An engine sounding a whistle of frequency $2000 \mathrm{~Hz}$ is receding from the stationary observer at $72 \mathrm{~km} / \mathrm{h}$. What is the apparent frequency of the observer? The velocity of sound in air is $340 \mathrm{~m} / \mathrm{s}$.

• Correct answer
  A

  $1889 \mathrm{~Hz}$
• B

  $2889 \mathrm{~Hz}$
• C

  $3889 \mathrm{~Hz}$
• D

  $4889 \mathrm{~Hz}$

Solution

Here a source is moving away from a stationary observer. So, frequency recorded at source will be lower than true frequency. Apparent frequency at observer, $f^{\prime}=f\left(\frac{v}{v+v_s}\right)$ where, $v=$ speed of sound $=340 \mathrm{~m} / \mathrm{s}$ and $v_s=72 \mathrm{~km} / \mathrm{h}=72 \times \frac{5}{18}=20 \mathrm{~ms}^{-1}$ $\therefore \quad f^{\prime}=2000\left(\frac{340}{340+20}\right)$ Hence, $\quad f^{\prime}=1889 \mathrm{~Hz}$

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