 GR 8677927796770177 | # 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 and respectively, then is

Lab Methods}Uncertainty

The general equation for uncertainty is given by , where and is the generalized standard deviation of quantity .

So, for this case, one has and thus its uncertainty is given by . (Basically, one has and .)

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 -> 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 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 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: petr1243
2008-03-09 18:13:35
Just a simple application of the general rules from error analysis:

For A = B +/- C :

=

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

=

Just treat as our

 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$produces . type this... to get...$\int_0^\infty\partial\Rightarrow\ddot{x},\dot{x}\sqrt{z}\langle my \rangle\left( abacadabra \right)_{me}\vec{E}\frac{a}{b}\$