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GR9677 #27
Problem
 GREPhysics.NET Official Solution Alternate Solutions
This problem is still being typed.
Lab Methods$\Rightarrow$}Log Graphs

Log graphs are good for exponential-related phenomenon. Thus (A), (C), and (E) are appropriate, thus eliminated. The stopping potential has a linear relation to the frequency, and thus choice (B) is eliminated. The remaining choice is (D).

Alternate Solutions
 Izaac2012-08-21 01:18:44 One can also simply remember that Bode plots (gain VS $\log\omega$) are semilog ones, so obviously D is inappropriate .Reply to this comment
Memol
2012-09-18 07:17:56
Can anyone help me with an study reference about these graph stuff?
Izaac
2012-08-21 01:18:44
One can also simply remember that Bode plots (gain VS $\log\omega$) are semilog ones, so obviously D is inappropriate .
keenanman
2007-10-16 12:38:05
In choice D, the graph gain vs 1/frequency is linear. The graph gain vs frequency is hyperbolic.
eshaghoulian
2007-10-02 04:11:09
Just to add a little bit as to why log graphs are good for exponential related phenomena, note that a power law in log-log coordinates is a line:

$y=ax^m \Rightarrow log(y) = log(ax^m) = mlog(ax) = mlog(x)+ mlog(a) \Rightarrow log(y) = mlog(x) + mlog(a)$which is of the form $y = mx+b$ (since $mlog(a)$ is just a constant (like $b$) and we identify $y$ with $log(y)$ and $x$ with $log(x)$, as these are our new axes in log-log coordinates). So the exponent in the power law becomes the slope in log-log coordinates. Testing this is a GRE favorite, as it is a major tool in experimental physics.
 tachyon7882009-10-06 11:48:47 You have a small math error in your use of logs. The equation should be: $log(y)=log(ax^m)=log(a)+log(x^m)=log(a)+mlog(x)$

One can also simply remember that Bode plots (gain VS $\log\omega$) are semilog ones, so obviously D is inappropriate .
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