Question

Question #a6987

Answer 1

when the speeds of source and the receiver relative to the air are lower than the velocity of sound in the air, the relationship between observed frequency `nu` and frequency of sound emitting from source `nu_0` is given by Doppler's effect as follows

`nu= (v+v_r)/(v+v_s)nu_0....[1]`,

where

• `v` is the velocity of sound in the air;

• `v_r` is the velocity of the receiver of sound relative to the air. It is positive if the receiver is moving towards the source and negative in the other direction;

• `v_s` is the velocity of the source relative to the air. It is positive if the source is moving away from the receiver and negative in the reverse direction.

• Let at an instant the line joining the position of 1st car (`A`) and position of the 2nd car (`B`) makes an angle `theta` with the direction of motion of the first car. So this line will make angle `(90-theta)^@` with the direction of motion of the 2nd car.

• So the component of velocity of the 1st car along `AB` will be `v_(AB)=v_1costheta`,where `v_1` is the velocity of the 1st car towards the crossing.

• And the component of velocity of the 2nd car along `BA` will be `v_(BA)=vcos(90-theta)=v_2sintheta`, where `v_2` is the velocity of the 2nd car towards the crossing.

So

`v_r=v_2sintheta`

and `v_s=-v_1costheta`

Now the equation [1] becomes

`nu= ((v+v_2sintheta)/(v-v_1costheta))nu_0`

Now

• `v_1=72km"/"hr=20m"/"s`

• `v_2=36km"/"hr=10m"/"s`

• `v==340m"/"s`

• `nu_0=280Hz`

So

`nu= ((340+10sintheta)/(340-20costheta))280` Hz

When `theta =45^@` we finally get the frequency of horn heard by the driver of 2nd car.

`nu= ((340+10sin45)/(340-20cos45))280~~298` Hz