A) 2.4 kHz
B) 0.24 kHz
C) 1.6 kHz
D) 1.2 kHz
Correct Answer: C
Solution :
The reflected sound appears to propagate in a direction opposite to that of moving engine. Thus, the source and the observer can be presumed to approach each other with same velocity. \[v'\frac{v\,(v+{{v}_{o}})}{(v-{{v}_{s}})}\] \[=v\left( \frac{v+{{v}_{s}}}{v-{{v}_{s}}} \right)\] \[[\because \,\,{{v}_{o}}={{v}_{s}}]\] \[\Rightarrow \] \[v'=1.2\left( \frac{350+50}{350-50} \right)\] \[=\frac{1.2\times 400}{300}\] = 1.6 kHz
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