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course Phy 121
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):
The sounds have a certain rhythm remains steady until the pearl pendulum stops moving and then the sound is gone.
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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):
The strikes are less frequently and the sounds get faster until the pendulum comes to a complete stop.
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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):
The pearl hit 8 times and I barely tilted the bracket forward to make the rhythm steady.
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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):
I think that the beginning position of the pendulum has the most sounds along with 90 degree turn on the left and right also had the most sounds. When the base was down hill it had fewer sounds and less hits.
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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):
I would put it the pendulum starting off not touching the base and then I think I would have the most regular beat.
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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):
.594
.500
.625
.594
.500
.422
.469
.484
I hit the timer when I released the pearl and then again on the second hit and then recorded my data.
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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):
.352, .766, .906
.672, .875, .953, 1.094
.797, .906, .961
.664, .875, 1.156, 1.297
.719, .859, .961
These are the results for the second, fourth, sixth, and eight hit.
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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):
65mm
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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):
.35, .67, .80, .66, .72
.77, .88, .91, 88
.91, .95, .96, 1.2, .86
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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):
There is .359 time interval between the release and the first hit of the bracket.
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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):
The time intervals between the first and second hit are .39 to .47 (.08 sec) for the first trial and .31 to .41 (.10sec) on the second trial. I just timed the first and second time of the bracket.
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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):
.77, .88, .91, 88 are the time intervals for the second hit which is .11sec, .03sec, and .03sec. The times are a lot faster than the difference between the first and second hit.
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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):
.91, .95, .96, 1.2, .86 are the time intervals for the fourth and sixth hit which is .04 sec, .01sec, .06sec, .34sec. These are even faster than the second and fourth hit.
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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):
Because it the second hit is at a faster rate than the second, fourth, or sixth hit. It is at the strongest velocity and acceleration of the cycle.
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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):
You expect the subsequent time intervals to decrease until a certain time and then the times would start to increase because the pendulum starts to slow down.
@& If the rhythm is steady then the intervals would remain the same.
It's difficult to achieve a steady rhythm, but it is at least theoretically possible to do so. The average speed of the 'pearl' decreases with every strike, but so does the distance it travels.
The amplitude of the swing of an ideal pendulum matches the average speed is such a way that the time required for every cycle is the same. When you timed the pendulum in a previous experiment your data would have showed that the rhythm of the pendulum was pretty much constant.*@
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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 shorter the pendulum is the faster the cycles and the shorter the swings. So with this information it would be faster than a longer pendulum.
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&#Good work on this lab exercise. See my notes and let me know if you have questions.
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