Query0Assignment0

course Phy 121

6/18 at 11:30 PM

ph1 query 0Most queries in this course will ask you questions about class notes, readings, text problems and experiments. Since the first two assignments have been lab-related, the first two queries are related to the those exercises. While the remaining queries in this course are in question-answer format, the first two will be in the form of open-ended questions. Interpret these questions and answer them as best you can.

Different first-semester courses address the issues of experimental precision, experimental error, reporting of results and analysis in different ways and at different levels. One purpose of these initial lab exercises is to familiarize your instructor with your work and you with the instructor 's expectations.

Comment on your experience with the three lab exercises you encountered in this assignment or in recent assignments.

*********************************************

Question: This question, related to the use of the TIMER program in an experimental situation, is posed in terms of a familiar first-semester system.

Suppose you use a computer timer to time a steel ball 1 inch in diameter rolling down a straight wooden incline about 50 cm long. If the computer timer indicates that on five trials the times of an object down an incline are 2.42sec, 2.56 sec, 2.38 sec, 2.47 sec and 2.31 sec, then to what extent do you think the discrepancies could be explained by each of the following:

• The lack of precision of the TIMER program.

To what extent to you think the discrepancies are explained by this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

I believe that lack of precision in the program is probably less likely to be a major factor in the discrepancies of the answer. Because it is computer-programmed, it has the ability to be very precise.

• The uncertain precision of human triggering (uncertainty associated with an actual human finger on a computer mouse)

To what extent to you think the discrepancies are explained by this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

This factor is more likely to cause discrepancy. No human can “program” themselves to release the object and click the mouse in the exact same time interval multiple times.

• Actual differences in the time required for the object to travel the same distance.

To what extent to you think the discrepancies are explained by this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

This would probably only be a factor if the wooden plank was not consistently smooth along the width of the plank. If there was a bump in one section of the wood, then it may slow the object down if it hits it.

• Differences in positioning the object prior to release.

To what extent to you think the discrepancies are explained by this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

If the student has not marked off an exact spot for the object’s starting point, then this may cause a discrepancy in the timing. If they start it further back one time, then it will take the object more seconds to roll across the entire plank.

• Human uncertainty in observing exactly when the object reached the end of the incline.

To what extent to you think the discrepancies are explained by this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

Similar to the starting point, if the student doesn’t have a marker at the specific location he/she considers the end, then there may be differences in the times of the experiment.

*********************************************

Question: How much uncertainty do you think each of the following would actually contribute to the uncertainty in timing a number of trials for the ball-down-an-incline lab?

• The lack of precision of the TIMER program.

To what extent to you think this factor would contribute to the uncertainty?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

The timer program is probably very precise considering it is programmed by a computer. Therefore, it probably is a very small factor in any uncertainty.

• The uncertain precision of human triggering (uncertainty associated with an actual human finger on a computer mouse)

To what extent to you think this factor would contribute to the uncertainty?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

There is a lot of uncertainty in the precision of human triggering. Although a person can attempt to click at the exact same time each trial, it is physically impossible for someone to click the mouse at exactly the same time each go-round.

• Actual differences in the time required for the object to travel the same distance.

To what extent to you think this factor would contribute to the uncertainty?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

This would cause uncertainty as to the amount of time it actually takes for the object to travel along the path. If you timed the experiment several times and averaged the times together, you could come up with a fairly accurate time.

• Differences in positioning the object prior to release.

To what extent to you think this factor would contribute to the uncertainty?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

This could contribute to uncertainty only if the surface along the incline wasn’t even along the whole width. Otherwise, it wouldn’t make much difference if it was towards the left or the right of the plank. If it was further back or forward however, there would be uncertainty in the answers.

• Human uncertainty in observing exactly when the object reached the end of the incline.

To what extent to you think this factor would contribute to the uncertainty?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

This could contribute a lot to the uncertainty because the person may not remember the same spot to stop the timer each time, and it would cause differences in all of the recorded times.

*********************************************

Question: What, if anything, could you do about the uncertainty due to each of the following? Address each specifically.

• The lack of precision of the TIMER program.

What do you think you could do about the uncertainty due to this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

If I were a computer programming wiz, I could program the system to be more precise. Otherwise, the only thing that would give more precision is to perform an experiment with a greater time length in between the start and finish points.

• The uncertain precision of human triggering (uncertainty associated with an actual human finger on a computer mouse)

What do you think you could do about the uncertainty due to this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

This uncertainty could be improved by putting a censor at the end of the incline connected to the timer program. This would automatically stop the timer when the rolling object reached the ending point.

• Actual differences in the time required for the object to travel the same distance.

What do you think you could do about the uncertainty due to this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

This could be improved by guaranteeing as close to perfect of a surface as possible. In the case of a knot in a wooden plank used as an incline, it could be sanded down to improve the evenness of the surface.

• Differences in positioning the object prior to release.

What do you think you could do about the uncertainty due to this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

To improve this, the student could mark a spot on the surface that would be the exact starting point for each trial.

• Human uncertainty in observing exactly when the object reached the end of the incline.

What do you think you could do about the uncertainty due to this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

Like the previous solution, the student could put a mark on the end of the incline that would indicate the end of the trial each time.

*********************************************

Question: If, as in the object-down-an-incline experiment, you know the distance an object rolls down an incline and the time required, explain how you will use this information to find the object 's average speed on the incline.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Because the average amount of speed is the number of units of measurement moved per every single unit of time, we can solve for it using the given information. If we measure the distance in cm, we would find the number of cm the object moves per second.

Confidence Assessment: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: If an object travels 40 centimeters down an incline in 5 seconds then what is its average velocity on the incline? Explain how your answer is connected to your experience.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

To find the average velocity we divide cm by seconds:

40 cm/ 5 sec= 8 cm/sec

The average velocity is 8 cm/sec. The average velocity for my experience was about 13 cm/sec. This means it was moving faster than this experiment.

Confidence Assessment: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: If the same object requires 3 second to reach the halfway point, what is its average velocity on the first half of the incline and what is its average velocity on the second half?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

If the object travels 40 cm total, the halfway point will be at 20 cm. The average velocity of the first half is:

20 cm/ 3 sec= 6.67 cm/sec

If it travels 5 seconds total, the last 20 cm would be travelled in 5-3= 2 seconds. The average velocity is: 20 cm/2 sec= 10 cm/ sec

Confidence Assessment: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `qAccording to the results of your introductory pendulum experiment, do you think doubling the length of the pendulum will result in half the frequency (frequency can be thought of as the number of cycles per minute), more than half or less than half?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Doubling the length will result in less than half of the frequency. For example, when the length was doubled from 14 cm to 28 cm, the frequency decreased from 84 cycles to only 58 cycles.

Confidence Assessment: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `qNote that for a graph of y vs. x, a point on the x axis has y coordinate zero and a point on the y axis has x coordinate zero. In your own words explain why this is so.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

If a point is on the x- axis, that means it hasn’t moved up or down on the coordinate plane, meaning y=0. If a point is on the y-axis, it has not moved to the left or right of 0 making x=0. This is because the y is measured vertically and the x is measured horizontally.

Confidence Assessment: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `qOn a graph of frequency vs. pendulum length (where frequency is on the vertical axis and length on the horizontal), what would it mean for the graph to intersect the vertical axis (i.e., what would it mean, in terms of the pendulum and its behavior, if the line or curve representing frequency vs. length goes through the vertical axis)? What would this tell you about the length and frequency of the pendulum?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

If the line intersects the vertical axis, it would mean that the length of the pendulum is 0. If the graph followed its same pattern, then the frequency would be at its highest point. However, it is physically impossible to still have a pendulum when its length equals 0 cm.

Confidence Assessment: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `qOn a graph of frequency vs. pendulum length, what would it mean for the graph to intersect the horizontal axis (i.e., what would it mean, in terms of the pendulum and its behavior, if the line or curve representing frequency vs. length goes through the horizontal axis)? What would this tell you about the length and frequency of the pendulum?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

If the curve intersected the horizontal axis, then the frequency would equal 0 cycles per second. This would mean that the pendulum is no longer moving. If the curve followed its usual pattern, it would also mean that the pendulum is at its longest length.

Confidence Assessment: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `qIf a ball rolls down between two points with an average velocity of 6 cm / sec, and if it takes 5 sec between the points, then how far apart are the points?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

We can find the amount of centimeters by using the given information to cancel out the seconds unit:

We do this by multiplying the velocity by the time:

6 cm/sec* 5 sec= 30 cm (because the seconds cancel each other out)

Confidence Assessment: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`aOn the average the ball moves 6 centimeters every second, so in 5 seconds it will move 30 cm.

The formal calculation goes like this:

• We know that vAve = `ds / `dt, where vAve is ave velocity, `ds is displacement and `dt is the time interval.

• It follows by algebraic rearrangement that `ds = vAve * `dt.

• We are told that vAve = 6 cm / sec and `dt = 5 sec. It therefore follows that

• `ds = 6 cm / sec * 5 sec = 30 (cm / sec) * sec = 30 cm.

The details of the algebraic rearrangement are as follows:

• vAve = `ds / `dt. We multiply both sides of the equation by `dt:

• vAve * `dt = `ds / `dt * `dt. We simplify to obtain

• vAve * `dt = `ds, which we then write as{}`ds = vAve *`dt

Be sure to address anything you do not fully understand in your self-critique.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I can also use the formal equation: vAve=`ds/`dt to solve this calculation.

Confidence Assessment: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `qYou were asked to read the text and some of the problems at the end of the section. Tell your instructor about something in the text you understood up to a point but didn't understand fully. Explain what you did understand, and ask the best question you can about what you didn't understand.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

One thing I did understand in the next was how to estimate uncertainty when it is not given. It is assumed that the uncertainty is +- one unit of the furthest decimal place. For example, the decimal 1.43 would have an uncertainty of +- .01. It would fall in between 1.42 and 1.44. In the section covering percent error on pg. 7, I am confused about why dividing 97 by 92 implys an accuracy of 1%.

97 and 92 are both close to 100; the uncertainty in either measurement is +-1, which is very nearly 1% of either quantity.

Why is it best to report the answer with 3 significant figures of the accuracy percent?

The number of significant figures reported depends on the accuracy of the data. 3 significant figures is pretty much a default figure; if you aren't sure about the limits on the uncertainty, report to 3 significant figures.

Confidence Assessment: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `qTell your instructor about something in the problems you understand up to a point but don't fully understand. Explain what you did understand, and ask the best question you can about what you didn't understand.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I understand how to solve for a requested variable using given numbers to cancel out units. For example, if you have the speed in cm/sec and the time in sec, you can solve for the distance in cm by multiplying the two together and canceling out the seconds. One thing I am uncertain about is the equation vAve=`ds/`dt. Is it ok to view it as simply as multiplying to cancel out units or do I always need to view it in this equation?

If `ds is measured in meters and `dt in seconds, then `ds / `dt will be in meters / seconds.

You should always do the algebra of the units. You'll see plenty of explanation and examples in the upcoming assigments.

SOME COMMON QUESTIONS:

*********************************************

QUESTION: I didn’t understand how to calculate uncertainty for a number such as 1.34. When given examples we had problems such as 1.34 ±0.5 and with that we had a formula (0.5/1.34)*100. So I do not understand how to compute uncertainty when no estimated uncertainty is given.

INSTRUCTOR RESPONSE:

The +- number is the uncertainty in the measurement.

The percent uncertainty is the uncertainty, expressed as a percent of the number being observed.

So the question in this case is simply, 'what percent of 1.34 is 0.5?'.

• 0.5 / 1.34 = .037, approximately. So 0.5 is .037 of 1.34.

• .037 is the same as 3.7%.

I recommend understanding the principles of ratio, proportion and percent as opposed to using a formula. These principles are part of the standard school curriculum, though it does not appear that these concepts have been well mastered by the majority of students who have completed the curriculum. However most students who have the prerequisites for this course do fine with these ideas, after a little review. It will in the long run save you time to do so.

There are numerous Web resources available for understanding these concepts. You should check out these resources and let me know if you have questions.

Please feel free to include additional comments or questions:

"

&#Your work looks good. See my notes. Let me know if you have any questions. &#