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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$\Rightarrow$}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).

Alternate Solutions
 LF2015-10-16 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.Reply to this comment
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.
 LF2015-10-16 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.
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:

$\sigma f = \sqrt{\sum \left(\frac{\partial f}{\partial x_i} \sigma x_i\right)^2}$
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 $\delta$

 note2008-08-21 23:12:47 There's a typo in your last equation, I think you meant dC/C

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$