Question

Q11:A police car in hot pursuit goes speeding past you. While the siren is approaching, the frequency of the sound you hear is 5500 Hz. When the siren is receding away from you, the frequency of the sound is 4500 Hz. Use the Doppler formula to determine the velocity of the police car. Use vsound

Answer 1

Answer:

   v_s = 34.269 m / s

Explanation:

This is a Doppler effect exercise, in this case the observer is fixed and the sound source is moving.

         f ’= f  $\frac{v}{v \mp v_s }$

where the negative sign is used for when the source approaches the observer and the positive sign for when the source moves away from the observer

In this case when f ’= 5500 Hz approaches and when f’ = 4500 Hz moves away, let's write the two expressions together

          5500 = f ($\frac{v}{v - v_s }$)

          4500 = f ( $\frac{v}{v + vs}$)

let's solve these two equations

           $\frac{5500}{4500} = \frac{v+v_s}{v-v_s}$

           1.222 (v-v_s) = v + v_s

            v_s (1+ 1.22) = v (1.222 -1)

            v_s = v   $\frac{0.222}{2.223}$

the speed of sound in air is v = 343 m / s

            v_s = 343 0.09990

 

             v_s = 34.269 m / s