Q108 / 160 · JIPMER 2012 · Physics

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

  • A
    2.4 kHz
  • B
    0.24 kHz
  • Correct answer
    C
    1.6 kHz
  • D
    1.2 kHz

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

Hence, the correct answer is option: 'C'.

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