An engine moving towards a wall with a velocity 50 ~m / s emits a note of 1.2 kHz . Speed of sound in air is 350 ~m / s . 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. v^ = v (v+v_o ) (v-v_s ) =v ( v+v_s v-v_s ) ( v_o=v_s ) v^ =1.2 ( 350+50 350-50 )= 1.2 400 300 =1.6 kHz