MCQ
A speeding motorcyclist sees traffic jam ahead him. He slows down to $36\,\, km\,\,hour^{-1}$ He finds that traffic has eased and a car moving ahead of him at $18 \,\, km\,\,hour^{-1}$ is honking at a frequency of $1392\,\, Hz.$ If the speed of sound is $343\, m s^{-1}$, the frequency of the honk as heard by him will be .... $Hz$
  • A
    $1332$
  • B
    $1372$
  • $1412$
  • D
    $1454$

Answer

Correct option: C.
$1412$
c
Here, speed of motorcyclist, $v_{m}=36 \mathrm{km}$ hour $^{-1}$

$=36 \times \frac{5}{18}=10 \mathrm{ms}^{-1}$

Speed of car,

$v_{c}=18 \mathrm{km} \text { hour }^{-1}=18 \times \frac{5}{18} \mathrm{ms}^{-1}=5 \mathrm{ms}^{-1}$

Frequency of source, $v_{0}=1392 \mathrm{Hz}$

Speed of sound, $v=343 \mathrm{ms}^{-1}$

The frequency of the honk heard by the motorcyclist is

$v^{\prime} =v_{0}\left(\frac{v+v_{m}}{v+v_{c}}\right)=1392\left(\frac{343+10}{343+5}\right)$

$=\frac{1392 \times 353}{348}=1412 \mathrm{Hz}$

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