Respuesta :
Explanation:
This is a good example of Doppler effect
Given
Source velocity Vs=20m/s
Apparent frequency Fa= 5.2 Hz
Recall speed of sound V=340m/s
The frequency of the horn Fh=?
But mathematically Fh=(V-Vs/V)*Fa
(340-20)/340*5.2=4.89Hz
Answer:
The frequency of the horn is 83.87 Hz
Explanation:
Given:
v = speed of the second horn = 20 m/s
vs = speed of the sound = 343 m/s
Δf = beat frequency = 5.2 Hz
The apparent frequency is equal to:
[tex]f_{ap} =f\frac{v_{s}-v_{o} }{v_{s}-v } \\f_{ap}=f\frac{343-0}{343-20} \\f_{ap}=1.062f[/tex]
The frequency of the horn is equal:
[tex]deltaf=f_{ap} -f\\deltaf=1.062f-f\\5.2=0.062f\\f=83.87Hz[/tex]
