Physics, 10.12.2019 05:31 ccarwile01

Apolice car sounding a siren with a frequency of 1650 hz is traveling at 120 km/h .

1)what frequencies does an observer standing next to the road hear as the car approaches and as it recedes?

2)what frequencies are heard in a car traveling at 90.0 km/h in the opposite direction before and after passing the police car?

3)the police car passes a car traveling in the same direction at 80.0 km/h. what two frequencies are heard in this car?

Answers: 2

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Answer from: lel11

1) 1829.3 and 1592.7 Hz, 2) 1963.8 and 1392.2 Hz, 3) 1948.9 Hz, 1404.5 Hz

Explanation:

This is a doppler effect exercise, the police car is the sound source with a speed ( $v_{s}$ ) and the observer is stopped (v₀ = 0),

In this configuration the equation that governs the doppler effect is

f’= (v + - v₀) / (v- + $v_{s}$ ) f

Where v₀ is the observer's velocity, the positive sign is for when it approaches the source and $v_{s}$ is the velocity of the source where the sign is for when it approaches the observer.

1) Let's calculate

The speed of sound in the air is v = 340 m / s

We reduce the speed of the car

vs = 120 km / h (1000m / 1 km) (1h / 3600s) = 33,333 m /s

Case 1. Police car approaching, still observer (v₀ = 0)

f’= (340 + 0) / (340 - 33.33) 1650

f’= 1.1087 1650

f’= 1829.3 Hz

Case 2. Police car moving away

f’= 340 / (340 + 33.33) 1650

f’= 0.9107 1650

f’= 1502.7 Hz

2) in this case both the source and the observed one move

Let's reduce the observer's car speed to SI units

v₀ = 90.0 km / h (1000m / 1 km) (1h / 3600s) = 25 m / s

Case 1. the cars approaching, we put the correct signs

f’= (340 + 25) / (340 - 33.33) 1650

f’= 1.19 1650

f’= 1963.8 Hz

Case 2 cars moving away

f’= (340 - 25) / (340 + 33.33) 1650

f’= 0.84376 1650

f’= 1392.2 Hz

3) in this case the two cars move, the observer's car (v₀)

v₀ = 80 km / h = 22.22 m / s

The cars initially approach as the police car goes faster, after a while the positive car passes to the other car and begins to move away

Case 1. Car approaching

f’= (340 + 22.22) / (340 - 33.33) 1650

f’= 1,181 1650

f’= 1948.9 Hz

Case 2. Cars moving away

f’= (340 - 22.22) / (340 + 33.33) 1650

f’= 0.8512 1650

f’= 1404.5 Hz

Answer from: Quest

d) physical means can be used to separate the components of air.

Answer from: Quest

it was given as a lesson and guide for prayer.

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