Question

The engine of a small and unmanned airplane produces a known and specific sound frequency (not given). A stationary sound detection device observes that the known emitted sound frequency is 90% of the perceived sound frequency. Relative to the detection device, in which direction and at what velocity is the plane moving

Answer 1

Answer:

the source (plane) moves away from the observer ,     v_{s} = 38.2 m / s

Explanation:

This is an exercise in the Doppler effect, which is the change in frequency due to the relative movement of the source and the observer; in this case the observer the sound detection system that is fixed and the source (plane) is mobile, in this case the relationship that describes the process is

            $f'= f ( \frac{v}{v \ \frac{-}{+} \ v_{s} })$

where negative sign is for the source moving towards the observer

It indicates that the frequency of the received sound is 90% of the emitted frequency

         f '/ f = 0.90

let's substitute

           0.9 = $\frac{v}{v \ \frac{-}{+} \ v_{s} }$

            $v \ \frac{-}{+} v_{s}$ = v / 0.9

           

The speed of sound is v = 344 m / s, let's substitute

           $344 \ \frac{-}{+} v_{s}$ = 344 / 0.9

           344 \ \frac{-}{+} v_{s} = 382.2

Let's analyze this expression, to fulfill the equation the speed of the source must be positive, therefore the source moves away from the observer

           344 +$v_{s}$= 382.2

           v_{s} = 382.2 -344

           v_{s} = 38.2 m / s