g Consider a beach ball bobbing up and down in the center of a swimming pool. Imagine that the ball remains at the same position at the center of the pool. The illustration below shows several consecutive wave crests at one instant of time. Suppose that observer A is moving toward the source and observer B is moving away from the source. Discuss with your group: does A or B observe higher frequency water wave vibrations, or do they observe the same frequency vibrations

This is an example of relative movement between the waves and the observer, let's first examine the observer A, they tell us that it is approaching the wave source (ball).

For waves there is a relationship between their speed, wavelength and frequency

v = λ f

the observed frequency if the relative velocity is zero

f₀ = v /λ₀

in this case the observer is approaching the source therefore the velocity is

v ’= v₀ + u

where u is the velocity of the observer

as the emission source remains fixed the wavelength does not change

v’= λ₀ f’

f ’= v’ /λ₀

f ’= [tex]\frac{v+u}{\lambda_o }[/tex]

we substitute the value of the wavelength

λ₀ = v / fo

f’= [tex]\frac{v+u}{v} \ f_o[/tex]

we can see that the frequency increases

for the case of observer B moving away from the source the relative velocity is

v ’= v-u

by substituting in the equation

f ’= v’ /λ₀

f ’= [tex]\frac{v-u}{\lambda_o}[/tex]v-u / lam

f ’= [tex]\frac{v-u}{v} \ f_o[/tex]

in this case the frequency decreases

In conclusion:

For the observer approaching the source, the frequency increases

for the observer who moves away from the source the frequency decreases

bhoopendrasisodiya34 bhoopendrasisodiya34 · Answer 2 · 2022-03-25T12:08:20+01:00

The observer A who is moving towards the source observe higher frequency water wave vibrations than the observer B.

What is frequency?

Frequency of wave is the number of waves, which is passed thorough a particular point at a unit time. It can be given as,

[tex]f=\dfrac{v}{\lambda}[/tex]

Here, (λ) is the wavelength of the wave and (v) is the speed of the wave.

The beach ball bobbing up and down in the center of a swimming pool. The ball remains at the same position at the center of the pool.

The observer A is moving toward the source. The frequency received by A will be,

[tex]f'=f_o\left(\dfrac{v+u}{v}\right)[/tex]

Here, [tex]f_o[/tex] is the natural sound source, u is the velocity of the observer and v is the velocity of sound propagation.

The observer B is moving away from the source. The frequency received by A will be,

[tex]f"=f_o\left(\dfrac{v-u}{v}\right)[/tex]

In this case the frequency of the observer who is approaching to the source will be increases and the frequency of the observer who is leaving the source will be decreases.

Thus, the observer A who is moving towards the source observe higher frequency water wave vibrations than the observer B.

Learn more about the frequency here;

https://brainly.com/question/1436295