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

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`

Step-by-step 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 [(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 [(v + v_o)/(v - v_s) ]`

`v_o = v_2`

`v_s = v_1`

`f_o = f_e`

Substituting value

`f_a = 2100 [(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 [(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 [(v - v_o)/(v + v_s) ]`

`v_o = v_2`

`v_s = v_1`

`f_o = f_e`

Substituting value

`f_a = 2100 [(343 - 7.22)/(343 + 13) ]`

`f_a = 1980.7 Hz`