Problem 4.101: A policeman on duty detects a drop of $30\: \%$ in the pitch of the horn of a motor-car as it crosses him. Calculate the speed of car, if the velocity of sound is $330m/s$.

Solution:

How to solve this problem? First try to visualize….There is a car and a policeman. The car is honking its horn. The policeman naturally hears the sound. But the car is moving towards him. He will hear an increase in pitch or frequency. This is due to Doppler effect. Now the car will go past him and will move away from him. He hears the sound with a lower frequency again due to Doppler effect.

The problem says there is a drop of $30\%$ in the pitch. This is the percentage difference between the high frequency heard while the car approached him and the lower frequency heard while the car move past him.

Let $f$ be the actual frequency of the sound. Policeman hear this as frequency $f_1$ as th car moves towards him and as $f_2$ as the car moves away from him. Then second frequency is $30\:\%$ less than the first frequency. It is only $70\:\%$ of $f_1$.

Mathematically, $f_2$ = $70\:\%$ of $f_1$.

Or, $f_2$ = $\frac{70}{100}\:f_1$.

Then, $\frac{f_2}{f_1}\:=\:\frac{70}{100}\:=\:\frac{7}{10}$------$(1)$.

Now we need to calculate $f_1$ and $f_2$ separately.

**Calculation of $f_1$.**

In Doppler effect the apparent frequency $f_1$ of a source moving towards a stationary observer is given by , $f_1\:=\:f\frac{V}{V\:-\:V_s}$------$(2)$.

**Calculation of $f_2$.**

The apparent frequency $f_2$ of a source moving away a stationary observer is given by , $f_2\:=\:f\frac{V}{V\:+\:V_s}$------$(3)$.

From $(1)$, $(2)$ and $(3)$,

$\frac{f_2}{f_1}\:=\:\frac{f\frac{V}{V\:+\:V_s}}{f\frac{V}{V\:-\:V_s}}\:=\:\frac{7}{10}$

Or, $\frac{V\:-\:V_s}{V\:+\:V_s}\:=\:\frac{7}{10}$.

Then, $10V\:-\:10V_s\:=\:7V\:+\:7V_s$.

Or, $7V_s\:+\:10V_s\:=\:10V\:-\:7V$.

That is, $17V_s\:=\:3V$

From this, $V_s\:=\:\frac{3V}{17}$.

Given that speed of sound in air is $330\:ms^{-1}$.

Therefore, speed of car,

$V_s\:=\:\frac{3\times 330}{17}\:=\:\frac{990}{17}\:=\:58.24\:ms^{-1}$.

So the car was traveling with a speed of $58.24\:ms^{-1}$.