GREPhysics.NET: GR9677 #27

GR9677 #27

Problem

GREPhysics.NET Official Solution

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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

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 .

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Comments

Memol
2012-09-18 07:17:56

Can anyone help me with an study reference about these graph stuff?

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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 .

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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.

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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:

$log(y)=log(ax^m)=log(a)+log(x^m)=log(a)+mlog(x)$

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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.

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