## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

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. $f_{+}=(1+\frac{v_{o}}{v})f_{0}=(1+\frac{25\,m/s}{340\,m/s})400\,Hz=429\,Hz$