Question

An engine is moving towards a wall with a velocity $50m{s}^{-1}$ emits a note of $1.2kHz$. Speed of sound in air = $350m{s}^{-1}$. The frequency of the note after reflection from the wall as heard by the driver of the engine is

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

Answer 1

Answer: (B)

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.
$v^{\prime }=\frac{v\left(v+v_0\right)}{\left(v-v_s\right)}$
$=v\left(\frac{v+v_s}{v-v_s}\right)\left[\because v_0=v_s\right]$
$\Rightarrow v^{\prime }=1.2\left(\frac{350+50}{350-50}\right)$
$=\frac{1.2\times 400}{300}$
$=1.6kHz$