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A train is traveling at 30.0 m/sm/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 HzHz. The speed of sound in air should be taken as 344 m/sm/s .

Respuesta :

frequency fapproach is heard by a passenger = 302.05 Hz

frequency frecede is heard by a passenger = 228.37433 Hz

Given :

speed of train to the ground = 30.0 m/s

frequency emitted by the train whistle = 262 Hz

speed of sound in air = 344 m/s

To Find :

(A) frequency fapproach (B) frequency frecede

Solution :

The frequency of approach is given by

f' = f v + v(relative)

v - v(air)

= 262x 344 + 18

344 - 30

f' = 302.05 Hz

The frequency of approach is 302.05 Hz

The frequency of recede is given by

f' = f v - v(relative)

v + v(air)

= 262 x 344 - 18

344 + 30

The frequency of recede is 228.37433 Hz

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