MCQ
Two engines pass each other moving in opposite directions with uniform speed of $30\,m/s$ . One of them is blowing a whistle of frequency $540\,Hz.$ Calculate the frequency heard by driver of second engine before they pass each other  ... $Hz$. Speed of sound is $330\,m/sec$
  • A
    $450$
  • B
    $540$
  • C
    $270$
  • $648$

Answer

Correct option: D.
$648$
d
We know that the apperent frequency

$\mathrm{f}^{\prime}=\left(\frac{\mathrm{v}-\mathrm{v}_{0}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{f}$ from Doppler's effect

where $v_{0}=v_{s}=30 \mathrm{m} / \mathrm{s},$ velocity of observer and source

Speed of sound $v=330 \mathrm{m} / \mathrm{s}$

$\because$ Frequency of whistle $(\mathrm{f})=540 \mathrm{Hz}$

$\therefore \mathrm{f}^{\prime}=\frac{330+30}{330-30} \times 540=648 \mathrm{Hz}$

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