Question

A policeman on duty detects a drop of $15 \%$ in the pitch of the horn of a motor car as it crosses him. If the velocity of sound is $330 \, m \, s^{-1}$ , calculate the speed of the car.

• A. $26.7 \, m \, s^{-1}$
• B. $27.6 \, m \, s^{-1}$
• C. $53.4 \, m \, s^{-1}$
• D. $54.3 \, m \, s^{-1}$

Answer 1

Answer: (A)

$26.7 \, m \, s^{-1}$

Answer 2

Here, velocity of sound, $v = 330 \, m \, s^{-1}$
Frequency of the horn = $\upsilon$
Let $\upsilon_c$ be the speed of the car.
The frequency of the horn of the car heard by the policeman before it crosses him is,
$\upsilon' = \upsilon \left( \frac{v}{v - v_c} \right)$ ....(i)
and after it crosses him is
$\upsilon" = \upsilon \left( \frac{v}{v + v_c} \right)$ ....(ii)
Divide (ii) by (i), we get
$\frac{\upsilon"}{\upsilon'} = \frac{\nu -\nu_{c}}{\nu+\nu_{c} } = \frac{330 - \nu_{c}}{330 + \nu_{c}}$
As the pitch (i.e. frequency) drops by $15 \%$ so
$\frac{ \upsilon"}{\upsilon'} = \frac{85}{100} = \frac{17}{20} , Thus , \frac{17}{20} = \frac{330 - \nu _{c}}{330 + \nu _{c}}$
$5610 + 17 \upsilon_{c} = 6600 - 20\upsilon_{c}$
$37 \upsilon_{c}=990 \, \, \therefore \, \, \nu_{c} = \frac{990}{37} = 26.7 m s^{-1}$