Answer :
Answer:
The pitch that he hears after the truck passes and is moving away is 819.6 Hz.
Explanation:
The pitch that he hears after the truck passes and is moving away can be calculated using the following equation:
[tex] f = f_{0}*\frac{v_{s} \pm v_{o}}{v_{s} \pm v_{f}} [/tex]
Where:
[tex]f[/tex]: is the perceived frequency
[tex]f_{0}[/tex]: is the emitted frequency
[tex]v_{s}[/tex]: is the speed of sound = 340 m/s
[tex]v_{o}[/tex]: is the speed of the observer = 0 (he is not moving)
[tex]v_{f}[/tex]: is the speed of the fire truck
First, we need to find the speed of the fire truck. When it approaches the observer we have:
[tex] f = f_{0}*\frac{v_{s} + v_{o}}{v_{s} - v_{f}} [/tex]
[tex] 950 Hz = 880 Hz*\frac{340 m/s}{340 m/s - v_{f}} [/tex]
[tex] v_{f} = 340 m/s - \frac{880 Hz*340 m/s}{950 Hz} [/tex]
[tex] v_{f} = 25.05 m/s [/tex]
Hence, the speed of the fire truck is 25.05 m/s.
Now, we can calculate the pitch that the observer hears after the truck passes:
[tex] f = f_{0}*\frac{v_{s} - v_{o}}{v_{s} + v_{f}} [/tex]
[tex] f = 880 Hz*\frac{340 m/s}{340 m/s + 25.05 m/s} [/tex]
[tex] f = 819.6 Hz [/tex]
Therefore, the pitch that he hears after the truck passes and is moving away is 819.6 Hz.
I hope it helps you!