|
GR8677 #39
|
|
|
|
|
Alternate Solutions |
ramparts
2009-10-30 15:26:13 | I *think* this is pretty much Yosun's solutions, but hers is a bit wordy and also poorly-TeXed, so here's my explanation:
A log-log graph is extremely useful because it turns a power law graph ( ) into a linear graph ( where y and x are the logs of the original y and x). This is a pretty common trick in experimental physics and observational astro.
So you notice the graph looks linear around the relevant point and rule out A in your head (don't even think about it). All you need to do is find the slope. I like the points Yosun used - the point you're given is at y=2 (called 10^2, but it's a log plot) and there's another at y=1 and x=6 (10^6). Now you estimate. It looks like the first point is roughly halfway between 5 and 6 on the x axis, so the slope is about 1/(1/2), or 2. Well, -2, because it's going down. So there you have it - E remains :) |  |
|
|
Comments |
ramparts
2009-10-30 15:26:13 | I *think* this is pretty much Yosun's solutions, but hers is a bit wordy and also poorly-TeXed, so here's my explanation:
A log-log graph is extremely useful because it turns a power law graph () into a linear graph ( where y and x are the logs of the original y and x). This is a pretty common trick in experimental physics and observational astro.
So you notice the graph looks linear around the relevant point and rule out A in your head (don't even think about it). All you need to do is find the slope. I like the points Yosun used - the point you're given is at y=2 (called 10^2, but it's a log plot) and there's another at y=1 and x=6 (10^6). Now you estimate. It looks like the first point is roughly halfway between 5 and 6 on the x axis, so the slope is about 1/(1/2), or 2. Well, -2, because it's going down. So there you have it - E remains :)
Albert
2009-11-01 04:29:00 |
ramparts wrote "I *think* this is pretty much Yosun's solutions, but hers is a bit wordy and also poorly-TeXed, so here's my explanation"
That's pure insult! And fella, how is your solution any better than hers? I didn't get anything new at all from you. Plus, yours is not even properly formatted and don't even get me started on 'poorly-TeXed'.
|
Prologue
2009-11-06 17:09:09 |
Although your solution could have done without the veiled insult it is definitely better because it gives background. I think it is important to realize that the person coming to this page is not sure of what the answer is, thus you need to spell it out for them, so that they are sure they understand it once they leave. After doing hundreds of solutions I am sure yosun gets tired and sees the easy way to do things and then just does them, but that is not the situation for the person visiting the site. They (I) need hand holding. So, give a complete background on the problem and everyone is happy.
|
| | PhyAnnie
2008-11-04 22:53:23 | Sorry, but you all seem wrong here. The graph itself is a log-log graph, i.e. it's ln(w).vs.ln(Gain). (you can easily tell by the non-uniform scalm of x-y axis). Having that in mind, we get : ln(Gain)=-2*ln(w), such the answer.
Cheers~
wittensdog
2009-07-25 19:42:57 |
I agree, but I also added in a non-zero y-intercept in the relation between the two logarithms. When you exponentiate both sides to find the Gain as a function of frequency, this is where the factor of K out front comes from. I found the value of negative two for the slope relating the two logarithms merely by eye-balling the graph. It's clearly negative, and it's too steep to be negative one. Also, I think A is a trap, because this plot clearly involves base 10 logarithms, not natural logarithms, so e shouldn't factor into the answer.
|
|  | sonnb
2008-06-07 11:16:48 | In your solution, I'm sure you meant ^n) instead of ^n) . Of course, the difference is insignificant. | | Furious
2007-08-31 09:26:41 | Hey, I'm probably confused. But I eliminated all choices but A by looking at the limiting behavior as  goes to zero. All choices including E, either go to zero or infinity as goes to zero, while A tapers out to a constant value. And I'm not sure, but that decay at the end doesn't look too bad for an exponential, on a log-log plot, does it?
D8less
2007-10-04 19:06:15 |
Yes, you are confused. The question just asks about how the function behaves at just one point. You are looking at the whole function but you should be thinking locally.
|
Richard
2007-11-02 18:22:34 |
I did the same thing...
I don't understand. How can it be anything but (A)?
...Unless of course, it blows up where you can't see it.
Then again, it just asks for which is most accurate near  .
|
naseermk
2008-09-22 17:02:08 |
If you apply the limiting condition of w approaching infinity, only D /E results in a gain that approaches zero as we see in the plot.
|
naseermk
2008-09-22 17:02:32 |
If you apply the limiting condition of w approaching infinity, only D /E results in a gain that approaches zero as we see in the plot.
|
| |
|
| Post A Comment! |
|
|
Bare Basic LaTeX Rosetta Stone
|
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces  .
|
| type this... |
to get... |
| $\int_0^\infty$ |
 |
| $\partial$ |
 |
| $\Rightarrow$ |
 |
| $\ddot{x},\dot{x}$ |
 |
| $\sqrt{z}$ |
 |
| $\langle my \rangle$ |
 |
| $\left( abacadabra \right)_{me}$ |
_{me}) |
| $\vec{E}$ |
 |
| $\frac{a}{b}$ |
 |
|
|
|
|
|
