An engine moving towards a wall with a velocity 50 ~ms ⁻¹ emits a note of 1.2 kHz . Speed of sound in air is 350 ~ms ⁻¹ . The frequency of the note after reflection from the wall as heard by the driver of the engine is

Step-by-step 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 aligned v^ &= v (v+v₀ ) (v-v_ s ) &=v ( v+v_ s v-v_ s ) [ v₀=v_ s ] v^ &=1.2 ( 350+50 350-50 ) &= 1.2 400 300 =1.6 kHz aligned