Question

A driver travels northbound on a highway at a speed of 20.0 m/s. A police car, traveling southbound at a speed of 43.0 m/s, approaches with its siren producing sound at a frequency of 2600 Hz. (a) What frequency does the driver observe as the police car approaches?
...................... Hz
(b) What frequency does the driver detect after the police car passes him?
..........................Hz
(c) Repeat parts (a) and (b) for the case when the police car is traveling northbound. while police car overtakes ....................Hz
after police car passes .....................Hz

Answer 1

Answer:

Explanation:

Let the velocity of observer the driver be V₀ and the velocity of police car the source be$v_s$. Let V be the velocity of sound in air.

The apparent  frequency f is given by the following relation

$\frac{f}{F} = \frac{V+V_0}{V-V_s}$

F is real frequency.

Now F = 2600,

$V_0 = 20 ms^{-1} and\\\ V_s= 40ms^{-1}$

Putting these values in the equation

$\frac{f}{2600} = \frac{343+20}{343-43}$

f = 3146 Hz.

b ) After the police passes , the sign of velocity of observer and source get reversed.

$\frac{f}{2600} = \frac{343-20}{343+43}$

f = 2175 Hz

c ) When the police car is also going northward and trying to overtaking the driver the frequency is calculated as follows .

$\frac{f}{2600} = \frac{343-20}{343-43}$

f = 2799 Hz

When the policy car already crosses the drives the frequency will be as follows

$\frac{f}{2600} = \frac{343+20}{343+43}$

f = 2445 Hz