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 $330m{s}^{-1}$ , calculate the speed of the car.

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

Answer 1

Answer: (A)

Here, velocity of sound, $v=330m{s}^{-1}$ 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,
${\upsilon }^{\prime }=\upsilon \left(\frac{v}{v-v_c}\right)$ ....(i)
and after it crosses him is
$\upsilon "=\upsilon \left(\frac{v}{v+v_c}\right)$ ....(ii)
Divide (ii) by (i), we get
$\frac{\upsilon "}{{\upsilon }^{\prime }}=\frac{\nu -{\nu }_c}{\nu +{\nu }_c}=\frac{330-{\nu }_c}{330+{\nu }_c}$
As the pitch (i.e. frequency) drops by $15\%$ so
$\frac{\upsilon "}{{\upsilon }^{\prime }}=\frac{85}{100}=\frac{17}{20},Thus,\frac{17}{20}=\frac{330-{\nu }_c}{330+{\nu }_c}$
$5610+17{\upsilon }_c=6600-20{\upsilon }_c$
$37{\upsilon }_c=990\therefore {\nu }_c=\frac{990}{37}=26.7m{s}^{-1}$