Answer
a) 432 Hz
b) 429 Hz
Work Step by Step
Given: $f_{0}=400\,Hz,\,v_{s}=25\,m/s$ and $v=340\,m/s$.
a) Recall that when the source is approaching,
$f_{+}=\frac{f_{0}}{1-\frac{v_{s}}{v}}$
Therefore the frequency we hear is,
$f_{+}=\frac{400\,Hz}{1-\frac{25\,m/s}{340\,m/s}}=432\,Hz$
b) In this case, the observer is approaching the source.
$f_{+}=(1+\frac{v_{o}}{v})f_{0}=(1+\frac{25\,m/s}{340\,m/s})400\,Hz=429\,Hz$
