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course Phy 121
5:30 pm June 9
The bead is referred to below as the 'pearl'.When the pearl is released it swings back to the bracket, bounces off the swings back again, repeatedly striking the bracket. The magnet can be used to clamp the thread so the length of the pendulum remains constant.
If you have just a plain bracket then you simply tilt the bracket in order to achieve a constant rhythm, as described below.
You should set the system up and allow the pearl to bounce off the bracket a few times. The bracket should be stationary; the pendulum is simply pulled back and released to bounce against the bracket.
Note whether the pearl strikes the bracket more and more frequently or less and less frequently with each bounce. If the pearl does not bounce off the bracket several times after being released, it might be because the copper wire below the pearl is getting in the way. If necessary you can clip some of the excess wire (being careful to leave enough to keep the bead from falling through).
If the bracket is tilted back a bit, as shown in the next figure below, the pearl will naturally rest against the bracket. Tilt the bracket back a little bit and, keeping the bracket stationary, release the pendulum.
Listen to the rhythm of the sounds made by the ball striking the bracket.
Do the sounds get closer together or further apart, or does the rhythm remain steady? I.e., does the rhythm get faster or slower, or does it remain constant?
Repeat a few times if necessary until you are sure of your answer.
Insert your answer into the space below, and give a good description of what you heard.
Your response (start in the next line):
When you tilt it back you do not get as many hits and the rhythm isn’t as long.
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If the bracket is tilted forward a bit, as shown in the figure below, the pearl will naturally hang away from the bracket. Tilt the bracket forward a little bit (not as much as shown in the figure, but enough that the pearl definitely hangs away from the bracket). Keep the bracket stationary and release the pendulum. Note whether the pearl strikes the bracket more and more frequently or less and less frequently with each bounce.
Again listen to the rhythm of the sounds made by the ball striking the bracket.
Do the sounds get closer together or further apart, or does the rhythm remain steady? I.e., does the rhythm get faster or slower, or does it remain constant?
Repeat a few times if necessary until you are sure of your answer.
Insert your answer into the box below, and give a good description of what you heard.
Your response (start in the next line):
When its tilted forward, the number of hits seems to decreases and the rhythm is also shorter.
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If the bracket is placed on a perfectly level surface, the pearl will hang straight down, just barely touching the bracket. However most surfaces on which you might place the bracket aren't perfectly level. Place the bracket on a smooth surface and if necessary tilt it a bit by placing a shim (for a shim you could for example use a thin coin, though on most surfaces you wouldn't need anything this thick; for a thinner shim you could use a tightly folded piece of paper) beneath one end or the other, adjusting the position and/or the thickness of the shim until the hanging pearl just barely touches the bracket. Pull the pearl back then release it.
If the rhythm of the pearl bouncing off the bracket speeds up or slows down, adjust the level of the bracket, either tilting it a bit forward or a bit backward, until the rhythm becomes steady.
Describe the process you used to make the rhythm steady, and describe just how steady the rhythm was, and how many times the pendulum hit the bracket..
Your response (start in the next line):
If the rhythm speeds up or slows down, you can adjust the level of the bracket by tilting it either back or forward. A domino under the bracket can help it hang more freely
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On a reasonably level surface, place one domino under each of the top left and right corners of your closed textbook, with the front cover upward. Place the bracket pendulum on the middle of the book, with the base of the bracket parallel to one of the sides of the book. Release the pendulum and observe whether the sounds get further apart or closer together. Note the orientation of the bracket and whether the sounds get further apart or closer together.
Now rotate the base of the bracket 45 degrees counterclockwise and repeat, being sure to note the orientation of the bracket and the progression of the sounds.
Rotate another 45 degrees and repeat.
Continue until you have rotated the bracket back to its original position.
Report your results in such a way that another student could read them and duplicate your experiment exactly. Try to report neither more nor less information than necessary to accomplish this goal. Use a new line to report the results of each new rotation.
Your response (start in the next line):
placed dominoes under to elevate it and keep it even
When you rotate it 45 degrees at a time. The part where the pearl hangs beat at a more regular sort of rhythm and the other side has faster beat rhythm.
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Describe how you would orient the bracket to obtain the most regular 'beat' of the pendulum.
Your response (start in the next line):
The bottom of the book is where the pearl should be aimed.
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Orient the bracket in this position and start the TIMER program. Adjust the pendulum to the maximum length at which it will still bounce regularly.
Practice the following procedure for a few minutes:
Pull the pendulum back, ready to release it, and place your finger on the button of your mouse. Have the mouse cursor over the Click to Time Event button. Concentrate on releasing the pendulum at the same instant you click the mouse, and release both. Do this until you are sure you are consistently releasing the pendulum and clicking the mouse at the same time.
Now you will repeat the same procedure, but you will time both the instant of release and the instant at which the pendulum 'hits' the bracket the second time. The order of events will be:
click and release the pendulum simultaneously
the pendulum will strike the bracket but you won't click
the pendulum will strike the bracket a second time and you will click at the same instant
We don't attempt to time the first 'hit', which occurs too soon after release for most people to time it accurately.
Practice until you can release the pendulum with one mouse click, then click again at the same instant as the second strike of the pendulum.
When you think you can conduct an accurate timing, initialize the timer and do it for real. Do a series of 8 trials, and record the 8 time intervals below, one interval to each line. You may round the time intervals to the nearest .001 second.
Starting in the 9th line, briefly describe what your numbers mean and how they were obtained.
Your response (start in the next line):
0..40
0.36
0.335
0.404
0.374
0.381
0.482
0.438
how long between the time the bead was released and the second bounce
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Finally, you will repeat once more, but you will time every second 'hit' until the pendulum stops swinging. That is, you will release, time the second 'hit', then time the fourth, the sixth, etc..
Practice until you think you are timing the events accurately, then do four trials.
Report your time intervals for each trial on a separate line, with commas between the intervals. For example look at the format shown below:
.925, .887, .938, .911
.925, .879, .941
etc.
In the example just given, the second trial only observed 3 intervals, while the first observed 4. This is possible. Just report what happens in the space below. Then on a new line give a brief description of what your results mean and how they were obtained.
Your response (start in the next line):
577, .433, .661
.450, .44, .719
.627, .684, .712
.514, .527, .639
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Now measure the length of the pendulum. (For the two-pearl system the length is measured from the bottom of the 'fixed' pearl (the one glued to the top of the bracket) to the middle of the 'swinging' pearl. For the system which uses a bolt and magnet at the top instead of the pearl, you would measure from the bottom of the bolt to the center of the pearl). Using a ruler marked in centimeters, you should be able to find this length to within the nearest millimeter.
What is the length of the pendulum?
Your response (start in the next line):
9 cm
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If you have timed these events accurately, you will see clearly that the time from release to the second 'hit' appears to be different than the time between the second 'hit' and the fourth 'hit'.
On the average,
how much time elapses between release and the second 'hit' of the pendulum,
how much time elapses between the second and fourth 'hit' and
how much time elapses between the fourth and sixth 'hit'?
Report your results as three numbers separated by commas, e.g.,
.63, .97, .94
Your response (start in the next line):
.49, .76, .73
.59, .67, .69
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A full cycle of a free pendulum is from extreme point to equilibrium to opposite extreme point then back to equilibrium and finally back to the original extreme point (or almost to the original extreme point, since the pendulum is losing energy as it swings)..
The pearl pendulum is released from an 'extreme point' and strikes the bracket at its equilibrium point, so it doesn't get to the opposite extreme point.
It an interval consists of motion from extreme point to equilibrium, or from equilibrium to extreme point, how many intervals occur between release and the first 'hit'?
Your response (start in the next line):
1 interval
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How many intervals, as the word was described above, occur between the first 'hit' and the second 'hit'? Explain how your description differs from that of the motion between release and the first 'hit'.
Your response (start in the next line):
I interval
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How many intervals occur between release and the second 'hit', and how does this differ from the motion between the second 'hit' and the fourth 'hit'?
Your response (start in the next line):
2
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How many intervals occur between the second 'hit' and the fourth 'hit', and how does this differ from a similar description of the motion between the fourth 'hit' and the sixth 'hit'?
Your response (start in the next line):
2
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Why would we expect that the time interval between release to 2d 'hit' should be shorter than the subsequent timed intervals (2d to 4th, 4th to 6th, etc.)?
Your response (start in the next line):
It’s going from rest, should be shorter
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Would we expect additional subsequent time intervals to increase, decrease or stay the same?
Your response (start in the next line):
Adding time should make it decrease.
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What evidence does this experiment provide for or against the hypothesis that the length of a pendulum's swing depends only on its length, and is independent of how far it actually swings?
Your response (start in the next line):
The length did not have to be changed so you cannot say in fact that it is the only dependent. Where a person decides to hold the pearl can alter results and how far it cold swing.
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Your instructor is trying to gauge the typical time spent by students on these experiments. Please answer the following question as accurately as you can, understanding that your answer will be used only for the stated purpose and has no bearing on your grades:
Approximately how long did it take you to complete this experiment?
Your response (start in the next line):
1 hour
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You appear to have good data, but your analysis of the motion of the pendulum isn't correct.
If an interval is regarded as motion from extreme point to equilibrium, or from equilibrium to extreme point:
How many intervals are there between release and the first 'hit'?
How many intervals are there between the first 'hit' and the second 'hit'?
How many intervals are there between the second 'hit' and the fourth 'hit'?
How many intervals are there between release and the second 'hit'?
What should be longer, the time from release to second 'hit' or the time from second 'hit' to fourth 'hit'?
Do your data provide evidence for your conclusion?
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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Question: State the given definition of the average rate of change of A with respect to B.
Briefly state what you think velocity is and how you think it is an example of a rate of change.
In reference to the definition of average rate of change, if you were to apply that definition to get an average velocity, what would you use for the A quantity and what would you use for the B quantity?
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Your solution:
The average rate of change A with respect to B is given by the change in A divided by the change in B.
Velocity gives you both the speed and direction of an object's motion. This is an example of a rate of change because it tells you how far something goes per unit time.
Quantity A would be distance units / time units, and quantity B would be units of time.
confidence rating #$&*: 2
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Given Solution:
A rate is a change in something divided by a change in something else.
This question concerns velocity, which is the rate of change of position: change in position divided by change in clock time. **
NOTE ON NOTATION
Students often quote a formula like v = d / t. It's best to avoid this formula completely.
The average velocity on an interval is defined as the average rate of change of position with respect to clock time. By the definition of average rate, then, the average velocity on the interval is v_ave = (change in position / change in clock time).
• One reason we might not want to use v = d / t: The symbol d doesn't look like a change in anything, nor does the symbol t. Also it's very to read 'd' and 'distance' rather than 'displacement'.
• Another reason: The symbol v doesn't distinguish between initial velocity, final velocity, average velocity, change in velocity and instantaneous velocity, all of which are important concepts that need to be associated with distinct symbols.
In this course we use `d to stand for the capital Greek symbol Delta, which universally indicates the change in a quantity. If we use d for distance, then the 'change in distance' would be denoted `dd. It's potentially confusing to have two different d's, with two different meanings, in the same expression.
We generally use s or x to stand for position, so `ds or `dx would stand for change in position. Change in clock time would be `dt. Thus
v_Ave = `ds / `dt
(or alternatively, if we use x for position, v_Ave = `dx / `dt).
With this notation we can tell that we are dividing change in position by change in clock time.
For University Physics students (calculus-based note):
If x is the position then velocity is dx/dt, the derivative of position with respect to clock time. This is the limiting value of the rate of change of position with respect to clock time. You need to think in these terms.
v stands for instantaneous velocity. v_Ave stands for the average velocity on an interval.
If you used d for position then you would have the formula v = dd / dt. The dd in the numerator doesn't make a lot of sense; one d indicates the infinitesimal change in the other d.
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Self-critique (if necessary):
I have no idea what the answers to the last two segments are based on this explanation. I'm finding myself more confused by the answer than the question. Can you please explain????
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Your answer was good, so if the given solution confuses you, ignore it.
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Question: Given average speed and time interval how do you find distance moved?
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Your solution:
The average speed will give you distance per time. If you multiply by the time interval, the time units cancel out, giving you the total distance moved.
confidence rating #$&*:
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Given Solution:
** You multiply average speed * time interval to find distance moved.
For example, 50 miles / hour * 3 hours = 150 miles. **
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Self-critique (if necessary): OK
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Self-critique rating: OK
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Question: Given average speed and distance moved how do you find the corresponding time interval?
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Your solution:
The average speed gives you distance per time. You solve this by cross multiplication.
(average speed) (corresponding numbers)
distance 1 / time 1 = distance 2 / time X
You would multiply time 1 and distance 2. You would divide that number by distance 1 to get the corresponding time interval (time X).
confidence rating #$&*: 3
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Given Solution:
** time interval = distance / average speed. For example if we travel 100 miles at 50 mph it takes 2 hours--we divide the distance by the speed.
In symbols, if `ds = vAve * `dt then `dt = `ds/vAve.
Also note that (cm/s ) / s = cm/s^2, not sec, whereas cm / (cm/s) = cm * s / cm = s, as appropriate in a calculation of `dt. **
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Self-critique (if necessary):
I did this incorrectly, but I understand what I'm supposed to do now.
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Self-critique rating: 3
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Question: Given time interval and distance moved how do you get average speed?
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Your solution:
You would divide the distance moved by the time interval.
confidence rating #$&*: 3
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Given Solution:
** Average speed = distance / change in clock time. This is the definition of average speed.
For example if we travel 300 miles in 5 hours we have been traveling at an average speed of 300 miles / 5 hours = 60 miles / hour. **
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Self-critique (if necessary): OK
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Self-critique rating: OK
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Question: A ball rolls from rest down a book, off that book and onto another book, where it picks up speed before rolling off the end of that book. Consider the interval that begins when the ball first encounters the second book, and ends when it rolls of the end of the book.
For this interval, place in order the quantities initial velocity (which we denote v_0), and final velocity (which we denote v_f), average velocity (which we denote v_Ave).
During this interval, the ball's velocity changes. It is possible for the change in its velocity to exceed the three quantities you just listed? Is it possible for all three of these quantities to exceed the change in the ball's velocity? Explain.
Note that the change in the ball's velocity is denoted `dv.
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Your solution:
It is only possible for it to exceed the quantities listed if an outside force acted to lower the initial or final velocity. Otherwise, it would continually slow down or pick up speed (assuming that the book is a flat surface.) It is possible got all of the quantities to exceed the change in velocity because the velocity might stay constant or barely change, thus giving a change in velocity of around zero.
confidence rating #$&*:3
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Question: If the velocity at the beginning of an interval is 4 m/s and at the end of the interval it is 10 m/s, then what is the average of these velocities, and what is the change in velocity?
List the four quantities initial velocity, final velocity, average of initial and final velocities, and change in velocity, in order from least to greatest.
Give an example of positive initial and final velocities for which the order of the four quantities would be different.
For positive initial and final velocities, is it possible for the change in velocity to exceed the other three quanities?
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Your solution:
The average of these velocities is 7 m/s, and the change in velocity is + 6 m/s.
Least to greatest:
Initial Velocity: 4 m/s
Change in Velocity: 6 m/s
Average if Initial and Final Velocities: 7 m/s
Final Velocity: 10 m/s
confidence rating #$&*: 3
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Question: If the position of an object changes by 5.2 meters, with an uncertainty of +-4%, during a time interval of 1.3 seconds, with an uncertainty of +-2%, then
What is the uncertainty in the change in position in meters>
What is the uncertainty in the time interval in seconds?
What is the average velocity of the object, and what do you think ia the uncertainty in the average velocity?
(this last question is required of University Physics students only, but other are welcome to answer): What is the percent uncertainty in the average velocity of the object, and what is the uncertainty as given in units of velocity?
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Your solution:
The uncertainty in meters is 0.208 m.
The uncertainty in seconds is 0.026 seconds.
The average velocity of the object is 5.2/1.3 = 4 meters / second. I think that the uncertainty of the velocity is +- .208 meters / 0.026 seconds, or 8 meters/second.
confidence rating #$&*: 2
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You only have 2% and 4% uncertainties in the given quantities.
An uncertaintly of 8 m/s on a 4 m/s result would be an uncertainty of 200%.
The small uncertainties in your given quantities won't result in that great an uncertaintly.
If you use the maximum possible distance, which would be 5.2 m + .208 m, and the minimum possible time interval 1.3 sec - .026 sec, what velocity do you get?
By how much does that velocity differ from the 4 m/s velocity you calculated based on the given quantities?
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Good overall, but check my notes on that last question and please submit a revision.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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