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Lab Methods}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).

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
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 .Alternate Solution - Unverified
Comments
Memol
2012-09-18 07:17:56
Can anyone help me with an study reference about these graph stuff?Help
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 .Alternate Solution - Unverified
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. NEC
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.
tachyon788
2009-10-06 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)
NEC

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

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