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3 years ago

Una ambulancia se aleja de una persona en línea recta a razón de 30 m/s. Si la sirena

Physics

1 answer:

saul85 [17]3 years ago

5 0

Answer:

$f_o=331.046Hz$

Explanation:

Use Doppler effect equation:

The Doppler effect is a physical phenomenon where an apparent change in wave frequency is presented by a sound source with respect to its observer when that same source is in motion. The general equation is given by:

$f_o=f_s\frac{v\pm v_o}{v\mp v_s} \\\\Where:\\\\f_s=Actual\hspace{3}frequency\hspace{3}of\hspace{3}the\hspace{3}sound\hspace{3}waves\\f_o=Observed\hspace{3}frequency\\v=Speed\hspace{3}of\hspace{3}the\hspace{3}sound\hspace{3}waves\\v_o=Velocity\hspace{3}of\hspace{3}the\hspace{3}observer\\v_s=Velocity\hspace{3}of\hspace{3}the\hspace{3}source$

When the observer moves towards the source $v_o$ is positive.

When the observer moves away from the source $v_o$ is negative.

When the source moves towards the observer $v_s$ is negative.

When the source moves away from the observer $v_s$ is positive.

Since the problem don't give us aditional information let's assume:

$v=343m/s$

Which is the speed of sound in air.

And using the information provided by the problem:

$v_s=30m/s\\v_o=0\\f_s=360Hz$

$f_o=f_s\frac{v}{v+v_s} =360*\frac{343}{343+30} =331.0455764\approx 331.046Hz$

The frequency perceived by the person is 331.046Hz

Translation:

Usa la ecuación del Efecto Doppler:

El efecto Doppler es un fenómeno físico en el que una fuente de sonido presenta un cambio aparente en la frecuencia de onda con respecto a su observador cuando esa misma fuente está en movimiento. La ecuación general viene dada por:

$f_o=f_s\frac{v\pm v_o}{v\mp v_s} \\\\Donde:\\\\f_s=Frecuencia\hspace{3}real\hspace{3}de\hspace{3}las\hspace{3}ondas\hspace{3}sonoras\\f_o=Frecuencia\hspace{3}observada(percibida)\\v=Velocidad\hspace{3}de\hspace{3}las\hspace{3}ondas\hspace{3}sonoras\\v_o=Velocidad\hspace{3}del\hspace{3}observador\\v_s=Velocidad\hspace{3}de\hspace{3}la\hspace{3}fuente$

Cuando el observador se mueve hacia la fuente $v_o$ es positivo.

Cuando el observador se aleja de la fuente es $v_o$ negativo.

Cuando la fuente se mueve hacia el observador $v_s$ es negativa.

Cuando la fuente se aleja del observador $v_s$ es positiva.

Como el problema no nos da información adicional, supongamos que:

$v=343m/s$

La cuál es la velocidad del sonido en el aire.

Y utilizando la información proporcionada por el problema:

$v_s=30m/s\\v_o=0\\f_s=360Hz$

$f_o=f_s\frac{v}{v+v_s} =360*\frac{343}{343+30} =331.0455764\approx 331.046Hz$

La frecuencia percibida por la persona es 331.046Hz

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