A convertible moves toward you and then passes you; all the while, its loudspeakers are producing a sound. The speed of the car is a constant 8.49 m/s, and the speed of sound is 345 m/s. What is the ratio of the frequency you hear while the car is approaching to the frequency you hear while the car is moving away?

Respuesta :

Answer:

[tex]\frac{f_s}{f_l}=(1.05)[/tex]

Explanation:

Due to Doppler effect:

[tex]f_l=f_s(\frac{v-v_0}{v+v_0} )[/tex]

f_l= frequency experienced by the observer

f_s= frequency of the source

v= velocity of sound( 345 m/s)

v_0= velocity of observer ( speed of car = 8.49 m/s)

therefore, ratio of frequency is

[tex]\frac{f_l}{f_s}=(\frac{345-8.49}{345+8.49} )[/tex]

[tex]\frac{f_l}{f_s}=(\frac{336.51}{353.49})[/tex]

[tex]\frac{f_s}{f_l}=(\frac{353.49}{336.51})[/tex]

=1.05