Answer:
a)[tex]f_1 = 2070.6 Hz[/tex]
b)[tex]f_2 = 1929.4 Hz[/tex]
Explanation:
Apparent frequency of the siren is given as
[tex]f = \frac{v_{rel}}{\lambda}[/tex]
as we know that the wavelength will remain the same as it is having originally
sowe know
[tex]\lambda = \frac{340}{2000}[/tex]
[tex]\lambda = 0.17 m[/tex]
a) Now when wind is blowing from source to official
so we have
[tex]v_{rel} = 340 + 12 = 352 m/s[/tex]
so we have
[tex]f_1 = \frac{352}{0.17}[/tex]
[tex]f_1 = 2070.6 Hz[/tex]
b) Now when wind is from official to source
so we have
[tex]v_{rel} = 340 - 12 = 328 m/s[/tex]
so we have
[tex]f_2 = \frac{328}{0.17}[/tex]
[tex]f_2 = 1929.4 Hz[/tex]
