GR9277 #17



Alternate Solutions 
behnamn90 20131014 17:09:46  The graph on the oscilloscope is a parametric plot with the parameter being angular frequency ranging from zero to 2pi. This is called a "Lissajous figure" on an oscilloscope used in XY mode. The question asks which one is possible? If we have y=cos(2w) and x=cos(w) you get (A). If you have y=sin(2w) and x=sin(w) you get (E) rotated 90 degrees so (E) isn't possible. Go on:
http://www.wolframalpha.com/input/?i=parametric+plot+y%3Dcos%282t%29+and+x%3Dcos%28t%29
The correct one is plotted in the link.   kstephe6 20121104 18:37:52  NO MATH HERE!!
This is old skool Electrical Engineer knowledge.
We have a 1:2 frequency ratio. This means the pattern has ONE "hump" in it.
**Extra info below****
(google phase shift chart to get a better idea of what I mean here)
All patterns for the 1:2 ratio look like a 3D wire frame of a Pringle's potato chip or horse saddle rotated about the vertical. (Again, ONE HUMP)
The answer here happens to be the simplest corresponding to a phase shift of 45 degrees which is a side view of the Pringle's potato chip where only the front have is visible and the back half of the chip is eclipsed.
Again, google the chart. You'll see.rnrn   nyuko 20091104 07:10:44  To solve this problem, all I did was to follow the trace and imagine a projection on X and Y. Then you see in (A), Y completes a round trip while X completes half a round trip. So the frequency in Y is twice that of X. Done!
(Sorry, I meant to label this as "user solution" yet didn't know it would be reset after preview)   jburkart 20071101 00:20:57  If we first assume , , then we can solve: , so , which then reduces to . This is not any of the figures listed, because this problem is extremely unreasonable. However, if we assume instead that and , then it resolves into , choice (A).   senatez 20061020 09:56:06  I think it is best to think of it as a vector function which traces out a curve:
f(t) = sin(t)i + sin(2t)j
This can immediately elliminate D and E which would be a vector of the form of an ellipse:
f(t) = cos(t)i + sin(t)j   ytsejam 20061017 11:16:56  The explanation given above is wrong. The oscilloscope is in XY mode. So the answer has to be a Lissajous figure. All the options are
Lissajous figures for various frequency ratios.
A) omega_Y = 2* omega_X
B)omega_Y = 4* omega_X
C)omega_Y = 3* omega_X
D)omega_Y = omega_X
E)omega_Y = 1/2 *omega_X  

Comments 
g7d 20141020 18:56:31  This demonstration should make things clearer. http://demonstrations.wolfram.com/VirtualOscilloscope/   salticus 20141013 11:19:44  It's a mistake to assume both oscillators are in phase.
What says the oscillators were "switched on" at the same time?
In general, not just .
Using :
produces an infinity shape,
produces a parabola opening upward or a "U" shape,
produces that infinity shape again,
and gives a parabola opening downward or a "frown."
Instead of calculating all that out, I like kstephe6 or nyuko approached it.
kstephe6: imagine the outline of a saddle (i.e. pringles potato chip shape) rotating around the zaxis.
nyuko: check the ratios, and pick the one that fits.
(see their comments for details.)   behnamn90 20131014 17:09:46  The graph on the oscilloscope is a parametric plot with the parameter being angular frequency ranging from zero to 2pi. This is called a "Lissajous figure" on an oscilloscope used in XY mode. The question asks which one is possible? If we have y=cos(2w) and x=cos(w) you get (A). If you have y=sin(2w) and x=sin(w) you get (E) rotated 90 degrees so (E) isn't possible. Go on:
http://www.wolframalpha.com/input/?i=parametric+plot+y%3Dcos%282t%29+and+x%3Dcos%28t%29
The correct one is plotted in the link.   kstephe6 20121104 18:37:52  NO MATH HERE!!
This is old skool Electrical Engineer knowledge.
We have a 1:2 frequency ratio. This means the pattern has ONE "hump" in it.
**Extra info below****
(google phase shift chart to get a better idea of what I mean here)
All patterns for the 1:2 ratio look like a 3D wire frame of a Pringle's potato chip or horse saddle rotated about the vertical. (Again, ONE HUMP)
The answer here happens to be the simplest corresponding to a phase shift of 45 degrees which is a side view of the Pringle's potato chip where only the front have is visible and the back half of the chip is eclipsed.
Again, google the chart. You'll see.rnrn
kstephe6 20121105 05:06:59 
In fact, E could be answer IF it were rotated 90 degrees in the plane of the page

memyselmineni 20141119 10:17:46 
I agree! No maths at all. Only frequency ratio is needed! Get help from this book, it nails the physics GRE! http://www.amazon.com/PhysicsMathematicaJudeNdubuisiOnicha/dp/1499691920/ref=sr_1_fkmr1_1?s=books&ie=UTF8&qid=1401627044&sr=11fkmr1&keywords=physics+mathematica++2nd+edition+by+onicha+jude

  alanalan 20120911 16:11:22  Something is terribly wrong here! If answer E is correct when X is twice the frequency of Y, then it should be the same but rotated by 90 degrees when Y is twice the frequency of X. You are just swapping the axis when the frequencies are swapped.
That means none of the answers are correct!   niux 20091106 10:39:39  ETS were nonsense in this answer. The condition that spot is initially at the center of the screen: IT IS THIS INITIAL CONDITION, that inmediatly discard (A), (B) and (D). NO MATTER WHAT THE RATIO OF FREQUENCY IS.
Then answer should be (E)
niux 20091106 10:45:36 
My last line is wrong, answer shuould not be (E) beacuse ratio of frequency is 1:2 (y:x) and not (2:1).
Anyway, as neither (C) satisfies this condition.
I Repeat: ETS were non sense in this one

neon37 20101110 11:36:45 
have you never used an oscilloscope? It refreshes everyone once in a while depending on what you set. The cursor being at the center means that the oscilloscope is calibrated or rather adjusted in such a way. This shapes that appear are the oscillations about the center and not spatially shifted.

  nyuko 20091104 07:10:44  To solve this problem, all I did was to follow the trace and imagine a projection on X and Y. Then you see in (A), Y completes a round trip while X completes half a round trip. So the frequency in Y is twice that of X. Done!
(Sorry, I meant to label this as "user solution" yet didn't know it would be reset after preview)   nyuko 20091104 07:09:36  To solve this problem, all I did was to follow the trace and imagine a projection on X and Y. Then you see in (A), Y completes a round trip while X completes half a round trip. So the frequency in Y is twice that of X. Done!   nyuko 20091030 09:50:21  I never remember a single expression for this kind of problem. I even didn't know the trace has a fancy name.
To solve this problem, all I did was to follow the trace and imagine a projection on X and Y. Then you see in (A), Y completes a round trip while X completes half a round trip. So the frequency in Y is twice that of X. Done!
  chrisfizzix 20081006 11:19:20  This is a totally dick question.
Just to be constructive about it and all.
GREmania 20081012 06:08:07 
yeah, it is. Haha

  jburkart 20071101 00:20:57  If we first assume , , then we can solve: , so , which then reduces to . This is not any of the figures listed, because this problem is extremely unreasonable. However, if we assume instead that and , then it resolves into , choice (A).   jburkart 20071101 00:20:39  If we first assume , , then we can solve: , so , which then reduces to . This is not any of the figures listed, because this problem is extremely unreasonable. However, if we assume instead that and , then it resolves into , choice (A).   dcb 20070916 14:51:46  While I initially thought it was E (a Lissajous curve), writing a simple example sheds some light.
Lets take a time interval from 0 to 8 (the scale is arbitrary for an easy integer example).
 In this interval, X will go from 0 to 4 and back to 0 (one oscillation).
 Y, with twice the frequency, will go from 0 to 4 to 0 to 4 to 0.
Simply, (using a linear example):
T X Y
0 0 0
1 1 2
2 2 4
3 3 2
4 4 0 <(at this point, X has gone through half a period,
5 3 2 Y has gone through a full period)
6 2 4
7 1 2
8 0 0
Plot X vs Y, It a simple triangular looking hump, similar to A. It is linear because I made the input waves linear for simplicity. However, it does not have to be.
danksci 20080908 14:29:09 
According to the question shouldnt 0 strart at the center of the screen?

taatliberty 20111011 14:54:51 
Yeah. Plus, Y should go negative as it continues through the oscillation.

  kwooley 20070406 19:08:43  Here is a link to a nice aplet that should help answer questions about this problem
http://ibiblio.org/enotes/Lis/Lissa.htm   hamood 20070401 16:23:46  The way I went about this was by just doing a rough plot
y = 1,0,1,0,1,0,1,0,1 (period 0f 4)
x=4,3,2,1,0,1,2,3,4 (period of 8)
The plotted graph looks likes A....   SonOfOle 20061102 21:27:32  x123. Check the link you supplied again. They give a Lissajous curve that looks exactly like (A) on that page where the ratio of the frequencies is 1:2.   senatez 20061020 09:56:06  I think it is best to think of it as a vector function which traces out a curve:
f(t) = sin(t)i + sin(2t)j
This can immediately elliminate D and E which would be a vector of the form of an ellipse:
f(t) = cos(t)i + sin(t)j   senatez 20061020 09:52:29    ytsejam 20061017 11:16:56  The explanation given above is wrong. The oscilloscope is in XY mode. So the answer has to be a Lissajous figure. All the options are
Lissajous figures for various frequency ratios.
A) omega_Y = 2* omega_X
B)omega_Y = 4* omega_X
C)omega_Y = 3* omega_X
D)omega_Y = omega_X
E)omega_Y = 1/2 *omega_X
D8less 20071002 17:30:06 
Thank you, I was waiting for someone to say this.

Poop Loops 20081005 13:41:47 
This is by far the easiest way to look at it.

kroner 20091104 23:35:31 
To be a nitpicky, (B) is actually just a nonsensical image. Notice at both extremes of X, Y seems to turn around at zero rather than reaching 1.

crossno 20101112 18:03:01 
I agree, this is the correct way to do it. except both B and C are omega_y=3*omega_x.....the only difference is the phase
the answer is A

  x123 20061017 08:56:46  I think the answer is (E), because it's a Lissajous curve.
http://en.wikipedia.org/wiki/Lissajous_curve
europa 20061101 22:11:01 
ETS says it is A but only 10 percent scored this answer correctly so go figure...

ducksquark 20071102 05:28:41 
It should be a Lissajous figure all right. But (E) shows the figure for the wrong freq. ratio. (X being twice the freq of Y  see wikipedia article  the example figure given there is (E) turned on is side.)

taatliberty 20111011 15:01:25 
Europe, where did you get the answer list?

 

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While I initially thought it was E (a Lissajous curve), writing a simple example sheds some light.
Lets take a time interval from 0 to 8 (the scale is arbitrary for an easy integer example).
 In this interval, X will go from 0 to 4 and back to 0 (one oscillation).
 Y, with twice the frequency, will go from 0 to 4 to 0 to 4 to 0.
Simply, (using a linear example):
T X Y
0 0 0
1 1 2
2 2 4
3 3 2
4 4 0 <(at this point, X has gone through half a period,
5 3 2 Y has gone through a full period)
6 2 4
7 1 2
8 0 0
Plot X vs Y, It a simple triangular looking hump, similar to A. It is linear because I made the input waves linear for simplicity. However, it does not have to be.

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