Answer:
a) When the train is approaching the frequency is 477.19 Hz and the wavelenght is 0.719 m
b) When the train is leaving the crossing the frequency is 367.152 Hz and the wavelenght is 0.934 m
Explanation:
Given data:
v = speed of the train = 44.7 m/s
V = speed of the sound = 343 m/s
f = frequency of sound = 415 Hz
a) If the train is approaching, the frequency is:
[tex]f_{ap} =f(\frac{V}{V-v} )=415(\frac{343}{343-44.7} )=477.19Hz[/tex]
The wavelength is:
[tex]w=\frac{V}{f_{ap} } =\frac{343}{477.19} =0.719m[/tex]
b) If the train leaves the crossing, the frequency is:
[tex]f_{le} =f(\frac{V}{V+v} )=415(\frac{343}{343+44.7} )=367.152Hz[/tex]
The wavelength is:
[tex]w=\frac{V}{f_{le} } =\frac{343}{367.152} =0.934m[/tex]
