Respuesta :
Answer:
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 ( [tex]v_{s}[/tex] ) and the observer is stopped (v₀ = 0),
In this configuration the equation that governs the doppler effect is
f’= (v + - v₀) / (v- + [tex]v_{s}[/tex]) f
Where v₀ is the observer's velocity, the positive sign is for when it approaches the source and [tex]v_{s}[/tex] 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
