Q21PE - College Physics (Urone)

Question

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?

Step-by-Step Solution

Answer

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

Step 1: Calculating 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 = $\mathrm{\delta t}=0.5 \mathrm{s}$

Step 2: Defining beats per minute

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.

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

Step 3: Calculating minimum time

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,-

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

Step 4: Calculating maximum beats per minute

The maximum number of beats per minute is,-

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

Step 5: Calculating maximum time

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,-

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

Step 6: Calculating minimum beats per minute

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

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

Step 7: Defining uncertainty beats per minute

The uncertainty beats per minute is defined as below.

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

Step 8: Calculating average uncertainty

The average uncertainty beats per minute is,-

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

Step 9: Deriving conclusions

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