# Homework Help: Propagation of errors? (lab phenomenon question)

1. Oct 24, 2004

### StonieJ

propagation of errors? (lab "phenomenon" question)

In our lab, we are measuring the acceleration due to gravity. We do this by dropping an object through a photogate, which records position vs. time data as the object falls. From this position vs. time data, the computer uses derivatives to calculate velocity vs. time data. So, at the end of one trial run, we have a table with position vs. time data and another table with velocity vs. time data. Using this data, we have the computer fit a best-fit line to each set of data. The position vs. time data obviously has a quadratic model function, and the velocity vs. time data has a linear function. Therefore, in order to solve for acceleration due to gravity, we simply do one of the following:

1) After fitting a quadratic curve to position vs. time data, take the second derivative of the curve.

2) After fitting a linear curve to velocity vs. time data, take the first derivative of the curve.

We continue to get better results using method 2. That is, we get results closer to 9.79 m/s/s, which we are supposed to use as our ideal value. One of the lab questions asks if one method appears to be better and if so why. Our group discussed it some, but the only thing we could really come up with was that propagation of errors played a role in it somehow. Considering that the position data is the only data that is really measured by the equipment (i.e. the computer calculates velocity data based on the position data) I'm having a hard time understanding how the velocity data could actually be better. In other words, it seems like the velocity data is "second-hand" and should be worse. Anyway, I was just wondering if anyone could help explain why this is so. Thanks.

2. Oct 24, 2004

### ZapperZ

Staff Emeritus
Could you explain a little bit more on the details of your measurement, in particular how the photogate is being used. Do you have a photogate at some height, and as the object passes through it, it start time until it hits another photogate, or until it hits a pad on the ground? So in this case, your "distance" is actually the distance between the top photogate and whatever's on the bottom to turn it off.

Of maybe you have a different arrangement... Without knowing exactly how you set this up, it is not easy to figure out why certain ways are more accurate than others.

Zz.

3. Oct 24, 2004

### StonieJ

The photogate is used in conjunction with a "picket fence." A picket fence is a thin, rectangular piece of plastic with alternating opaque and clear bands of equal width. As this picket fence is dropped through the photogate, there will be times when an opaque band is blocking the light beam on the photogate, and this duration of time is recorded by the computer. Then, as the object falls further, a clear band takes its place, allowing the beam to pass through to the sensor. As soon as the clear band passes, another opaque band takes its place and again blocks the light beam for a certain amount of time.

Because the computer has been pre-programmed with the widths of the opaque and clear bands, it can use the it can use the amounts of time that the light beam was blocked (one for each opaque band on the picket fence) to calculate position vs. time.

4. Oct 24, 2004

### ZapperZ

Staff Emeritus
Ah, to make sure I get the picture correctly, does this mean that in one single drop, you get both the height vs. time curve, and also the velocity vs. time curve? And how sure are you that the vel. vs. time curve was obtained from the derivative of the height vs. time curve?

Zz.

5. Oct 24, 2004

### StonieJ

Correct. Dropping the picket fence through the photogate one time produces both position vs. time data and velocity vs. time data. The only difference is that the position vs. time data is measured directly by the photogate, and the velocity vs. time data is derived from the position data. Here is a little schematic that may help.

[PLAIN [Broken]][/URL]
https://webspace.utexas.edu/youngba2/www/Steps.gif [Broken]
[/url]

Last edited by a moderator: May 1, 2017
6. Oct 24, 2004

### ZapperZ

Staff Emeritus
Unfortunately, the link doesn't tell me how the computer actually obtained the vel. vs. time curve, i.e. did it get it directly from the "infinitesimal" change Delta(h)/Delta(t) of those pickets, or did it actually do a derivative of the height vs. time. If it does the latter, then it makes no sense that you got two different values of "g" at the very end. This is because if you take the first derivative of your height vs. time curve, then you should get the vel. vs. time curve that the computer computed.

However, if it did it directly, then it could end up with a slightly different values, depending on how wide the openings on the pickets are to accurately represent the "instantaneous" velocity.

This is a good problem to think of. I haven't come across such a problem before.

Zz.

Last edited by a moderator: May 1, 2017
7. Oct 24, 2004

### StonieJ

Interestingly enough, we are not told how exactly the computer computes the velocity. In fact, one of our questions is stated as such:

The computer records position as a function of time, but you found it also displays velocity and even acceleration information. A common method for measuring instantaneous velocity and acceleration given some position and time is to employ the first derivative approximation. The instantaneous velocity of a particle is given by the derivative of the position with respect to time. For small time intervals (Δt -> 0), this derivative can be approximated by,

v(ti) = dx(ti)/dt

= limΔt -> 0 (x(ti + Δt/2) - x(ti - Δt/2)) / Δt

v(ti) ≈ (x(ti+1) - x(ti-1)) / ti+1 - ti-1

Use the tables you printed out to determine if this is how your computer found the velocities. If not, what might be another possible method the computer used?

And then the next question is...

Which set of data (position vs. time or velocity vs. time) gave the best measurement of g? Why do you think this is?

Last edited: Oct 24, 2004
8. Oct 24, 2004

### ZapperZ

Staff Emeritus
Ah ha! This then would be a very good check. Do a first derivative of both your fit and a numerical derivative of your data directly (not sure if you are able to do the latter). Check if either of them corresponds to the vel. vs. time curve that the computer spew out. If neither of these are idential to that vel. vs time curve, then it is likely that the computer obtained this differently.

I'm guessing that the width of the opening between the pickets is Delta(x), and the time that this opening passes through the photodiode is Delta(t). The computer found the instantaneous velocity as Delta(x)/Delta(t). This increases as the object drops since Delta(t) gets progressively smaller.

Zz.

9. Oct 24, 2004

### StonieJ

Well, this is slightly interesting (but I think I just had a pretty major breakthrough in understanding how this computer got velocity). First of all, here is my data:

Code (Text):

Time            Position
0.7921          0.050
0.8208          0.100
0.8458          0.150
0.8682          0.200
0.8887          0.250
0.9077          0.300

Time            Velocity
0.7749          1.430
0.8065          1.742
0.8333          2.000
0.8570          2.232
0.8785          2.439
0.8982          2.632
[/indent]

Now, the model function for my position data is:
x(t) = 0.050 + 1.60t + 4.85t2

and the model function for my velocity data is:
v(t) = -6.10 + 9.72t

The first thing is that the derivative of the position function is not the velocity function. Specifically, the middle term of x(t) should equal the first term of v(t). (Correct?) However, this is the interesting part. If you solve for the velocities by applying the following method, this is what you get:

v(t) = x2 - x1 / t2 - t1

Starting with the second velocity...
v(t2) = (0.100 - 0.050) / (0.8208 - 0.7921) = 1.742

Which is exactly what the computer generated as the second velocity. Doing this for the rest gives the same results as the computer generated. However, I'm not sure how v(t1) was generated, since there is not a previous value for it to use in the equation. It's also interesting that the times for the velocity data are the midpoint times for the position data.

So, in conclusion, I believe that your statement in the last post was correct. Nevertheless, going back to the original problem of "which method is better and why?", I'm still not sure.​

10. Oct 25, 2004

### ZapperZ

Staff Emeritus
Great! So at least we solved that mystery. Now that we know what is going on, it gives us (at least me) a clearer picture on what to think of. I will have to get back to you on this one....

Zz.

11. Oct 27, 2004

### ZapperZ

Staff Emeritus
I did think about this one... but I don't think I can come up with anything intelligent to explain the discrepancy. It probably would help to actually get to see the setup, especially the pickets.

Did you ever find out the plausible explanation given for this? I am very much interesting in finding out. I'm a sucker for methodology issues such as this.....

Zz.