Question

g Two cars, car 1 and car 2 are traveling in opposite directions, car 1 with a magnitude of velocity v1=13.0 m/s and car 2 v2= 7.22 m/s. If car 1’s exhaust system is loud enough to be heard by car 2 and the frequency fe produced from the exhaust is 2.10 kHz. What frequencies would be heard by car 2 when the cars are approaching, passing, and retreating from one another?

Answer 1

Answer:

When they are approaching each other

    $f_a = 2228.7 \ Hz$

When they are passing  each other

    $f_a = 2100Hz$

 When they are retreating  from each other

     $f_a = 1980.7 Hz$

Explanation:

From the question we are told that

     The velocity of car one is  $v_1 = 13.0 m/s$

      The velocity of car two is  $v_2 = 7.22 m/s$

     The frequency of sound from car one is  $f_e = 2.10 kHz$

Generally the speed of sound at normal temperature is  $v = 343 m/s$

  Now as the cars move relative to each other doppler effect is created and this  can be represented  mathematically  as

              $f_a = f_o [\frac{v \pm v_o}{v \pm v_s} ]$

Where $v_s$ is the velocity of the source of sound

            $v_o$ is the velocity of the observer of the sound

            $f_o$ is the actual frequence

             $f_a$  is the apparent frequency

Considering the case when they are approaching each other

        $f_a = f_o [\frac{v + v_o}{v - v_s} ]$

          $v_o = v_2$  

         $v_s = v_1$

         $f_o = f_e$

Substituting value

            $f_a = 2100 [\frac{343 + 7.22}{ 343 - 13} ]$

              $f_a = 2228.7 \ Hz$

Considering the case when they are passing  each other    

At that instant

                  $v_o = v_s = 0m/s$

                   $f_o = f_e$

               $f_a = f_o [\frac{v }{v } ]$

              $f_a = f_o$

Substituting value

             $f_a = 2100Hz$

Considering the case when they are retreating  from each other    

                $f_a = f_o [\frac{v - v_o}{v + v_s} ]$

          $v_o = v_2$  

         $v_s = v_1$

         $f_o = f_e$      

Substituting value

         $f_a = 2100 [\frac{343 - 7.22}{343 + 13} ]$    

          $f_a = 1980.7 Hz$