Extended Experimental Investigation.

In summary: No, I don't think so. I just wanted to show that I had the data and that I was looking for help. I don't know how to do the calculations and I don't know the units. In summary, Aaron is looking for help with a physics task that he has forgotten to measure his displacement. He has a graph and data for the average acceleration of the tape for each section. However, he is not sure on a few things about the task and is confused. If anyone could help him, that would be much appreciated.
  • #1
Arizzle
21
0
My teacher has given me a Extended Experimental Investigation to do that requires physics. I know a bit about it (the very basics) and I'm not sure on a few things about it. For my task I have forgotten to measure my displacement and I was wondering if anybody knew a way I could still find it. I'll write down what I've got so far (I'm using a ticker timer and my task is to investigate the effect of a load on displacement, velocity and acceleration on a trolley):

I have a graph showing my section of the tape, mass, time, length, Average acceleration of 5 intervals, Average acceleration for section and Average Acceleration for tape.

If anybody could help me, that would be very much appreciated.

Thanks,
Aaron.
 
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  • #2
The experimental data need to be analyzed since it is an experimental investigation. It seems you have got only the data for the average acceleration of the tape. Do you still have access to the tape itself?
 
  • #3
No, the tape is not on me at the moment, but I have the measurements of it written down if that helps. =)
 
  • #4
What measurements do you have?
 
  • #5
Tape 1.
Sections:
1)2cm
2)5.7 cm
3)9.8 cm
4)14.0 cm
5)17.4 cm
6)21.0 cm
7)19.4 cm

Tape 2.
Sections:
1)2.1 cm
2)6.1 cm
3)10.3 cm
4)14.4 cm
5)18.3 cm
6)19.8 cm
7)20.4 cm

Tape 3.
Sections:
1)2.1 cm
2)5.5 cm
3)9.9 cm
4)14 cm
5)14.2 cm
6)20 cm
7)19.8 cm

Tape 4.
Sections:
1)1.3 cm
2)5 cm
3)9.2 cm
4)13.7 cm
5)17.3 cm
6)21.1 cm
7)20.7 cm

Tape 5.
Sections:
1)2.5 cm
2)6.8 cm
3)11 cm
4)15.5 cm
5)19.6 cm
6)20.4 cm
7)19.4 cm

If it helps, I had the trolley run down a ramp but I'm also unsure of the angle degree, sorry. This is the first time I've ever done physics because I've missed half of this semmester because I was in hospital. I'm not lazy. =P
 
  • #6
So you kept the angle constant and recorded the same motion five times?
 
  • #7
Yep. But I was increasing the mass each time.

Tape 1: 0g
Tape 2: 250g
Tape 3: 500g
Tape 4: 750g
Tape 5: 1kg
 
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  • #8
I assume the the measurements above in each section is the distances between successive marks on the tape?
 
  • #9
Yeah I have divideded the dots into groups of 5 intervals and measured them in cm.
 
  • #10
So each distance (in cm) is the "length" of a group of five dots?
 
  • #11
That is correct.
 
  • #12
We seem to be getting somewhere. The average acceleration values that you've got are then for each such group of five dots? How/from where did you get those values?
 
  • #13
"The average acceleration values that you've got are then for each such group of five dots?" I don't mean do be annoying or anything, but what do you mean?

I don't think I actually mentioned this but I'm using a ticker timer. My ticker tape is roughly 1m long on each of the tapes.

I used the ticker time and got all the dots on my ticker tape, then I divided the dots on the tape into groups of 5 intervals each (giving me sections 1-7 on each tape).
 
  • #14
Well this is what you said in your first post:

Arizzle said:
My teacher has given me a Extended Experimental Investigation to do that requires physics. I know a bit about it (the very basics) and I'm not sure on a few things about it. For my task I have forgotten to measure my displacement and I was wondering if anybody knew a way I could still find it. I'll write down what I've got so far (I'm using a ticker timer and my task is to investigate the effect of a load on displacement, velocity and acceleration on a trolley):

I have a graph showing my section of the tape, mass, time, length, Average acceleration of 5 intervals, Average acceleration for section and Average Acceleration for tape.

If anybody could help me, that would be very much appreciated.

Thanks,
Aaron.
 
  • #15
I'm not really following, sorry. I'm just getting really confused now.

I'm not sure what you meant about the average acceleration and I'm also not sure as to why you quotes what I said.

I attatched my work so far if you want to have a look at it. I'm probably confusing you ALOT.
 

Attachments

  • Extended Experimental Investigation.doc
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  • #16
No, you are not confusing me, but your document could take quite a while to get approved before I can view it. I am not sure if you want to wait that long. I just thought that you somehow got the accelerations since you mentioned it in your first post. You do not know by any means what the length of each time interval of the ticker timer is? Usually it is the rate at which the local electric power are running - 50 or 60 hertz.
 
  • #17
The average acceleration for tape 1:
Sections:
1)200
2)370
3)410
4)420
5)340
6)360
7)-160

The average acceleration for the whole of tape 1 is 277.14
 
  • #18
So the question is again: How did you arrive at these numerical values and what are the units for the acceleration? Maybe show how one of these values were calculated.
 
  • #19
Well I have 0.1 seconds for time (cant remember how I got it but it was divided by something off memory). The length of my first section of the first tape is 2cm. From that I've gotten average acceleration of 200cm/s. I multiplied the 2cm by something to get the 200cm/s. I done it a while ago and reading over it, I have no clue how I've done it. =S

I think I went 0.1s x 10 = 1s. Then 2cm x 10 = 20cm/s. Then I think I've went x 10 again to get 200cm/s for my average acceleration.
 
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  • #20
I think I know how you got there...

Your power is running at 50 hertz. This means that the ticks are

[tex]\frac{1}{50} = 0.02\ seconds[/tex]

apart. So that elapsed time for five ticks should actually be

[tex]4 \times 0.02 = 0.08\ seconds[/tex]

but you took it as

[tex]5 \times 0.02 = 0.1\ seconds[/tex]

so you did for the acceleration

[tex]\frac{2}{0.1 \times 0.1} = 200 cm/s^2[/tex]
 
  • #21
I understand everything you've just written there. Yep that seems right to me.
 
  • #22
Okay, so here's what you could do now with your data...

Lets start with the average displacement first. Draw a graph of average displacement versus time. On this graph you are going to have five curves fot the five different masses. This will then indicate how the average dislacement depends on the loads.

How did you group your dots when you measure their length like this:

. . . . . . . . . . . . . . . . . . . .
----- ----- ----- -----

or did you continue the group from the last one in the previous group?
 
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  • #23
I'm trying to find the average displacement. Or do I have it already? =S

I measured it so that I had 5 spaces inbetween the dots. So every 6 dots I started a new section. I continued the next section from the ending section of the last if that makes any sense.
 
  • #24
Yes, that makes (good) sense. So for the no load (0 gram) curve the displacement coordinates for the first two points will be

[tex](0.1\ s,\ 2.0\ cm)[/tex]

and

[tex](0.2\ s,\ 7.7\ cm)[/tex]
 
  • #25
For the second section I've got 5.7cm>57cm/s>370cm/s
I went 57 - 20 = 37 x 10 = 370cm/s
 
  • #26
I assume you are constructing the points for the average velocity graph. This is done by calculating the gradient of the displacement graph using pairs of data points on the average displacement graph. The first average velocity is then 57 cm/s as you said. Its x-coordinate will then be 0.15 seconds. The second point will be 98 cm/s at 0.25 seconds ...
 
  • #27
The second point is 98 cm/s, yes, but how do you get the 1.15 seconds? How does this help tell me the displacement?
 
  • #28
The displacement graph tells you what the displacement of the trolley was at what time during the recording. When you determine the velocity it is the gradient of two successeive data points on the displacement graph that you use. This average velocity will be at the time in the middle of the two points that you used to determine the gradient with. So the 57 cm/s will be in between 0.1 and 0.2 seconds in the recording, that is it is the average velocity at 0.15 seconds in the recording.
 
  • #29
The displacement graph? I haven't done a displacement graph though. If I had a displacement graph, that would tell me what I want to know. I think. =S
 
  • #30
Do you understand how these two points were generated for such a displacement graph?

andrevdh said:
Yes, that makes (good) sense. So for the no load (0 gram) curve the displacement coordinates for the first two points will be

[tex](0.1\ s,\ 2.0\ cm)[/tex]

and

[tex](0.2\ s,\ 7.7\ cm)[/tex]
 
  • #31
No. I don't really have any clue what your talking about there, sorry.
 
  • #32
On order to draw the graph you need to generate the coordinates for the datapoints using the data. For the no load graph (0 gram) the displacement was 2.0 cm after 0.1 seconds (time for 5 intervals). Another 0.1 seconds later is was 5.7 centimeters further away, so the coordinates is 0.2 seconds and 5.7 + 2.0 cm = 7.7 cm
 
  • #33
So..
0.1 seconds = 2cm
0.2 seconds = 7.7cm (5.7cm + 2.0cm)
0.3 seconds = 17.5cm (9.8cm + 5.7cm + 2.0cm)
0.4 seconds = 31.5cm (14.0cm + 9.8cm + 5.7cm + 2.0cm)
0.5 seconds = 48.9cm (17.4cm + 14.0cm + 9.8cm + 5.7cm + 2.0cm)
0.6 seconds = 69.9cm (21.0cm + 17.4cm + 14.0cm + 9.8cm + 5.7cm + 2.0cm)
0.7 seconds = 89.3cm (19.4cm + 21.0cm + 17.4cm + 14.0cm + 9.8cm + 5.7cm + 2.0cm)

Does that look right?
 
  • #34
Correct. To generate the average velocity graph you would determine the gradient of this graph by using pairs of successive data points on the displacement graph. This means that the delta y's will be the additional distance added to get the displacement and the delta x will be 0.1 second each time. We say that the average velocity was in the middle of each of these intervals, so that would be at 0.15 seconds, 0.25 seconds. This means that the first two point will be

[tex](0.15\ s,\ 57\ cm/s)[/tex]

and

[tex](0.25\ s,\ 98\ cm/s)[/tex]
 
  • #35
I don't get why we start at 0.15 s and not at like 0.05 s. Plus, how did you get 57cm/s?

(Try to dumb things down ALOT hehe) I don't really know what half of this stuff is such as delta y's, gradient and delta x.
 

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