GREPhysics.NET: GR8677 #39

GR8677 #39

Problem

GREPhysics.NET Official Solution

Lab Methods $\Rightarrow$ }Log-Log Graph

The key phrase is most accurately expressed. Choice (E) works because it applies decently well to the following two key points, one of them given, and the other one right next to the given: $(\omega=3 E 5,1E2)$ and $(1E6,1E1)$

Try a power-law relation: $Gain=K\omega^n\Rightarrow$ $1E2=K(3E5)^n$ and $1E1=K(1E6)^n$ . Divide the two equations to get $1E1=(3E-1)^n Rightarrow 10=(1/3)^n$ . The closest answer would be $n=-2$ , which is case (E).

(Can't be an exponential relation, since the Gain would decrease exponentially as $\omega$ increases. It's not decaying nearly as fast.)

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Comments

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~

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sonnb
2008-06-07 11:16:48

In your solution, I'm sure you meant $(3/10)^n$ instead of $(1/3)^n$ . Of course, the difference is insignificant.

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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 $\omega$ goes to zero. All choices including E, either go to zero or infinity as $\omega$ 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 $\omega =3\times 10^5$ .

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.

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