Answer:
a) 877.76 Hz
b) 734.88 Hz
Explanation:
This kind of exercise is about explaining the Doppler efect.
The doppler effect is the change in frequency of a wave in relation to an observer who is moving relative to the wave source. In this case, a person watching an ambulance.
This shifting can be calculated using the following expression:
f' = (V - Vo / V - Vs) * f
Where:
V: speed of the sound in the medium (345 m/s in this case)
Vo: speed of the observer, in this case 0, cause the person is standing still.
Vs: speed of the source (in this case 110 km/h)
Let's convert the Vs from km/h to m/s
Vs = 110 * 1000 m/km * 1h/3600s = 30.56 m/s
Now that we have the data, let's calculate the frequency before the ambulance pass, and then, after.
a) using the above expression, we have the following:
f' = (345 - 0 / 345 - 30.56) * 800
f' = (1.0972) * 800
f' = 877.76 Hz
b) In this part, we'll use the same expression, with the difference that, as the ambulance has passed the person, the sign of the Vs change. This is because before, the ambulance was behind the person, and now it's in front passing. So, the frequency here is:
f = (345 - 0 / 345 + 30.56) * 800
f = 0.9186 * 800
f' = 734.88 Hz
