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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

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1332 Hz

1372 Hz

1412 Hz

1454 Hz

answer is C.

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Detailed Solution

Here speed of Motorcyclist speed is

$v_{m} = 36 km {hour}^{- 1} = 36 \times \frac{5}{18} m s^{- 1} = 10 m s^{- 1}$

Speed of car,

$v_{c} = 18 km {hour}^{- 1} = 18 \times \frac{5}{18} m s^{- 1} = 5 m s^{- 1}$

Frequency of source, $ν_{0}$ = 1392 Hz

Speed of sound, $ν =$ 343 m $s^{- 1}$

The frequency of the honk heard by the motorcyclist is

$v^{'} = v_{0} (\frac{v + v_{m}}{v + v_{c}}) = 1392 (\frac{343 + 10}{343 + 5}) = \frac{1392 \times 353}{348} = 1412 Hz$

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