot it. Now using a straightedge, draw a straight line through your graph points and extend this line to the vertical (y) axis; also extend it a little ways past your last data point. If your line doesn't exactly fit every point, make it come as close as possible, on the average, to your graph points. Describe your graph and how well the straight line fits it.......!!!!!!!!...................................
RESPONSE --> I don't have an exact straight line for my graph, but I also did it by hand so it could be that my calculations are right but my graphing isn't as accurate as it would be on a calculator (i don't own a graphing calculator). My straight line does fit the points well, but just not perfectly, it cuts in between a few of them as opposed to following them all in a straight line as the answer said it should.
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00:16:10 `q012. By measuring the rise and the run between the left and right endpoints of your line (not between two points from your table), determine the slope of the line. Call this value p.
Determine the y-intercept of your straight line (i.e., the value of log(T) where your line meets the vertical axis). Call this value y0. Give your values of p and y0.......!!!!!!!!...................................
RESPONSE --> p = rise/run = (.21-.037) / (1.43-0.35) = 0.5 is the slope of the line. My handwritten graph is huge and not very accurate and I couldn't figure out the rise/run by looking at it so I went ahead and used the endpoints from my table to get a more precise number. I know the directions said to use my graph but I'm not sure I could have ever gotten a decent answer from it.
The y-intercept, y0, is 0.03 according to my graph..................................................
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00:17:58 Your pendulum model will be the function T = A * L^p, where p is the slope you determined in a recent step and A = 10^y0. Use your calculator to determine 10^y0 for your value of y0. What do you get for your value of A?
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RESPONSE --> A = 10 ^ .03 = 1.072 is my value for A.
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00:19:10 q013. Your function model is T = A * L^p. Substitute the values of A and p into this form. What is your function?
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RESPONSE --> T = 1.072 * L ^ (.05)
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00:22:51 Compare your function T = A * L^p with your data, using the values you have found for A and p, as follows:
For every pendulum length L, calculate the period Tpred predicted by this function. That is, plug in each length L and get your predicted value of T, then call this value Tpred. What values do you get for Tpred?......!!!!!!!!...................................
RESPONSE --> Tpred values for the function T = A * L^p are L Tpred 12 1.214 15 1.227 18 1.234 21 1.248 24 1.257 27 1.264
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16:59:09 `q014. For each pendulum length, calculate the difference T - Tpred.
Express each of these differences as a percent of Tpred. Give a table with columns for L, T, Tpred, T-Tpred, and percent difference. What is your maximum percent difference, and how nearly do you conclude that your function predicts the actual behavior of the pendulum? Does the function do a good job of predicting the period of your pendulum?......!!!!!!!!...................................
RESPONSE --> I can't figure out what T stands for so I subtracted L, the pendulum length minus the Tpred numbers. I'm assuming that's what the question is referring to. Length Tpred L-Tpred Tpred/(L-Tpred) %Diff 12 1.214 10.786 0.11255331 11.26 15 1.227 13.773 0.089087345 8.91 18 1.234 16.766 0.073601336 7.36 21 1.248 19.752 0.063183475 6.32 24 1.257 22.743 0.055269753 5.53 27 1.264 25.736 0.049114081 4.91
My maximum percent difference is 11.26% from the very first test at 12 inches. I think this function predicts the behavior of the pendulum well because the largest decrease in oscillations occurred when the string went from 12 to 15 inches and the function shows that decrease clearly. And like previously mentioned, the number of oscillations decreased by less every increase in string length, and the function shows this by the percentages decreases accordingly..................................................
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17:11:42 The 'correct' function for T vs. L is one of the following two equivalant functions:
If L was measured in feet and T in seconds, the ideal function would be about T = 1.1 L^.5. If L was measured in centimeters and T in seconds, then the ideal function would be about T = .2 L^.5. How close was your function to the ideal function? Figure out the periods corresponding to the lengths you observed. How close are these 'ideal' periods to the periods you observed?......!!!!!!!!...................................
RESPONSE --> My function was right with the 1.1, only that I didn't round mine and left it at 1.072 just to cut down on rounding errors with the later calculations.
My periods that correspond to the lengths i observed are kind of close to the "ideal" periods from the function. The largest difference I had between the periods was at the longest length, where my calculations say 1.62sec/cycle and the ideal period says 1.264sec/cycle. Outside of that, the differences between my periods from the pendulum and the function itself averaged to be 0.02 and I feel that's pretty accurate. If I had rounded my inital T value up to 1.1 instead of leaving it at 1.072, the periods would more than likely have been even closer..................................................