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\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).

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Comments
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

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