Answer:
a) 30.82 m/s
b) 183.33 Hz
Step-by-step explanation:
a)
V = speed of the sound = 339 m/s
v = speed of the train = ?
f' = observed frequency by the observer = 220 Hz
f = actual frequency of the observer = 200 Hz
using the equation
![f' = (fV)/(V - v)](https://img.qammunity.org/2020/formulas/physics/college/3oa8mjltk4a0s0kardbjzk309h09ve71gt.png)
![220 = ((200)(339))/(339 - v)](https://img.qammunity.org/2020/formulas/physics/college/zrlhzu5y3exr2nfbz2h775ni8puquykpjk.png)
v = 30.82 m/s
b)
V = speed of the sound = 339 m/s
v = speed of the train = 30.82 m/s
f'' = observed frequency by the observer as train moves away = ?
f = actual frequency of the observer = 200 Hz
using the equation
![f'' = (fV)/(V + v)](https://img.qammunity.org/2020/formulas/physics/college/tv6nzg3azah0fyypssf1u80larok7117gt.png)
![f'' = ((200)(339))/(339 + 30.82)](https://img.qammunity.org/2020/formulas/physics/college/y8gv4hzl0tgfedr0nnn11jpvpzcw8chgoh.png)
f'' = 183.33 Hz