Problem 1-21: Counting heart rate with uncertainties in number of beats and time

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?

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SOLUTION:

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

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

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

\begin{align*} \text{\%\:uncertainty} & =\left( \frac{1\:\text{beat}}{40\:\text{beats}}\times 100\% \right)+ \left(\frac{0.5\:\text{s}}{30.0\:\text{s}}\times 100\% \right)\\ &=2.5\%+1.7\% \\ & =4.2\% \\ \end{align*}

Based on this percent uncertainty, we compute for the tolerance

\begin{align*} \delta _A & =\frac{\text{\%\:uncertainty}}{100\:\%}\times A \\ & = \frac{4.2 \%}{100 \%} \times 80 \ \text{beats/min} \\ & =3.4\:\text{beats/min}\\ \end{align*}

Therefore, the heart rate is

\displaystyle 80\pm 3\:\text{beats/min} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)