Question

Two cars are moving towards each other and sound emitted by first car with real frequency of 3000 hertz is detected by a person in second with apparent frequency of 3400 Hertz what was the speed of cars

Answer 1

v ’= 21.44 m / s

Step-by-step explanation:

This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s

f ’= f (v + v₀) / (v-
`v_(s)`)

where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer

in this exercise both the source and the observer are moving, we will assume that both have the same speed,

v₀ = v_{s} = v ’

we substitute

f ’= f (v + v’) / (v - v ’)

f ’/ f (v-v’) = v + v ’

v (f ’/ f -1) = v’ (1 + f ’/ f)

v ’= (f’ / f-1) / (1 + f ’/ f) v

v ’= (f’-f) / (f + f’) v

let's calculate

v ’= (3400 -3000) / (3000 +3400) 343

v ’= 400/6400 343

v ’= 21.44 m / s