A person measures his or her heart rate by counting the number of beats in 30 s. If$\mathbf{40}\mathbf{\pm }\mathbf{1}\mathbf{}\mathbf{beats}$ are counted in$\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{s}$, what is the heart rate and its uncertainty in beats per minute?

Minute heart rate (heart rate), the number of heartbeats obtained within one minute.

Given Data

Consider the given data as below.

Counting the number of beats, t = 30 s

Number of beats, N = 40 beats

Uncertainty in beats,$\mathrm{\delta N}=1$

Uncertainty in time =$\mathrm{\delta t}=0.5\mathrm{s}$

The expression for the number of beats per minute is

$\mathrm{K}=\raisebox{1ex}{$\mathrm{N}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{t}$}\right.$

Substitute known values in the above equation.

$$\begin{gathered} \mathrm{K}=\frac{40\mathrm{beats}}{30\sec }\times \frac{60\sec }{1\min } \\ =80\mathrm{beats}/\min \end{gathered}$$

The maximum number of beats is

$$\begin{gathered} {\mathrm{N}}_{\max }=\mathrm{N}+\mathrm{\delta N} \\ =40+1 \\ =41\mathrm{beats} \end{gathered}$$

The minimum time is,-

$$\begin{gathered} {\mathrm{t}}_{\min }=\mathrm{t}-\mathrm{\delta t} \\ =30-0.5 \\ =29.5\sec \end{gathered}$$

The maximum number of beats per minute is,-

$$\begin{gathered} {\mathrm{K}}_{\max }=\frac{{\mathrm{N}}_{\max }}{{\mathrm{t}}_{\min }} \\ =\frac{41\mathrm{beats}}{29.5\sec }\times \frac{60\sec }{1\min } \\ =83.4\mathrm{beats}/\min \end{gathered}$$

The minimum number of beats is,-

$$\begin{gathered} {\mathrm{N}}_{\min }=\mathrm{N}-\mathrm{\delta N} \\ =40-1 \\ =39\mathrm{beats} \end{gathered}$$

The maximum time is,-

$$\begin{gathered} {\mathrm{t}}_{\max }=\mathrm{t}+\mathrm{\delta t} \\ =30+0.5 \\ =30.5\sec \end{gathered}$$

The minimum number of beats per minute is define as follows.

$$\begin{gathered} {\mathrm{K}}_{\min }=\frac{{\mathrm{N}}_{\max }}{{\mathrm{t}}_{\min }} \\ =\frac{39\mathrm{beats}}{30.5\sec }\times \frac{60\sec }{1\min } \\ =76.8\mathrm{beats}/\min \end{gathered}$$

The uncertainty beats per minute is defined as below.

$$\begin{gathered} {\mathrm{\delta K}}_1={\mathrm{K}}_{\max }-\mathrm{K} \\ =83.4-80 \\ =3.4\mathrm{beats}/\min \\ {\mathrm{\delta K}}_2=\mathrm{K}-{\mathrm{K}}_{\min } \\ =80-76.8 \\ =3.2\mathrm{beats}/\min \end{gathered}$$

The average uncertainty beats per minute is,-

$$\begin{gathered} {\mathrm{\delta K}}_{\mathrm{avg}}=\frac{{\mathrm{\delta K}}_1+{\mathrm{\delta K}}_2}{2} \\ =\frac{3.4+3.2}{2} \\ =3.3\mathrm{beats}/\min \\ \approx 3\mathrm{beats}/\min \end{gathered}$$

Therefore, the heart rate is $\mathrm{K}=80\mathrm{beats}/\min$and average uncertainty beats per min is ${\mathrm{\delta K}}_{\mathrm{avg}}=3\mathrm{beats}/\min$.