course Phy 121
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13:55:12 `q001. In this experiment you will construct a pendulum using a thread and one of the washers that came with your initial pack of CDs. If you have to use something heavier than a thread you will need to report what the string was made of and how thick the string is.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.
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RESPONSE --> it took about 1 1/2 sec. for a complete oscillation so to figure out how manyoscillations there would be in a min. 60sec in a min. so 60/1 1/2=40 times
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13:58:29 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.
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14:58:54 10-28-2005 14:58:54 `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). This corresponds to the traditional way of graphing y vs. x, with y vertical and x horizontal. Summarize these conventions in your own words. And do note them since you will be using these conventions later in this experiment.
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NOTES -------> freq. of the pendulum is defined as f while the length of it is described as L. f is dependant while Lis independant. when making atable with the two, the independant(L) goes in the first column, and the dependant(f) in the second column. when making a graph of the two L will go across the page, while the f will go up and down.
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15:02:44 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?
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RESPONSE --> it will take longer and longer because the length of the pendulum increases causing the distance of the swing to grow longer as well. If it takes 2 sec. the average length then if the string was twice as long it would take 4 seconds because the length is increased, so if it was half aslong as the normal it would take 1 sec. because the length is shorter
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15:05:40 `q003. A pendulum of length 1 foot will require about 1 second to complete 1 oscillation (from one extreme point in its motion to the other and back). How many oscillations would you therefore expect to count in a minute?
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RESPONSE --> you would expect 60 oscillations because there are 60 seconds in one minute 60*1=60
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15:06:20 There are 60 seconds in a minute. If the pendulum completes one oscillation every second then you will count 60 complete oscillations in a minute. Note that the actual time isn't exactly one second, and you will not count exacly 60 complete oscillations in a minute, but when you make your count you will be fairly close to 60.
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15:13:14 `q004. Suppose for the moment that a pendulum of length 1 foot does in fact make exactly 60 oscillations in a minute, as it would if Earth's density and/or radius were just a little different. Then based on your experience of pendulums, how many oscillations do you think a pendulum of length 2 feet would make in a minute? How many oscillations do you think would be made in a minute by pendulums of length 3 feet and 4 feet?
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RESPONSE --> the pendulum of 2ft will make 120 oscillations because it is double the length of the first 60*2=120, the 3ft would be 180oscillations, because it is triple of the original 60*3=180, and the 4ft would make 240 oscillations because it is 4x the original. 60*4=240
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15:13:37 You should have specified three estimates, one each for pendulum lengths 2, 3 and 4 feet. If you didn't, do so now.
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RESPONSE --> i did
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15:19:02 `q005. Suppose that a pendulum of a certain length made 120 oscillations in a minute. Then how much time, in seconds or fractions of a second, would be required for a single oscillation? If a pendulum of a certain length required 1.5 seconds for a complete oscillation then how many oscillations would it complete in a minute? Three different pendulums complete 20, 30 and 40 cycles, respectively, in a minute. How many seconds are required for a single oscillation of each pendulum?
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RESPONSE --> 120/60(secpermin)=2sec for each oscillation 60(secpermin) /1.5sec=40 oscillations 60(spm)/20=3sec 60(spm)/30=2sec 60(spm)/40=1.5sec
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15:20:26 If a pendulum completes 120 oscillations in 60 seconds then it completes two oscillations in a second. An oscillation therefore requires 1/2 second, or .5 second. If a pendulum requires 1.5 seconds to complete an oscillation, then in 60 seconds it will complete (60 seconds) / (1.5 second / cycle) = 40 cycles. Note that since it takes more than one second to complete an cycle, fewer than 60 cycles will be completed in a minute. This is your cue to divide rather than multiply 60 by 1.5. If a pendulum completes 20 cycles in 60 seconds, then it should be clear that 3 seconds are required for each cycle. The formal calculation is 60 seconds / (20 cycles) = 3 seconds / cycle. It should also be clear that 30 cycles in 60 seconds implies 2 seconds for each cycle. Formally, 60 seconds / (30 cycles) = 2 seconds/cycle. The answer for 40 cycles in a minute might not be as clear, so the formal calculation is useful. Just as in the previous examples we need to divide the number of seconds by the number of cycles. This time we obtain 60 seconds / (40 cycles) = 1.5 seconds/cycle. Each cycle takes 1.5 seconds.
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RESPONSE --> ok
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15:24:58 `q006. According to your previous predictions, will it require more or less than 1 second for a 2-foot pendulum to complete a cycle? Will the time required for a cycle increase or decrease as the pendulum gets longer? According to your predictions of the number of cycles expected in a minute, how long would it require for a pendulum of each length 2, 3 and 4 feet to complete a single oscillation?
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RESPONSE --> It will take 2 sec. because it takes 1 sec. for one foot pend. and so it would be double, from this you can see the time required will increase as the pen gets longer. because it takes 1 sec. for a one ft. to complete one oscillation, then at 2 ft it will take2 sec. for a 2ft. 2x1=2,3 sec for three ft 3x1=3, and 4 sec for 4 ft. 4x1=4
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15:32:56 You should in each case have divided 60 seconds by the predicted number of oscillations, as was done in the preceding problem. If you did not do so, do so now and enter your corrected response.
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RESPONSE --> 60/2=30 60/3=20 60/4=15
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10:59:45 `q007. Now using the conventions outlined earlier, sketch a graph of the time required per cycle vs. the length of the pendulum. Be careful about which axis you use for the time per cycle and which axis for the length of the pendulum. Using the conventions outlined in the 'Describing Graphs' exercise (i.e., using the language 'increasing at an increasing rate', 'increasing at a decreasing rate' etc.) give a description of your graph.
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RESPONSE --> on my graph I plotted the 2,3,&4 on the x-axis, and the 30,20,&15 on the y-axis, I drew a line connecating the following points, (2,30); (3,20); (4,15) the graph was decreasing at a decreasing rate this shows thaqt as the pendulum's length increases the number of oscillations decreases
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11:00:28 If you assumed, 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, as many people also do, 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.
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14:53:52 `q008. Construct a pendulum 1 foot long (that's 30.5 cm in case you don't have a ruler that measures feet) by tying a light string or thread to the washer in your initial materials. Holding the string as nearly stationary as possible, with a small nudge allow the pendulum to swing a few inches, no more, between its extremes observe the number of cycles in a minute. Get the most accurate count you can, estimating if possible even fractions of a cycle. Make at least three one-minute counts to see if your results are consistent. Give your results and state just how accurately you think you have counted (e.g., plus or minus 2 cycles per minute, or plus or minus 1/2 cycle per minute).
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RESPONSE --> I couldn't find any thread in my house, so I had to use a foot long silver necklace, and the large washer out of the lab kit, for a timer I used my microwave timer, the first time I did it i got 53.5 sec, the second timeI got 54 sec., and the third time I got 53.5sec. (plus or minus 1sec. each time)
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15:00:30 The accurate number of cycles in a minute is easily calculated from a certain formula. It turns out that the number is very close to 60 / sqrt(1.22). Calculate this number and compare your result to this result. How close did you come?
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RESPONSE --> 60/sqrt(1.22)= 60/1.10=54.5 I came within one second of this number and given my guess for hw much ai might have been under in counting, I might have actually hit it just right
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15:05:59 `q009. How does your 1-minute count of an actual pendulum change your estimates of the counts for lengths of 2, 3 and 4 feet? What are your revised estimates?
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RESPONSE --> It shows me that it will take a little less time than what was orginally expected, so this would aalso be true on the 2,3,&4 foot pendulums for the 2ft. 54.5x2=109 for the 3ft 54.5x3=163.5 for the 4ft 54.5x4=218
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15:09:32 Your count should have been around the ideal value of 54, rounded to the nearest whole number. This is 54 / 60 = .9 of the 60 counts we assumed originally. It would therefore be reasonable to multiply estimates for 2, 3 and 4 feet by .9 to take account of the observations you have now made.
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RESPONSE --> so 2x.9=.18 3x.9=.27 4x.9=.36
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16:06:50 `q010. Observe the number of cycles in a minute for different lengths: Measure different lengths, from 1 foot to 4 feet in increments of 1 foot. Increase the length by the same amount each time, and time the pendulum for a minute at each length. Be sure to let the pendulum swing only a few inches from one extreme position to another. 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, and give the table in your response. Let the first column be the length, the second the frequency. Label this table Data Set 1.
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RESPONSE --> I. Data Set One L f 1 53.5 2 38 3 31.25 4 26.5
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16:08:20 Analyzing The Experiment Look at your data table for f vs. L 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?
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RESPONSE --> they change slower and slower with the different lengths
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16:10:17 `q011. 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.
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RESPONSE --> i'm not sure what kind of response you want to this
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16:12:33 Describe the graph using the same type of language you used earlier. How does the graph reflect the behavior of the numbers on the table?
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RESPONSE --> the graph is decreasing at decreasing rate, the highest point is much higher than the others which slowly decline, they are all straight lines
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16:14:54 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, a 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.
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RESPONSE --> yep
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16:39:34 `q012. Now sketch an accurate frequency vs. length graph of your results by marking a scale and plotting points, and compare this graph with your rough sketch. How does the shape of your graph compare with the shape of the rough sketch?
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RESPONSE --> the graphs were actually quite accurate ,my distances were almost exact, the lines were just more steep on the actual graph compared to the freehand one
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17:31:24 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.
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RESPONSE --> the period increases with length, I think the graph wouldwould increase at an increasing rate, the graph increases at an increasing rate with the 4ft pendlum as the highest and the1ft at the lowest
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17:39:02 `q013. From your data, figure out the period associated with each length. Make a table of period vs. length, and label it clearly.
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RESPONSE --> II. Length vs. Period L p 1 1.12 2 1.58 3 1.92 4 2.26
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17:39:30 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 give the corrected table in your response. 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 from the graph you sketched before you made the table.
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RESPONSE --> I did it right
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17:46:49 `q014. Sketch your graph of period vs. length. Describe the graph (i.e., is it increasing or decreasing; and at an increasing or a decreasing rate?) .
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RESPONSE --> They are increasing at an increasing rate. The straight lines are increaing at a steadily increasing rate, however the lines aren't as far apart as the other graph of length vs.freq.
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17:50:43 The graph of period vs. length is increasing--the longer the pendulum the longer it takes to complete a cycle. The graph is increasing at a decreasing rate, since with every foot of increased length the period increases by a smaller and smaller amount. If your graph didn't show these characteristic then you should go back conceived where your observations and/or calculations when wrong and report any revised results here.
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RESPONSE --> that is how my graph is I justmissedtheincreasing at a decreasing rate, but I get it now
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21:20:32 `q015. The formula for the period of the pendulum is very close to T = .2 `sqrt(L), with T in seconds when L is in centimeters. The exact formula is T = 2 `pi `sqrt( L / g), where L is the length of the pendulum and g the acceleration of gravity; however this formula will be developed later in the course and for now we will use T = .2 `sqrt(L), remembering that L must be measured in cm. What should be the period of the pendulum, according to this formula, for lengths of 1, 2, 3 and 4 feet (remember that to use T = .2 `sqrt(L) the length L must be in cm and that the period T will be in seconds)?
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RESPONSE --> for 1ft; T=.2*sqrt(30.48 cm) T=.2*5.52=1.10cm/sec for 2ft; t=.2*sqrt(60.96 cm) T=.2*7.81=1.56cm/sec for 3ft. T=.2*sqrt(91.44 cm) T=.2*9.56=1.91cm/sec for 4ft. T=.2*sqrt(121.92cm) T=.2*11.04=2.20cm/sec
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21:20:53 The periods will be 1.10 sec, 1.56 sec, 1.91 sec and 2.21 sec. If you didn't obtain these results then go back and substitute again in see if you do get the correct results. If not explain in detail how you are using the formula T = .2 `sqrt(L).
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RESPONSE -->
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