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

Options

A. `2.4` kHz
B. `0.24` kHz
C. `1.6` kHz
D. `1.2` kHz

Answer

C

Explanation

The reflected sound appears to propagate in a direction opposite to that of direct sound of engine.Thus the source and the observer can be presumed to approach each other with same velocity.
`upsilon' = upsilon((v + v_(0))/(v-v_(s)))= upsilon((v + v_(s))/(v - v_(s))) [because v_(0) = v_(s)]`
or `upsilon' = 1.2((350 + 50)/(350-50)) = (1.2 xx 400)/(300) = 1.6` kHz.