Question

A fire truck emits an 880Hz siren. As the truck approaches an observer on the sidewalk, he perceives the pitch to be 950Hz. Approximately what pitch does he hear after the truck passes and is moving away?

Answer 1

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:    

$f = f_{0}*\frac{v_{s} \pm v_{o}}{v_{s} \pm v_{f}}$

Where:

$f$: is the perceived frequency

$f_{0}$: is the emitted frequency

$v_{s}$: is the speed of sound = 340 m/s

$v_{o}$: is the speed of the observer = 0 (he is not moving)

$v_{f}$: 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:

$f = f_{0}*\frac{v_{s} + v_{o}}{v_{s} - v_{f}}$

$950 Hz = 880 Hz*\frac{340 m/s}{340 m/s - v_{f}}$

$v_{f} = 340 m/s - \frac{880 Hz*340 m/s}{950 Hz}$

$v_{f} = 25.05 m/s$

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:

$f = f_{0}*\frac{v_{s} - v_{o}}{v_{s} + v_{f}}$

$f = 880 Hz*\frac{340 m/s}{340 m/s + 25.05 m/s}$

$f = 819.6 Hz$

Therefore, the pitch that he hears after the truck passes and is moving away is 819.6 Hz.

I hope it helps you!