Question

The speed limit for vehicles on a road is $ 100km/h $ . A drop of $ 20\% $ in the pitch of a car horn is detected by a policeman when the car passes him. Does the car driver cross the speed limits? Velocity of sound in air $ = 340m/s $

Answer 1

Hint
Here the source of sound is the car. The policeman is the receiver. This change in frequency of a moving source is explained by a phenomenon called Doppler Effect. When there is a relative motion of the source or the listener or both, there is an apparent change in the frequency of the sound. This phenomenon is called the Doppler Effect.
When the source moves towards the listener
 $\Rightarrow {f'} = \left( {\dfrac{V}{{V - {V_S}}}} \right)f $
When the source moves away from the listener
 $\Rightarrow {f''} = \left( {\dfrac{V}{{V + {V_S}}}} \right)f $
(Where, $ {f'} $ is the changed frequency after the source moves, $ V $ stands for the velocity of the sound, $ {V_S} $ stands for the velocity of source and $ f $ stands for the frequency of the sound)

Complete step by step answer
As the car passes the policeman, initially the source is moving towards the listener and then the source moves away from the listener.
We know that there is a $ 20\% $ drop in the pitch as the car passes the policeman,
 $\Rightarrow \therefore {f''} = {f'} - 0.2{f'} = 0.8{f'} $
By Doppler Effect,
We know that,
When the source moves towards the listener
 $\Rightarrow {f'} = \left( {\dfrac{V}{{V - {V_S}}}} \right)f $ …………………..
When the source moves away from the listener
 $\Rightarrow {f''} = \left( {\dfrac{V}{{V + {V_S}}}} \right)f $ ……………………
Dividing equation by equation
We get,
 $\Rightarrow \dfrac{{{f'}}}{{{f{''}}}} = \left( {\dfrac{{\dfrac{V}{{V - {V_S}}}}}{{\dfrac{V}{{V + {V_S}}}}}} \right) = \dfrac{{V + {V_S}}}{{V - {V_S}}} $
The velocity of sound is given by, $ 340m/s $
There is a $ 20\% $ drop in the pitch,
We can write,
 $ \therefore {f{''}} = {f'} - 0.2{f'} = 0.8{f'} $
Now we can substitute these values in the above equation, we get
 $\Rightarrow \dfrac{1}{{0.8}} = \dfrac{{340 + {V_S}}}{{340 - {V_S}}} $
Cross multiplying, we get
 $\Rightarrow 340 - {V_S} = \left( {340 - {V_S}} \right)0.8 $
 $\Rightarrow 340 - {V_S} = 272 + 0.8{V_S} $
Rearranging the above equation, we get
 $\Rightarrow 340 - 272 = 0.8{V_S} + {V_S} $
 $\Rightarrow 68 = 1.8{V_S} $
From this, we get the velocity of the source as,
 $\Rightarrow {V_S} = \dfrac{{68}}{{1.8}} = 37.78m/s $
Converting this velocity into $ km/h $ by multiplying with $ \dfrac{{18}}{5} $ , we get
 $\Rightarrow {V_S} = 37.78 \times \dfrac{{18}}{5} = 136km/h $
So, the driver is driving the car at a speed of $ 136km/h $ .
The speed limit is given as, $ 100km/h $ . Hence we can say that the driver has crossed the speed limit.

Note
Doppler Effect has many real-life applications. It is used to estimate the speed of submarines, airplanes, automobiles, etc. It can be used for tracking artificial satellites. Doppler Effect is used to estimate the velocity and rotation of stars. Doppler Effect can be observed in light also.