College Physics Problem 1.21


A person measures his or her heart rate by counting the number of beats in 30 s. If 40±1  beats are counted in 30±0.5 s, what is the heart rate and its uncertainty in beats per minute?


Solution:

In order to compute for the heart rate in beats per minute, we need to solve for the base. The base is

\displaystyle \frac{40\:beats}{30\:sec\:}\times \frac{60\:sec}{1\:min}=80\:beats/min

Then we compute for the percent uncertainty by combining the uncertainties of the number of beats and time. That is

\displaystyle \%\:uncertainty=\frac{1\:beat}{40\:beats}\times 100\%+\frac{0.5\:s}{30.0\:s}\times 100\%

\displaystyle \%\:uncertainty=2.5\%+1.7\%

\displaystyle \%\:uncertainty=4.2\%

Based on this percent uncertainty, we compute for the tolerance

\displaystyle \delta _A=\frac{\%\:uncertainty}{100\:\%}\times A

\displaystyle \delta _A=3.3\:beats/min

Therefore, the heart rate is

\displaystyle 80\pm 3\:beats/min          ☚