GREPhysics.NET
GR | #
<< Previous | Next >>
Login | Register
   
  GR9277 #48
Problem
GREPhysics.NET Official Solution    Alternate Solutions
\prob{48}
The magnitude of the force F on an object can be determined by measuring both the mass m of an object and the magnitude of its acceleration a, where F=ma. Assume that these measurements are uncorrelated and normally distributed. if the standard deviations of the measurements of the mass and acceleration are $\sigma_m$ and $\sigma_a$ respectively, then $\sigma_F/F$ is

  1. $(\frac{\sigma_m}{m})^2+(\frac{\sigma_a}{a})^2$
  2. $(\frac{\sigma_m}{m}+\frac{\sigma_a}{a})^{1/2}$
  3. $((\frac{\sigma_m}{m})^2+(\frac{\sigma_a}{a})^2)^{1/2}$
  4. $\frac{\sigma_m \sigma_a}{ma}$
  5. $\sigma_m/m + \sigma_a/a$
Lab Methods}Uncertainty

The general equation for uncertainty is given by \delta f/f=\sqrt{\sum_i (\frac{\delta x_i}{x_i})^2}, where f=\Pi_i x_i and \delta x_i is the generalized standard deviation of quantity x_i.

So, for this case, one has f=ma and thus its uncertainty is given by \delta f/f = \sqrt{(\frac{\delta m}{m})^2+(\frac{\delta a}{a})^2}. (Basically, one has x_1=m and x_2=a.)

This is choice (C).
See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
LF
2015-10-16 02:47:49		It\'s possible to guess the answer even without knowing anything about error analysis. \r\n\r\nThere should be nothing special about the equation F = ma, so a priori, there should be no restriction on m or a being negative. However, the standard deviation is always positive. With these two observations we eliminate A, B and E:\r\n\r\nA is eliminated because if m or a are negative, then F (=ma) is negative. However option A is always positive, so sigma_F/F = [something positive] -> sigma_F must be negative, something that\'s not possible.\r\n\r\nB is eliminated because if both a and m are negative, then we have a negative number within the square root and the standard deviation can\'t be imaginary.\r\n\r\nE is eliminated because if m is negative and really large, while a is positive and really small, then the second term dominates the first and we have a positive sum. But F (=ma) is again negative, so sigma_F is again negative, and that is impossible.\r\n\r\nThat leaves C and D. Now we can use another observation: if we can measure either m or a with infinite precision, then its standard deviation would be zero. However because of the other variable, we do not know F with infinite precision, and therefore sigma_F is not zero. Plugging in this condition eliminates D, leaving C as the only possible answer.Alternate Solution - Unverified
Comments
LF
2015-10-16 02:47:49		It\'s possible to guess the answer even without knowing anything about error analysis. \r\n\r\nThere should be nothing special about the equation F = ma, so a priori, there should be no restriction on m or a being negative. However, the standard deviation is always positive. With these two observations we eliminate A, B and E:\r\n\r\nA is eliminated because if m or a are negative, then F (=ma) is negative. However option A is always positive, so sigma_F/F = [something positive] -> sigma_F must be negative, something that\'s not possible.\r\n\r\nB is eliminated because if both a and m are negative, then we have a negative number within the square root and the standard deviation can\'t be imaginary.\r\n\r\nE is eliminated because if m is negative and really large, while a is positive and really small, then the second term dominates the first and we have a positive sum. But F (=ma) is again negative, so sigma_F is again negative, and that is impossible.\r\n\r\nThat leaves C and D. Now we can use another observation: if we can measure either m or a with infinite precision, then its standard deviation would be zero. However because of the other variable, we do not know F with infinite precision, and therefore sigma_F is not zero. Plugging in this condition eliminates D, leaving C as the only possible answer.
LF
2015-10-16 03:46:36		 Attempting to fix the broken code. It looks OK in the preview, hope it displays right as well.\r\n---\r\nIt\'s possible to guess the answer even without knowing anything about error analysis. \r\n\r\nThere should be nothing special about the equation F = ma, so a priori, there should be no restriction on m or a being negative. However, the standard deviation is always positive. With these two observations we eliminate A, B and E:\r\n\r\nA is eliminated because if m or a are negative, then F (=ma) is negative. However option A is always positive, so \\sigma_{F}/F = [something positive] -> \\sigma_{F} must be negative, something that\'s not possible.\r\n\r\nB is eliminated because if both a and m are negative, then we have a negative number within the square root and the standard deviation can\'t be imaginary.\r\n\r\nE is eliminated because if m is negative and really large, while a is positive and really small, then the second term dominates the first and we have a positive sum. But F (=ma) is again negative, so \\sigma_{F} is again negative, and that is impossible.\r\n\r\nThat leaves C and D. Now we can use another observation: if we can measure either m or a with infinite precision, then its standard deviation would be zero. However because of the other variable, we do not know F with infinite precision, and therefore \\sigma_{F} is not zero. Plugging in this condition eliminates D, leaving C as the only possible answer.
Alternate Solution - Unverified
MuffinSpawn
2009-09-24 18:57:56		If you don't remember the product rule, you can always derive it quickly using the general function rule:

\sigma f = \sqrt{\sum \left(\frac{\partial f}{\partial x_i} \sigma x_i\right)^2}		NEC
petr1243
2008-03-09 18:13:35		Just a simple application of the general rules from error analysis:

For A = B +/- C :

\delta A = \sqrt{(\delta B)^2 + (\delta C)^2}

For A = B*C or A = B/C:

\frac{\delta A}{A} = \sqrt{(\frac{\delta B }{B})^2 + (\frac{\delta A }{A})^2}

Just treat \sigma as our \delta

note
2008-08-21 23:12:47		 There's a typo in your last equation, I think you meant dC/C
NEC

Post A Comment!
You are replying to:
Just a simple application of the general rules from error analysis: For A = B +/- C : \delta A = \sqrt{(\delta B)^2 + (\delta C)^2} For A = B*C or A = B/C: \frac{\delta A}{A} = \sqrt{(\frac{\delta B }{B})^2 + (\frac{\delta A }{A})^2} Just treat \sigma as our \delta

Username:
Password:
Click here to register.
This comment is best classified as a: (mouseover)
 
Mouseover the respective type above for an explanation of each type.

Bare Basic LaTeX Rosetta Stone
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces .
type this... to get...
$\int_0^\infty$
$\partial$
$\Rightarrow$
$\ddot{x},\dot{x}$
$\sqrt{z}$
$\langle my \rangle$
$\left( abacadabra \right)_{me}$
$\vec{E}$
$\frac{a}{b}$
 
The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...

All-Solutions List | Latest Comments | Unanswered Questions (Cries for Help!) |::| About

© 2005-2007 by Yosun Chang, All rights reserved.

This site is not affiliated with ETS, nor is it officially endorsed (yet) by ETS. This site is the work of an individual named Yosun Chang. The solutions, sans user comments and ETS questions, are © 2005 by Yosun Chang except where noted. All rights reserved. / The comments appending particular solutions are due to the respective users. / Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 1997 and © 2001 by ETS.