Question

a policeman on duty detects a drop of 10% in the pitch of the horn of motion of car as it crosses him if the velocity
of sound is 330 M per second calculate the speed of the car

Answer 1

according to question, police man detects 10% drop in the pitch when the car crosses him. So, the apparent frequency when the car moves away from him is 90% of the apparent frequency when the car approaches him.
now use Doppler's effect of sound,
therefore , when car approaches him,
$\nu_1=\nu_0\left(\frac{V}{V-V_s}\right)$

The apparent frequency , when the car moves away
$\nu_2=\nu\left(\frac{V}{V+V_s}\right)$

from question,
$v_2=0.9v_1$

so, $\frac{V}{V+V_s}=0.9\frac{V}{V-V_s}$

9(V+Vs) = 10(V - Vs)

19Vs = V

Vs = V/19 = 330/19 = 17m/s

Answer 2

Given:

Drop 10 %

Velocity = 330 M

To find:

Speed.

Solution:

90 % of the pitch of the horn heard.

By Doppler effect,

When the car is moving towards the policeman,

Velocity ( toward ) = Initial velocity * ( Velocity of the sound / Velocity of the sound + speed )

When the car crosses the policeman,

Velocity ( away ) = Initial velocity * ( Velocity of the sound / Velocity of the sound - speed )

Here,

Velocity ( away ) = 0.9 Velocity ( toward )

Substituting,

We get,

Initial velocity * ( Velocity of the sound / Velocity of the sound * speed ) = 0.9  Initial velocity * ( Velocity of the sound / Velocity of the sound * speed )

Cancelling common terms and grouping,

9 ( Velocity of the sound / Velocity of the sound + speed ) = 10 ( Velocity of the sound / Velocity of the sound - speed )

19 * Speed = Velocity

Substituting,

19 * Speed = 330

Hence, Speed = 330 / 19

Speed = 17 m / s