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To measure her speed, a skydiver carries a buzzer emitting a steady tone at 1 800 Hz. A friend on the ground at the landing site directly below listens to the amplified sound he receives. Assume the air is calm and the speed of sound is independent of altitude. While the skydiver is falling at terminal speed, her friend on the ground receives waves of frequency 2 130 Hz. (Use 343 m/s as the speed of sound.)

a) What is the skydiver's speed of descent?
b) Suppose the skydiver can hear the sound of the buzzer reflected from the ground. What frequency does she receive?

Respuesta :

Answer:

Explanation:

The original frequency of sound being emitted f₀ = 1800

Its velocity towards the observer v ( let )

Apparent frequency f = 2130

velocity of sound = V

 [tex]f=f_0\times\frac{V}{(V - v)}[/tex]

Placing the given values

[tex]2130=1800\times\frac{343}{(343 - v)}[/tex]

1.1833 = [tex]\frac{343}{343 - v}[/tex]

1.1833 v = 62.87

v = 53.13 m /s .

b ) In the second case

formula for apparent frequency

[tex]f=f_0\times\frac{V+v}{(V - v)}[/tex]

Substituting the values

[tex]f=1800\times\frac{343+53}{(343 - 53)}[/tex]

= 2458 Hz .

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