To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,
[tex]f = f_0 (\frac{v_0}{v_0-v})[/tex]
Here,
[tex]f_0[/tex] = Frequency of Source
[tex]v_s[/tex] = Speed of sound
f = Frequency heard before slowing down
f' = Frequency heard after slowing down
v = Speed of the train before slowing down
So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,
The first equation is,
[tex]f = f_0 (\frac{v_0}{v_0-v})[/tex]
[tex]300 = f_0 (\frac{343}{343-v})[/tex]
[tex](300*343) - 300v = 343f_0[/tex]
Now the second expression will be,
[tex]f' = f_0 (\frac{v_0}{v_0-v/2})[/tex]
[tex]290 = (343)(\frac{v_0}{343-v/2})[/tex]
[tex]290*343-145v = 343f_0[/tex]
Dividing the two expression we have,
[tex]\frac{(300*343) - 300v}{290*343-145v} = 1[/tex]
Solving for v, we have,
[tex]v = 22.12m/s[/tex]
Therefore the speed of the train before and after slowing down is 22.12m/s
