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

contexto.120

3 years ago

The correct option is C

Apparent frequency can be calculated as,

$f^{\prime}=f_{0}\left(\frac{v \pm v_{0}}{v \pm v_{s}}\right)$

$\mathrm{f}^{\prime}=\mathrm{f}_{0}\left(\frac{\mathrm{v}+\mathrm{v}_{0}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right)$

$=1392\left(\frac{343+10}{343+5}\right)$

$=1412 \mathrm{HZ}$