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How fast, in meters per second, does an observer need to approach a stationary sound source in order to observe a 7.1 % increase in the emitted frequency?

Respuesta :

Answer:

v0 = 24.42 m/s (Approx)

Explanation:

Given:

Increase in frequency = 7.1% =

Computation:

Assume n = 100%

n1 = [(v+v0)/(v+v1)]n

[100 + 7.1] =  [(344+v0)/(344+0)]100

107.1 =   [(344+v0)/(344)]100

v0 = 24.42 m/s (Approx)

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