Answer:
[tex]f_o = 132.91 Hz[/tex]
Explanation:
Frequency of the train whistle which is approaching him is given as
[tex]f_1 = (\frac{v}{v - v_s}) f_o[/tex]
[tex]f_1 = (\frac{343}{343 - 8.40}) f_o[/tex]
now similarly the frequency of the train whistle which is moving away from him is given as
[tex]f_2 = (\frac{v}{v + v_s})f_o[/tex]
[tex]f_2 = (\frac{343}{343 + 8.40})f_o[/tex]
now we know that
[tex]f_{beat} = f_1 - f_2[/tex]
so we have
[tex]6.50 = (\frac{343}{343 - 8.40}) f_o - (\frac{343}{343 + 8.40}) f_o[/tex]
[tex]6.50 = 1.025 f_o - 0.976 f_o[/tex]
[tex]f_o = 132.91 Hz[/tex]