Answer :
Let at velocity v the frequency is 330 Hz.
At velocity v/2 the frequency is 320 Hz.
Using Doppler effect,
[tex]f_1=f_o\frac{v_s}{v_s-v}[/tex][tex]f_2=f_o\frac{v_s}{v_s-\frac{v}{2}}[/tex]Dividing the equations,
[tex]\begin{gathered} \frac{f_1}{f_2}=\frac{f_o(\frac{v_s}{v_s-v})}{f_o(\frac{v_s}{v_s-\frac{v}{2}_{}_{}})_{}_{}} \\ \Rightarrow v=v_s\frac{(\frac{f_1}{f_2}-1)}{(\frac{f_1}{f_2}-\frac{1}{2})} \end{gathered}[/tex]Putting the values we have,
[tex]\begin{gathered} v=\frac{20.17m}{s} \\ \frac{v}{2}=10.08\text{ m/s} \end{gathered}[/tex]Thus, the speed of the train before slowing down is 20.17 m/s
The speed of the train after slowing down is 10.08 m/s
