A police car moving at 75m/s is initially behind a truck moving in the same direction at 14m/s. The natural frequency of the car's siren is 1000Hz.
What is the change in frequency observed by the truck driver as the car overtakes him?
To find the change in frequency observed by the truck driver as the car overtakes him, we can use the Doppler effect formula:
Δf = [(v + vr) / (v + vs)] * f
where:
- Δf is the change in frequency observed
- v is the speed of sound
- vr is the velocity of the receiver (truck)
- vs is the velocity of the source (car)
- f is the natural frequency of the car's siren
First, let's convert the speeds from m/s to Hz using the following conversion:
1 m/s = 343 Hz (speed of sound)
So, the speed of the car (vs) is:
vs = 75 m/s * 343 Hz/m/s = 25725 Hz
And the speed of the truck (vr) is:
vr = 14 m/s * 343 Hz/m/s = 4802 Hz
Now we can calculate the change in frequency observed by the truck driver:
Δf = [(v + vr) / (v + vs)] * f
= [(343 Hz + 4802 Hz) / (343 Hz + 25725 Hz)] * 1000 Hz
= (5145 Hz / 26068 Hz) * 1000 Hz
≈ 197.69 Hz
Therefore, the change in frequency observed by the truck driver as the car overtakes him is approximately 197.69 Hz.
Fr = (Vs+Vt)/(Vs-Vc)*Fc =
(443+14)/(443-75) * 1000 = 457/368 * 1000 = 1242 Hz. = Freq.
heard by the receiver(Truck driver).
Correction: Vs = 343 m/s.
(343+14)/(343-75) * 1000 = 357/268 * 1000 = 1332 Hz.