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?

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 ( ) and the observer is stopped (v₀ = 0),

In this configuration the equation that governs the doppler effect is

     f’= (v + - v₀) / (v- +  ) f

Where v₀ is the observer's velocity, the positive sign is for when it approaches the source and   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

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