To solve this problem we will apply the concepts related to the Doppler Effect, defined as the change in apparent frequency of a wave produced by the relative movement of the source with respect to its observer. Mathematically it can be written as
[tex]f_{obs} = f(\frac{v_w}{v_w-v_s})[/tex]
Here,
[tex]f_s[/tex]= Frequency of the source
[tex]v_w[/tex] = Speed of the sound
[tex]v_s[/tex]= Speed of source
Now the velocity we have that
[tex]v_s = 116km/h (\frac{1000m}{1km})(\frac{1h}{3600s})[/tex]
[tex]v_s = 32.22m/s[/tex]
Then replacing our values,
[tex]f_{obs} = (950Hz) (\frac{345m/s}{345m/s-32.22m/s})[/tex]
[tex]f_{obs} = 1047.86Hz[/tex]
Therefore the frequency of the observer is 1047.86Hz
