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

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

Step-by-step 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 ( (v)/(v \ (-)/(+) \ 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 =
`(v)/(v \ (-)/(+) \ v_(s) )`

`v \ (-)/(+) v_(s)` = v / 0.9

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

`344 \ (-)/(+) 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