#$&* Phy 121
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#$&* 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 &&&&&&&&&&&&&&&&&&
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The sounds get closer together as we tilt the bracket. #$&* If the bracket is placed on a perfectly level surface, the pearl will hang straight down, just barely touching the bracket. However most surfaces on which you might place the bracket aren't perfectly level. Place the bracket on a smooth surface and if necessary tilt it a bit by placing a shim (for a shim you could for example use a thin coin, though on most surfaces you wouldn't need anything this thick; for a thinner shim you could use a tightly folded piece of paper) beneath one end or the other, adjusting the position and/or the thickness of the shim until the hanging pearl just barely touches the bracket. Pull the pearl back then release it. If the rhythm of the pearl bouncing off the bracket speeds up or slows down, adjust the level of the bracket, either tilting it a bit forward or a bit backward, until the rhythm becomes steady. Describe the process you used to make the rhythm steady, and describe just how steady the rhythm was, and how many times the pendulum hit the bracket..your response &&&&&&&&&&&&&&&&&&
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I got the brackets on as flat of a surface as possible. The pendulum hit the bracket twelve times and was steadily reducing time between impacts for the first 8 hits then sped up more as there was less distance to travel. #$&* 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 &&&&&&&&&&&&&&&&&&
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The further the pearl hangs from the bracket initially, based upon the slant, the less frequently and slower it will hit the bracket when released. #$&* Describe how you would orient the bracket to obtain the most regular 'beat' of the pendulum.your response &&&&&&&&&&&&&&&&&&
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I believe that the most consisting pattern is the one I get when the pearl is facing south at a 180 degree or 0 degree position. From her the pearl will not actually be touching the bracket at rest so there are no sudden increases in speed near the end of the experiment. #$&* 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 &&&&&&&&&&&&&&&&&&
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.562 .613 .484 .468 .546 .484 .484 .484 These results were obtained by timing the release and second it of the pearl. #$&* Finally, you will repeat once more, but you will time every second 'hit' until the pendulum stops swinging. That is, you will release, time the second 'hit', then time the fourth, the sixth, etc.. Practice until you think you are timing the events accurately, then do four trials. Report your time intervals for each trial on a separate line, with commas between the intervals. For example look at the format shown below: .925, .887, .938, .911 .925, .879, .941 etc. In the example just given, the second trial only observed 3 intervals, while the first observed 4. This is possible. Just report what happens in the space below. Then on a new line give a brief description of what your results mean and how they were obtained.your response &&&&&&&&&&&&&&&&&&
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.5, .734, .843, .828,.921,1.109, 1.203 .515.859.875 .5, .843,.921, 1.015 .578,.734.812.812.894 #$&* 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 &&&&&&&&&&&&&&&&&&
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130 mm #$&* If you have timed these events accurately, you will see clearly that the time from release to the second 'hit' appears to be different than the time between the second 'hit' and the fourth 'hit'. On the average, • how much time elapses between release and the second 'hit' of the pendulum, • how much time elapses between the second and fourth 'hit' and • how much time elapses between the fourth and sixth 'hit'? Report your results as three numbers separated by commas, e.g., .63, .97, .94your response &&&&&&&&&&&&&&&&&&
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.523, .27, .45 #$&* 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 &&&&&&&&&&&&&&&&&&
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1/4 #$&* How many intervals, as the word was described above, occur between the first 'hit' and the second 'hit'? Explain how your description differs from that of the motion between release and the first 'hit'.your response &&&&&&&&&&&&&&&&&&
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½, it is twice the amount of the first hit because it has to go back to the extreme point and then the equilibrium again. #$&* 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 &&&&&&&&&&&&&&&&&&
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¾,this is different because on the second and fourth hit it is already at equilibrium and has further to travel for a complete hit than when released from the extreme point. #$&* How many intervals occur between the second 'hit' and the fourth 'hit', and how does this differ from a similar description of the motion between the fourth 'hit' and the sixth 'hit'?your response &&&&&&&&&&&&&&&&&&
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1, the only difference in motion is that the extreme point will be closer to the equilibrium due to a loss of force. #$&* 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 &&&&&&&&&&&&&&&&&&
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We would think this because it has more force behind it and loses power and speed over time. #$&* Would we expect additional subsequent time intervals to increase, decrease or stay the same?your response &&&&&&&&&&&&&&&&&&
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I would expect later hits to occur more frequently because there is less distance to travel. #$&*your response &&&&&&&&&&&&&&&&&&
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It doesn’t provide any because there are other factors such as the weight and size of the item or how extreme the release point is. #$&*your response &&&&&&&&&&&&&&&&&&
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1 hour and ten minutes. #$&* *#&!Revision isn't requested, but if you do choose to submit revisions, clarifications or questions, please insert them into a copy of this document, 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.