Question 41

The siren of a police car traveling away from an intersection with a velocity [tex]$v_c$[/tex] emits a sound wave with a frequency of [tex]$f_e$[/tex]. If the speed of sound in air is [tex][tex]$v_a$[/tex][/tex], then the frequency heard by a person standing in the intersection is given by which one of the following expressions?

A. [tex]f_c\left(\frac{v_1}{v_s+v_c}\right)[/tex]

B. [tex]f_c\left(\frac{\nu_s}{\nu_s+\nu_c}\right)[/tex]

C. [tex]f_c\left(\frac{v_c}{v_s-v_c}\right)[/tex]

D. [tex]f_c\left(\frac{\nu_s}{\nu_a-\nu_c}\right)[/tex]



Answer :

To determine the frequency heard by a person standing in the intersection when the police car is moving away, we need to use the Doppler effect formula. The Doppler effect describes the change in the frequency of a wave in relation to an observer who is moving relative to the wave source.

When the source of the wave is moving away from the observer, the frequency heard by the observer is given by:

[tex]\[ f_h = f_e \left( \frac{v_a}{v_a + v_c} \right) \][/tex]

In our scenario:
- [tex]\( f_e \)[/tex] is the emitted frequency of the sound from the police car.
- [tex]\( v_a \)[/tex] is the speed of sound in air.
- [tex]\( v_c \)[/tex] is the velocity of the police car, which is moving away from the intersection.

Given this relationship, we can map the given multiple-choice options to our derived formula. Let's analyze each option step by step:

A. [tex]\( f_c \left( \frac{v_1}{v_s + v_c} \right) \)[/tex]
- This can be written in terms of our variables as [tex]\( f_e \left( \frac{v_1}{v_s + v_c} \right) \)[/tex].
- Here, [tex]\( v_1 \)[/tex] and [tex]\( v_s \)[/tex] do not directly relate to our given variables [tex]\( v_a \)[/tex] and [tex]\( v_c \)[/tex].

B. [tex]\( f_c \left( \frac{\nu_s}{\nu_s + \nu_c} \right) \)[/tex]
- This can be written as [tex]\( f_e \left( \frac{\nu_s}{\nu_s + \nu_c} \right) \)[/tex].
- [tex]\( \nu_s \)[/tex] and [tex]\( \nu_c \)[/tex] again are not our variables [tex]\( v_a \)[/tex] and [tex]\( v_c \)[/tex].

C. [tex]\( f_c \left( \frac{v_c}{v_s - v_c} \right) \)[/tex]
- This becomes [tex]\( f_e \left( \frac{v_c}{v_s - v_c} \left). - It is incorrect as it implies that the observer is moving towards the source. D. \( f_c \left( \frac{\nu_s}{\nu_a - \nu_c} \right) \)[/tex]
- Rewritten as [tex]\( f_e \left( \frac{\nu_s}{\nu_a - \nu_c} \right) \)[/tex].
- Using the variables in this option directly corresponds to our derived formula where [tex]\( \nu_s \)[/tex] (speed of sound in air) matches [tex]\( v_a \)[/tex] and [tex]\( \nu_c \)[/tex] (velocity of the police car) matches [tex]\( v_c \)[/tex].

On comparison, the correct expression that matches our derived formula for the Doppler effect is option D:

[tex]\[ f_e \left( \frac{\nu_{ s }}{\nu_{ a } - \nu_{ c }} \right) \][/tex]

So, the correct answer is:

D. [tex]\( f_c \left( \frac{\nu_{ s }}{\nu_{ a } - \nu_{ c }} \right) \)[/tex]