Physics: If a person is 18 % shorter than average, what is the ratio of his walking pace (that is, the frequency f of his motion) to the walking pace of a person of average height? Assume that a person s leg swings like a pendulum and that the angular amplitude of everybody s stride is about the same. f(short)/f(avg)=?

If a person is 18 % shorter than average, what

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07.09.2021

We have that the ratio of his walking pace to the walking pace of a person of average height is

$\frac{V_2}{V_1}=1.10$

given the assumption and the calculation given below

From the question we are told that:

Consider a person 18\% shorter than average

Let average height of a person be 10m

Therefore

The height of an $18\%$ shorter man is mathematically given as

H=10*0.18

H=8.2m

Generally, the equation for velocity is mathematically given by

$v=\frac{1}{2\pi} \sqrt{{g}{l}}$

Where we have the Assumption that a person's leg swings like a pendulum and that the angular amplitude of everybody's stride is about the same

Therefore

$\frac{V_1}{V_2}=\frac{l_1}{l_2}$

$\frac{V_1}{V_2}={82}{100}$

$\frac{V_2}{V_1}=1.10$

In conclusion

The ratio of his walking pace (that is, the frequency 'f' of his motion) to the walking pace of a person of average height is

$\frac{V_2}{V_1}=1.10$

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