Question

A person measures his or her heart rate by counting the number of beats in 30s. If 40±1 beats are counted in 30.0±0.5s, what is the heart rate and its uncertainty in beats per minute?

Answer 1

Answer:

Heart rate=80 beats per minute

Uncertainty=3 beats per minute

Explanation:

We are given that a person measures heart beat by counting the number of beats in 30 s.

If number of beats=$40\pm 1$

Time=$30.0\pm 0.5$

We have to find the heart rate and its uncertainty in beats per minute

Maximum number of beats=40+1=41

Minimum time=30.0-0.5=29.5 s

Because when heart beat increases then time reduces.

We know that 1 minute =60 seconds

Therefore, time =$\frac{29.5}{60}=0.492 minutes$

Average number of beats=40

Average time=30 seconds=$\frac{30}{60}=0.5 minutes$

Heart rate =$\frac{number\;of\;beats}{time}$

Heart  rate =$\frac{40}{0.5}=80 beats per minute$

Maximum heart rate=$\frac{41}{0.492}=83.33 beats per minute$

When minimum number of beats=40-1=39

Maximum time=30.0+0.5=30.5 seconds

Because when heart beat slow then it takes more times

Time=$\frac{30.5}{60}=0.508 minutes$

Minimum heart rate=$\frac{39}{0.508}=76.77 beats per minute$

Uncertainty when heart rate is maximum

Uncertainty=83.33-80=3.33 beats per minute

Uncertainty when heart rate is minimum

Uncertainty=80-76.77=3.23 beats per minute

Average uncertainty=$\frac{3.33+3.23}{2}$

Average uncertainty=3.28 $\approx 3$beats per minute

Hence, heart rate=80 beats per minute

Uncertainty=3 beats per minute