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Waves Part 2

Section: Physics

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Question : 33 of 54

Marks: +1, -0

A car P travelling at

20 \text{ms}^{-1}

sounds its horn at a frequency of

400 \text{Hz}

Q

is travelling behind the first car in the same direction with a velocity

40 \text{ms}^{-1}

. The frequency heard by the passenger of the car

Q

is approximately [Take, velocity of sound

=360 \text{ms}^{-1}

[11-Apr-2023 shift 2]

$471 \text{Hz}$
$514 \text{Hz}$
$421 \text{Hz}$
$485 \text{Hz}$

Solution:

\; V_{c}=20 \text{ms}^{-1}

\; f=400 \text{Hz}

f_{app}= \left[ \frac{V s - (-V_Q)}{V_S - (-V_p)} \right] f

\; = \left( \frac{360+40}{360+20} \right) \times 400 = \frac{400}{380} \times 400 = 421 \text{Hz}

Ans. (3)

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