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course Phy 242
6/30 3
Experiment kinmodel_01: The Distribution of Atomic SpeedsWhen the speed of the simulation is moderate it is possible to watch a specific particle (the red particle or the blue particle in the default simulation) and obtain an intuitive feeling for the relative frequencies of various speeds.
Run the simulation billiard simulation at the default settings.
Observe the simulation long enough to get a feel for the maximum velocity you are likely to see.
Then estimate how much time it spends at slow (less than 1/3 of max vel.), medium (between 1/3 and 2/3 of max. vel.) and fast (more than 2/3 of max. vel.) velocities.
Express your estimates in percents of the total time spent in the three different velocity ranges.
Draw a histogram (a bar graph) of your estimates. Describe your histogram in your writeup.
Answer****
I would estimate that the 0-1/3 max velocity spends about 35% of the time, 1/3 - 2/3 spends about 40% and 2/3 - max spends about 15% of the time.
My histogram looks skewed to the left or slower velocities. The largest area is 1/3 - 2/3, the second largest area is 0 - 1/3 and the smallest is the 2/3 - max velocity.
Now suppose you had estimated the percent of time spent in each of 10 velocity ranges (i.e., from 0 to .1 of max. vel., .1 to .2 of max. vel., etc, up to max. vel.).
From your previous estimates, without further viewing the simulation, make a reasonably consistent estimate of the proportion of time spent in each of these ranges.
Answer****
I would estimate (0 - .1)....(.9 - max) percents go as followed:
10%, 13%, 17% 20%, 14%, 10%, 7%, 5%, 3%, 2%
Sketch a histogram of your estimates and describe the graph in your writeup.
Answer****
My graph is skewed to the left of the mid-point. It tapers off much more on the right than the left side.
Sketch the smooth curve you think best represents the distribution, with the curve being highest at the most likely speed, near the horizontal axis for speeds you very seldom observe.
According to your sketch, which speed is the most likely?
Answer****
According to my sketch, the .3-.4 (max) speed has the greatest area under the curve, but is closer to .3 (max), which is about 3.6 on a max speed scale of 12. So I would estimate at a speed of 4.
What percent of the area under your curve corresponds to speeds within one unit of your most likely speed (e.g., if your most likely speed was 3, you would estimate the area under the curve between speed 3 - 1 = 2 and speed 3 + 1 = 4).
Answer****
Between speeds 2 and 4, there is about 30% of the area.
Watch the green particle for long enough to estimate the percent of time it spends at speeds more than 2 units greater than the most likely speed, but not more than 4 units greater.
What percent of the time do you estimate that the green particle is moving at less than half its most likely speed?
Answer****
It seems its moving greater than the most likely speed about 70% of the time and less than half its most likely speed about 10% of the time
Watch the number corresponding to the speed of the green particle.
Close your eyes for a few seconds at a time and open them suddenly, and each time write down the velocity of the particle as you see it immediately after your eyes open.
Record about 100 velocities in this manner.
Answer****
4, 5, 10, 6, 4, 3, 3, 7, 7, 11,
13, 9, 4, 6, 6, 5, 3, 4, 3, 2,
4, 1, 6, 7, 6, 8, 6, 2, 7, 8,
8, 3, 0, 0, 5, 9, 6, 6, 10,
5, 9, 2, 4, 7, 9, 2, 6, 4, 2,
5, 7, 3, 3, 4, 2, 2, 6, 5, 3,
3, 7, 3, 9, 8, 6, 5, 8, 2, 10
8, 6, 8, 8, 3, 3, 8, 2, 4, 7,
6, 5, 3, 8, 5, 9, 10, 12, 13, 6,
1, 6, 5, 9, 5, 5, 3, 3, 7, 6,
Tally your velocities to see how many of the 100 velocities were 0, how many were 1, how many were 2, etc.
Construct a histogram of your results and compare to the histograms you predicted earlier.
Answer****
0: 2
1: 2
2: 9
3: 16
4: 9
5: 12
6: 16
7: 8
8: 9
9: 7
10: 4
11: 1
12: 1
13: 2
The histogram is not as smooth as I predicted. There is still a skew to the left, but there is a gap between 3 and 5 where 3 and 5 are tall and 4 has a dip. Also the ends did not taper out as smoothly as I predicted. There is more of a sudden drop after 4 and a sudden increase after 1.
Experiment kinmodel_02: Mean free path; mean time between collisions
It is possible to observe the mean free path of the green particle between collisions.
First observe the particle for a few minutes and try to get a feel for how the distances traveled between collisions with other particles are distributed.
Make your best estimate of what percent of the time the particle travels less than 1 inch between collisions, the percent of the time the distance is between 1 and 2 inches, the percent of the time the distance is between 2 and 3 inches, etc.. When the particle collides with a 'wall', it doesn't count as a collision and distance keeps accumulating until it collides with another particle.
Sketch a histogram of your estimates, and also document the distance on your monitor between the 'walls' that confine the particles.
Now take some data.
Using the 'pause' and 'restart' buttons, stop and start the particle motion as required in order to observe the distances traveled by the green particle between collisions.
Use a ruler to measure distances traveled.
Don't leave any distances out, because this would bias the sample.
Observe at least 100 distances.
Answer****
9.3
4.8
6.7
3.5
4.5
3.0
0.7
3.9
3.0
5.8
13.3
1.4
0.7
3.5
4.4
3.4
3.0
3.3
7.5
3.3
2.8
1.7
7.5
3.9
4.7
6.4
2.1
2.0
1.7
2.6
1.7
7.8
5.0
3.7
1.5
1.0
6.9
0.8
7.5
7.5
2.8
9.5
7.1
6.0
1.4
1.5
7.2
5.0
3.8
8.4
9.0
3.9
2.6
2.5
1.9
1.4
0.3
7.0
1.6
1.0
2.5
1.8
4.3
2.8
2.9
1.8
0.5
7.0
3.2
0.6
4.5
1.7
1.6
4.0
3.4
5.4
4.2
1.4
0.4
0.3
0.6
0.7
8.3
2.2
3.7
1.8
4.8
1.8
2.9
1.1
2.5
5.5
2.4
3.5
3.1
1.0
2.0
3.5
3.2
Describe how you obtained your data and report your data as a frequency distribution (i.e., the number of observations for which the distance rounded to 0, 1, 2, 3, ..., inches).
Answer****
I paused the screen and measured in cm with a ruler against the screen every straight line in between collisions.
Frequency distribution:
0:3
1:15
2:18
3:18
4:13
5:6
6:4
7:6
8:6
9:2
10:1
11:0
12:0
13:1
Sketch a histogram of your results.
Sketch the histogram you would expect from a large number of observations.
Describe your histograms, and how they compare with your previous predictions.
Answer****
In my prediction histogram, the curve is smooth but skewed to the value of about 2.5.
In my actual histogram, there is a trend, however the curve isn't smooth and there seems to be an ourlier. The curve is skewed toward the values between 1 and 5 but is choppy.
@&
Very good.
You would need a much greater number of observations in order to get a smooth curve. Your results are as good as could be expected based on 100 obervations.
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Experiment kinmodel_03: Equipartition of energy and the direction of disorder to (increasing or decreasing)
ALTERNATIVE
Start the program using default values. Let it run for several seconds, then start observing the green particle. Keep track of whether it is moving more in the x or more in the y direction. Just say to yourself 'x x x y y y y y x x y x y y y ... ', according to what you see. Do this at a steady but comfortable pace. Continue this for a minute or so.
Then take a pencil and paper, or alternatively open a text editor in a separate window, and start writing down or typing your x and y observations. I just did this and in about a minute or two I got the following: xxyyyyxyyxxyxyyxxyxxxyyyxxyyxxyyxyxxyyyxyyyxyyxy. I haven't done this before and found this a little confusing. Every time the particle got hit I wanted to type a letter right away, but I hadn't had time to figure out in what direction it was headed. With practice I began to get over that. You will experience different glitches in the process, but with a few minutes of practice you'll be able to do a reasonably good job. I suspect I also had some tendency to type one of the letters in preference to the other (e.g., x in preference to y, or maybe y in preference to x). I don't recommend fighting this sort of tendency but just noticing it and gently trying to improve. I didn't do this with pencil and paper, and it would be interesting to see if the tendencies are the same when writing as opposed to typing. However that's not our purpose here.
At whatever pace you prefer, write or type about 50 observations of x or y. List them here.
Answer****
xyxyxxxyyxyyxxyyyxyyyxyyyxxyyxyxyxyyxxyxyyxxyyyyxy
Now notice the KEx and KEy values represented toward the right-hand part of the program's window, just a little ways below the middle of the screen. KEx represents the total x component of the kinetic energies of all the particles and KEy the total y component.
Using the Pause and Restart buttons, stop and start the program and with each stop record the KEx and KEy. After each observation quickly hit 'Restart' then 'Pause', and record another. Record about 50 observations.
Answer****
KEx KEy
825 1647 y
1162 1337 y
1042 1457 y
1186 1313 y
1128 1371 y
1174 1325 y
1208 1291 y
1114 1385 y
1283 1217 x
1401 1099 x
1350 1150 x
1173 1327 y
1353 1146 x
1498 1102 x
1408 1091 x
1365 1134 X
1407 1092 X
1515 984 X
1106 1494 Y
1115 1384 Y
946 1553 Y
956 1543 Y
1118 1381 Y
1099 1401 Y
924 1576 Y
1099 1401 Y
1080 1420 Y
965 1535 Y
906 1593 Y
1083 1417 Y
1285 1214 X
1164 1335 Y
1421 1078 X
1448 1051 X
1601 899 X
1565 934 X
1344 1156 X
992 1508 Y
1338 1161 X
1588 912 X
1228 972 X
1662 838 X
1657 842 X
1318 1181 X
1460 1040 X
1369 1131 X
872 1627 Y
1092 1408 Y
1208 1291 Y
1266 1234 X
Having recorded the 50 KEx and KEy values, write 'x' next to each pair for which the x value is greater, 'y' next to each pair for which the y value is greater. List your x's and y's in sequence here (don't list your values for the KE).
Answer****
x/y
YYYYYYYYXXXYXXXXXXYYYYYYYYYYYYXYXXXXXYXXXXXXXXYYYX
What is the greatest KEx value you observed and what is the least?
Answer****
1662, 825
What is the greatest KEy value you observed and what is the least?
Answer****
1647, 838
On a 50-trial sample of a normal distribution, the mean would be expected to occur about halfway between the least and greatest values observed, and the expected standard deviation would be very roughly 1/5 of the difference between the least and greatest values. According to this (very approximate) rule, what would be the mean and standard deviation of your KEx values, and what would be the mean and standard deviation of your KEy values?
Answer****
The mean for KEx values would be 1244 +/- 167.4
The mean for the KEy values is 1243 +/- 161.8
Do you think the mean KEx value differs significantly from the mean KEy value? There is a difference. By 'significantly', we mean a difference that seems greater than what would naturally occur by chance statistical variations.
Answer****
I do not think the KEx and KEy values differ significantly. The means are only 2 values away even though their standard deviations are much larger.
Experiment kinmodel_04: The improbability of all particles being segregated on one side of the viewing area (order vs. disorder)
Any selected region of the screen can be selected for viewing by masking the rest of the screen. The viewer can estimate the probability of this region being vacated within an hour, within a day, within a year, ..., within the age of the universe. Results will differ with the size of the region, the number of particles and the speed of the simulation.
Cut out a 1-inch square and watch the simulation for 2 minutes on the middle default speed. Observe how many times the square becomes 'empty' of particles. Estimate what percent of the time this square is empty.
Answer****
it is empty about 13 times. I estimate the square to be empty about 15% of the time.
Enlarge the square to a 1-inch by 2-inch rectangle and repeat.
Answer****
this area was only vacant 2 times in 2 minutes. I estimate about 1-3% of the time it was empty.
Enlarge to a 2-inch by 2-inch square and repeat.
Answer****
there was never a time when the square was empty in 2 minutes.
Enlarge this square to a 2-inch by 4-inch rectangle and repeat.
Answer****
there was never a time when the square was empty in 2 minutes.
Enlarge to a 4-inch by 4-inch square and repeat.
Answer****
there was never a time when the square was empty in 2 minutes.
Mask all but 1/4 of the screen and repeat.
Answer****
there was never a time when the square was empty in 2 minutes.
How long do you think it would take, on the average, for 1/4 of the screen to become completely empty of particles?
Answer****
Since the probability is so low, it would probably take hours to days for this to happen.
How long do you think it would take, on the average, for 1/2 of the screen to become completely empty of particles?
Answer****
Since the probability is so low, it would probably take days to weeks for this to happen, if it would at all.
A typical closet is about 100 million times as far across as the distance represented by the screen. Ignoring for the moment that the closet is three-dimensional and hence contains many more air molecules than would be represented by a 2-dimensional simulation, how long do you think you would have to wait for all the molecules to move to one side of the closet?
Answer****
if the molecules are as dense as they are in this model, just 100 million more of them, I think it is near impossible for all of the molecules to move to one side of an area that large of an area. I could only guess it would take infinity time for this rare phenonemon to occur.
Experiment kinmodel_06: The connection between relative particle mass and average speed; equality of average kinetic energies
Using default settings, answer the following:
What do you think is the average speed of the dark blue particles as a percent of the average speed of the green particles? (you might, for example, observe how long, on the average, it takes a particle of each color to move a distance equal to that across the screen)
Answer****
I think on average, the blue particle is about 80% slower than the green particle.
What do you think is the average speed of the red particle as a percent of the average speed of the green particles?
Answer****
Calculating the speed of the red particle and green particle, I estimate that the red particle is 100% slower than the green particle.
A blue particle is 4 times more massive than a green particle. How do you think its average KE therefore compares with the average KE of the green particles?
Answer****
I calculated the average KE of the green particle to be 10 times larger than the KE of the blue particle.
A red particle is 64 times more massive than a green particle. How do you think its average KE therefore compares with the average KE of the green particles?
Answer****
I calculated the average KE of the green particle to be 100 times larger than the KE of the red particle.
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Very well done and insightful.
Check my one note.
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Very good work.
Check my notes.
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